Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 215614 11 pageshttpdxdoiorg1011552013215614

Research ArticleCusps of Bishop Spherical Indicatrixes and Their Visualizations

Haiming Liu12 and Donghe Pei1

1 School of Mathematics and Statistics Northeast Normal University Changchun 130024 China2 School of Science Mudanjiang Normal University Mudanjiang 157011 China

Correspondence should be addressed to Donghe Pei peidh340nenueducn

Received 15 October 2013 Accepted 9 December 2013

Academic Editor Bo Shen

Copyright copy 2013 H Liu and D Pei This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The main result of this paper is using Bishop Frame and ldquoType-2 Bishop Framerdquo to study the cusps of Bishop spherical imagesand type-2 Bishop spherical images which are deeply related to a space curve and to make them visualized by computer We findthat the singular points of the Bishop spherical images and type-2 Bishop spherical images correspond to the point where Bishopcurvatures and type-2 Bishop curvatures vanished and their derivatives are not equal to zero As applications and illustration of themain results two examples are given

1 Introduction

This paper is written as one of the research projects onvisualization of singularities of submanifolds generated byregular curves embedded in Euclidean 3-space by usingBishop Frame It is well known that Bishop introducedthe Relatively Parallel Adapted Frame of regular curvesembedded in Euclidean 3-space and this frame was widelyapplied in the area of Biology and Computer Graphics see[1ndash5] For instance in [6] we introduced some detailedapplications of Bishop Frame so we omit them here Inspiredby the work of Bishop in [7] the authors introduced anew version of Bishop Frame using a common vector fieldas binormal vector field of a regular curve and called thisframe ldquoType-2 Bishop FramerdquoThey also introduced two newspherical images and called them type-2 Bishop sphericalimages We know that the properties of geometric objectsare independent of the choice of the coordinate systemsBut the researchers [1 7ndash13] found that when they adoptedthese frames there would be some new geometric objectssuch as Bishop spherical images Bishop Daroux image andtype-2 Bishop spherical images Although Bishop sphericalimages and type-2 Bishop spherical images have been wellstudied from the standpoint of differential geometry whenthey are regular spherical curves there are little papers ontheir singularities Actually sometimes they are singular for

example [7 10] Two questions are what about their singu-larities and how to recognize the types of the singularitiesThus the current study hopes to answer these questions andit is inspired by the works of Bishop [1] Yılmaz et al [7] Peiand Sano [14] and Wang et al [15] On the other hand forthe reason that the vector parameterized equations of Bishopspherical images and type-2 Bishop spherical images are verycomplicated (see [7 10]) it is very hard to recognize theirsingular points by normal way In this paper we will give asimple sufficient condition to describe their singular points byusing the technique of singularity theory To do this we hopethat Bishop spherical images and type-2 Bishop sphericalimages can be seen as the discriminants of unfolding ofsome functions In this paper adopting Bishop Frame [1] andldquoType-2 Bishop Framerdquo [7] as the basic tools we constructBishop normal indicatrix height functions (denoted by 119867

119894

119868 times 1198782rarr R 119867

119894(119904 k) = ⟨N

119894(119904) k⟩ where 119894 = 1 2 and

N119894(119904) is the first Bishop spherical indicatrix or the second

Bishop spherical indicatrix and 119867119895 119868 times 119878

2rarr R

119867119895(119904 k) = ⟨120577

119895(119904) k⟩ where 119895 = 3 4 and 120577

119895(119904) is the first type-

2 Bishop spherical indicatrix or the second type-2 Bishopspherical indicatrix) locally around the point (119904

0 k0) These

functions are the unfolding of these singularities in the localneighbourhood of (119904

0 k0) and depend only on the germ that

they are unfolding We create these functions by varyinga fixed point k in these Bishop normal indicatrix height

2 Mathematical Problems in Engineering

functions to get some families of functions We show thatthese singularities are versally unfolded by the families ofBishop normal indicatrix height functions If the singularityof ℎ119894V = 119867

119894(119904 k) and ℎ

119895V = 119867119895(119904 k) is 119860

119896-type (119896 =

1 2) and the corresponding 2-parameter unfolding is versalthen applying the theory of singularity [14 15] we knowthat discriminant set of the 2-parameter unfolding is locallydiffeomorphic to a line or a cusp thus we finished theclassification of singularities of the Bishop spherical imagesand type-2 Bishop spherical images because the discriminantsets of the unfolding are precisely the Bishop spherical imagesand type-2 Bishop spherical images Moreover we see thatthe 119860

119896-singularity (119896 = 1 2) of ℎ

119894V and ℎ119895V is closely

related to Bishop curvatures and type-2 Bishop curvaturesThe singular points of the Bishop spherical images and type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero Thus we get themain results in this paper which are stated in Theorem 1 Itis important that the properties of those functions relatedwith space curve 120574(119904) needed to be generic [15] Once weproved that the properties of those functions were genericwe could deduce that all singularities were stable under smallperturbations for our family of functions By consideringtransversality we prove that these properties given by us aregeneric As applications of our main results we give twoexamples

The rest of this paper is organized as follows Firstlywe introduce some basic concepts and the main results inthe next two sections Then we introduce four differentfamilies of functions on 120574 that will be useful to the study ofBishop spherical images and type-2 Bishop spherical imagesAfterwards some general results on the singularity theory areused for families of function germs and the main results areproved Finally we give two examples to illustrate the mainresults and the conclusion of the work is drawn

2 Preliminaries and Notations

In this section wewill introduce the notions of Bishop FrameldquoType-2 Bishop Framerdquo Bishop spherical images and type-2Bishop spherical images of unit speed regular curve Let 120574 =120574(119904) be a regular unit speed Frenet curve in E3 We know thatthere exist accompanying three frames called Frenet frame forFrenet curve Denote by (T(119904)N(119904)B(119904)) the moving Frenetframe along the unit speed Frenet curve 120574(119904)Then the Frenetformulas are given by

(

T1015840 (119904)N1015840 (119904)B1015840 (119904)

) = (

0 119896 (119904) 0

minus119896 (119904) 0 120591 (119904)

0 minus120591 (119904) 0

)(

T (119904)N (119904)B (119904)

) (1)

Here 119896(119904) and 120591(119904) are called curvature and torsion respec-tively [16] The Bishop Frame of the 120574(119904) is expressed by thealternative frame equations

(

T1015840 (119904)N10158401(119904)

N10158402(119904)

) = (

0 1198961 (119904) 1198962 (119904)

minus1198961 (119904) 0 0

minus1198962 (119904) 0 0

)(

T (119904)N1 (119904)

N2 (119904)

) (2)

Here we will call the set (T(119904)N1(119904)N2(119904)) as Bishop Frame

and 1198961(119904) = ⟨T1015840(119904)N

1(119904)⟩ and 119896

2(119904) = ⟨T1015840(119904)N

2(119904)⟩ as

Bishop curvatures The relation matrix can be expressed as

(

T (119904)N (119904)B (119904)

) = (

1 0 0

0 cos 120579 (119904) sin 120579 (119904)0 minus sin 120579 (119904) cos 120579 (119904)

)(

T (119904)N1 (119904)

N2 (119904)

) (3)

One can show that

119896 (119904) = radic1198962

1(119904) + 119896

2

2(119904)

120579 (119904) = arctan(1198962 (119904)1198961 (119904)

) where 1198961 (119904) = 0 120591 (119904) =

119889120579 (119904)

119889119904

(4)

so that 1198961(119904) and 119896

2(119904) effectively correspond to a Cartesian

coordinate system for the polar coordinates 119896(119904) and 120579(119904)with120579 = int 120591(119904)119889119904 Here Bishop curvatures are also defined by

1198961 (119904) = 119896 (119904) cos 120579 (119904)

1198962 (119904) = 119896 (119904) sin 120579 (119904)

(5)

The orientation of the parallel transport frame includes thearbitrary choice of integration constant 120579

0 which disappears

from 120591 (and hence from the Frenet frame) due to thedifferentiation [1]

The ldquoType-2 Bishop Framerdquo of the 120574(119904) is defined by thealternative frame equations see [7]

(

1205771015840

1(119904)

1205771015840

2(119904)

B1015840 (119904)) = (

0 0 minus1205981 (119904)

0 0 minus1205982 (119904)

1205981 (119904) 1205982 (119904) 0

)(

1205771 (119904)

1205772 (119904)

B (119904)) (6)

The relation matrix between Frenet-Serret and ldquoType-2Bishop Framerdquo can be expressed as

(

T (119904)N (119904)B (119904)

) = (

sin 120579 (119904) minus cos 120579 (119904) 0cos 120579 (119904) sin 120579 (119904) 0

0 0 1

)(

1205771 (119904)

1205772 (119904)

B (119904)) (7)

Here the type-2 Bishop curvatures are defined by

1205981 (119904) = minus120591 (119904) cos 120579 (119904)

1205982 (119904) = minus120591 (119904) sin 120579 (119904)

(8)

We will call the set (1205771(119904) 1205772(119904)B(119904)) ldquoType-2 Bishop Framerdquo

which is properly oriented and 1205981(119904) = ⟨B1015840(119904) 120577

1(119904)⟩ and

1205982(119904) = ⟨B1015840(119904) 120577

2(119904)⟩ type-2 Bishop curvatures One also can

show that

120581 (119904) =119889120579 (119904)

119889119904 (9)

so that 1205981(119904) and 120598

2(119904) also effectively correspond to a

Cartesian coordinate system for the polar coordinates 120591(119904)120579(119904) with 120579 = int 120581(119904)119889119904

The following notions are the main objects in this paperThe unit sphere with center in the origin in the space E3 isdefined by

1198782= x isin E

3| ⟨x x⟩ = 1 (10)

Mathematical Problems in Engineering 3

Translating framersquos vector fields to the center of unit spherewe obtain Bishop spherical images or Bishop spherical indi-catrixes The first Bishop spherical indicatrix and the secondBishop spherical indicatrix are denoted byFBN(119904) = N

1(119904)

and SBN(119904) = N2(119904) separately The first type-2 Bishop

spherical indicatrix and the second type-2 Bishop sphericalindicatrix in [7] is denoted by FNBN(119904) = 120577

1(119904) and

SNBN(119904) = 1205772(119904) separately

3 Cusps of Bishop Spherical Indicatrixes andTheir Visualizations

Themain results of this paper are in the following theorem

Theorem 1 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(1) Suppose that 1198962(119904) = 0 Then one have the following

claims

(a) The first Bishop spherical image N1(119904) is locally

diffeomorphic to a line 0 timesR at 1199040if 1198961(1199040) = 0

(b) The first Bishop spherical image N1(119904) is locally

diffeomorphic to the cusp119862 at 1199040if 1198961(1199040) = 0 and

1198961015840

1(1199040) = 0

(2) Suppose that 1198961(119904) = 0 Then one has the following

claims

(a) The second Bishop spherical image N2(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1198962(1199040) = 0

(b) The second Bishop spherical image N2(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1198962(1199040) =

0 and 11989610158402(1199040) = 0

(3) Suppose that 1205982(119904) = 0 Then one has the following

claims

(a) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1205981(1199040) = 0

(b) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205981(1199040) =

0 and 12059810158401(1199040) = 0

(4) Suppose that 1205981(119904) = 0 Then one has the following

claims

(a) The second type-2 Bishop spherical image 1205772(119904)

is locally diffeomorphic to a line 0 times R at 1199040if

1205982(1199040) = 0

(b) The second type-2 Bishop spherical image 1205772(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205982(1199040) =

0 and 12059810158402(1199040) = 0 where the ordinary cusp is 119862 =

(1199091 1199092) | 1199092

1= 1199093

2 The picture of cusp will be

seen in Figure 1

minus05 minus04 minus03 minus02 minus01 0 01 02 03 04 050

01

02

03

04

05

06

07

Figure 1 Cusp

4 Bishop Normal Indicatrix Height Functions

In this section we will introduce four different families offunctions on 120574 that will be useful to study the singular pointsof the Bishop spherical images and type-2 Bishop sphericalimages of unit speed regular curve Let 120574 119868 rarr E3 bea regular unit speed curve Now we define two families ofsmooth functions on 119868 as follows

119867119894 119868 times 119878

2997888rarr R by 119867

119894 (119904 k) = ⟨N119894 (119904) k⟩ (11)

where 119894 = 1 2 We call it the first (second) Bishop normalindicatrix height function for the case 119894 = 1 (119894 = 2) Forany k isin 1198782 we denote ℎ

119894V(119904) = 119867119894(119904 k) We also define two

families of smooth functions on 119868 as follows

119867119895 119868 times 119878

2997888rarr R by 119867

119895 (119904 k) = ⟨120577119895 (119904) k⟩ (12)

where 119895 = 3 4 We call it the first (second) type-2 Bishopnormal indicatrix height function for the case 119895 = 3 (119895 = 4)For any k isin 1198782 we denote ℎ

119895V(119904) = 119867119895(119904 k) Then we have thefollowing proposition

Proposition 2 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(A) Suppose that 1198961(119904) = 0 Then one has the following

claims

(1) ℎ1V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that v = 120582T(119904)+120583N2(119904) and 1205822+1205832 =

1(2) ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)(3) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0 if and only if k =∓N2(119904) and 119896

2(119904) = 0

(4) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = ℎ(3)

1V (119904) = 0 if and onlyif k = ∓N

2(119904) and 119896

2(119904) = 119896

1015840

2(119904) = 0

4 Mathematical Problems in Engineering

(B) Suppose that 1198962(119904) = 0 Then one has the following

claims

(1) ℎ2V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 120582T(119904)+120583N1(119904) and 1205822+1205832 =

1(2) ℎ2V(119904) = ℎ

1015840

2V(119904) = 0 if and only if k = ∓N1(119904)(3) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = 0 if and only if k =∓N1(119904) and 119896

1(119904) = 0

(4) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = ℎ(3)

2V (119904) = 0 if and onlyif k = ∓N

1(119904) and 119896

1(119904) = 119896

1015840

1(119904) = 0

(C) Suppose that 1205981(119904) = 0 Then one has the following

claims

(1) ℎ3V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205772(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ3V(119904) = ℎ

1015840

3V(119904) = 0 if and only if k = ∓1205772(119904)(3) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = 0 if and only if k =∓1205772(119904) and 120598

2(119904) = 0

(4) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = ℎ(3)

3V (119904) = 0 if and onlyif k = ∓120577

2(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0

(D) Suppose that 1205982(119904) = 0 Then one has the following

claims

(1) ℎ4V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205771(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ4V(119904) = ℎ

1015840

4V(119904) = 0 if and only if k = ∓1205771(119904)(3) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = 0 if and only if k =∓1205771(119904) and 120598

1(119904) = 0

(4) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = ℎ(3)

4V (119904) = 0 if and onlyif k = ∓120577

1(119904) and 120598

1(119904) = 120598

1015840

1(119904) = 0

Proof (A) (1) If ℎ1V(119904) = ⟨N

1(119904) k⟩ = 0 then we have that

there are real numbers 120582 and 120583 such that k = 120582T(119904) + 120583N2(119904)

Moreover in combinationwith k isin 1198782 whichmeans 1205822+1205832 =1 it follows that ℎ

1V(119904) = 0 if and only if k = 120582T(119904) + 120583N2(119904)

and 1205822 + 1205832 = 1(2) When ℎ

1V(119904) = 0 the assertion (2) follows from thefact that ℎ1015840

1V(119904) = ⟨1198961(119904)T(119904) k⟩ = 1198961(119904)120582 and 1198961(119904) = 0 Thus

we get that ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)

(3) When ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 the assertion (3) followsfrom the fact that ℎ10158401015840

1V(119904) = ⟨1198961015840

1(119904)T(119904) + 119896

2

1(119904)N1(119904) +

1198961(119904)1198962(119904)N2(119904) ∓N

2(119904)⟩ = ∓119896

1(119904)1198962(119904) and 119896

1(119904) = 0

(4) Under the condition that ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0this derivative is computed as follows ℎ(3)

1V (119904) = ⟨(11989610158401015840

1(119904) +

1198963

1(119904) + 119896

1(119904)1198962

2(119904))T(119904) + 3119896

1(119904)1198961015840

1(119904)N1(119904) + (2119896

1015840

1(119904)1198962(119904) +

1198961(119904)1198961015840

2(119904))N2(119904) ∓N

2(s)⟩ = 2119896

1015840

1(119904)1198962(119904) + 119896

1(119904)1198961015840

2(119904) =

1198961(119904)1198961015840

2(119904) Since 119896

1(119904) = 0 we get that ℎ(3)

1V (119904) = 0 is equivalentto condition 1198961015840

2(119904) = 0 The assertion (4) follows

(B) Using the same computation as the proof of (A) wecan get (B)

(C) (1) Since ℎ3V(119904) = ⟨120577

1(119904) k⟩ = 0 we have that there

are real numbers 120582 and 120583 such that k = 120582B(119904) + 1205831205772(119904) By

the condition that k isin 1198782 we get 1205822 + 1205832 = 1 The conversedirection also holds

(2) Since ℎ10158403V(119904) = ⟨minus1205981(119904)B(119904) k⟩ = 0 by the conditions

of assertion (2) and 1205981(119904) = 0 we have that 120598

1(119904)120582 = 0 It

follows from the fact that 1205981(119904)120582 = 0 and 120598

1(119904) = 0 that this is

equivalent to condition 120582 = 0 Therefore we have k = plusmn1205772(119904)

The converse direction also holds(3) Since ℎ

10158401015840

3V(119904) = ⟨minus1205981015840

1(119904)B(119904) + 120598

2

1(119904)1205771(119904) +

12059811205982(119904)1205772(119904) k⟩ = 0 we have 120598

1(119904)1205982(119904) = 0 by the

conditions of assertion (3) and 1205981(119904) = 0 that this is equivalent

to condition 1205982(119904) = 0 Therefore we have k = plusmn120577

2(119904) and

1205982(119904) = 0 The converse direction also holds(4) Since ℎ(3)

3V (119904) = ⟨(1205981015840

1(119904)1205981(119904))1205771(119904) + 120598

1(119904)1205981015840

2(119904)1205772(119904) minus

(1205983

1(119904) + 120598

1(119904)1205982(119904)2minus 12059810158401015840

1(119904))B(119904) k⟩ = 0 by the conditions

of assertion (4) and 1205981(119904) = 0 we have 120598

1(119904)1205981015840

2(119904) = 0 This is

equivalent to condition 12059810158402(119904) = 0 Therefore we have k =

plusmn1205772(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0 The converse direction also

holds(D) Using the same computation as the proof of (C) we

can get (D)

Proposition 3 Let 120574 119868 rarr E3 be a regular unit speed curveOne has the following claims

(1) Suppose 1198961(119904) = 0 Then 119896

2(119904) = 0 if and only if each

of k = ∓N2(119904) is a constant vector

(2) Suppose 1198962(119904) = 0 Then 119896

1(119904) = 0 if and only if each

of k = ∓N1(119904) is a constant vector

(3) Suppose 1205981(119904) = 0 Then 120598

2(119904) = 0 if and only if each

of k = ∓1205772(119904) is a constant vector

(4) Suppose 1205982(119904) = 0 Then 120598

1(119904) = 0 if and only if each

of k = ∓1205771(119904) is a constant vector

Proof (1) Suppose 1198961(119904) = 0 By straightforward calculations

we have

k1015840 (119904) = plusmn1198962 (119904)T (119904) (13)

Thus k1015840(119904) equiv 0 if and only if 1198962(119904) = 0

(2) Using the same computation as the proof of (1) we canget (2)

(3) Suppose 1205981(119904) = 0 By straightforward calculations we

have

k1015840 (119904) = plusmn1205982 (119904)B (119904) (14)

Thus k1015840(119904) equiv 0 if and only if 1205982(119904) = 0

(4) Using the same computation as the proof of (3) wecan get (4)

5 Versal Unfolding and Proofof the Main Result

In this section we use some general results on the singularitytheory for families of function germs Detailed descriptionscan be found in the book [16] Let 119865 (R times R119903 (119904

0 1199090)) rarr

Mathematical Problems in Engineering 5

R be a function germ We call 119865 an 119903-parameter unfoldingof 119891(119904) where 119891(119904) = 119865(119904 119909

0) We say that 119891(119904) have 119860

119896-

singularity at 1199040if 119891(119901)(119904

0) = 0 for all 1 le 119901 le 119896 and

119891(119896+1)

(1199040) = 0 We also say that 119891(119904) has 119860

ge119896-singularity at 119904

0

if 119891(119901)(1199040) = 0 for all 1 le 119901 le 119896 Let 119865 be an unfolding

of 119891(119904) and let 119891(119904) has 119860119896-singularity (119896 ge 1) at 119904

0 We

denote the (119896 minus 1)-jet of the partial derivative 120597119865120597119909119894at 1199040by

119895(119896minus1)

((120597119865120597119909119894)(119909 1199040))(1199040) = sum

119896minus1

119895=1119886119895119894119904119895 for 119894 = 1 119903 Then

119865 is called a versal unfolding if the 119896times 119903matrix of coefficients(1198860119894 119886119895119894) has rank 119896 (119896 le 119903) where 119886

0119894= (120597119865120597119909

119894)(1199090 1199040) We

now introduce an important set concerning the unfoldingThe discriminant set of 119865 is the set

D119865= 119909 isin R

119903| 119865 (119904 119909) =

120597119865

120597119904(119904 119909) = 0 (15)

Then we have the following well-known result (cf [16])

Theorem4 Let119865 (RtimesR119903 (1199040 1199090)) rarr R be an 119903-parameter

unfolding of 119891(119904) which has the 119860119896singularity at 119904

0

Suppose that 119865 is a (119901) versal unfolding Then one has thefollowing

(a) If 119896 = 1 thenD119865is locally diffeomorphic to 0 timesR119903minus1

(b) If 119896 = 2 thenD119865is locally diffeomorphic to 119862 timesR119903minus2

By Proposition 2 and the definition of discriminant setwe have the following proposition

Proposition 5 (1) The discriminant sets of 1198671and 119867

2are

respectively

D1198671

= k = ∓N2 (119904) | 119904 isin 119868

D1198672

= k = ∓N1 (119904) | 119904 isin 119868

(16)

(2) The discriminant sets of1198673and119867

4are respectively

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868

(17)

For the Bishop normal indicatrix height functions we canconsider the following propositions

Proposition 6 (1) If ℎ1V0

(119904) has 119860119896-singularity (119896 = 1 2) at

1199040 then119867

1(119904 k) is a (119901) versal unfolding of ℎ

1V0

(119904)(2) If ℎ

2V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198672(119904 k) is a (119901) versal unfolding of ℎ

2V0

(119904)(3) If ℎ

3V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198673(119904 k) is a (119901) versal unfolding of ℎ

3V0

(119904)(4) If ℎ

4V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198674(119904 k) is a (119901) versal unfolding of ℎ

4V0

(119904)

Proof (1) We denote that

N1 (119904) = (11989911 (119904) 11989912 (119904) 11989913 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(18)

Under this notation we have that

1198671 (119904 k) = 11989911 (119904) V1 + 11989912 (119904) V2 plusmn 11989913 (119904)radic1 minus V2

1minus V22 (19)

Thus we have that

1205971198671

120597V119894

= 1198991119894 (119904) ∓

V11989411989913 (119904)

radic1 minus V21minus V22

119894 = 1 2 (20)

We also have that

120597

120597119904

1205971198671

120597V119894

= 1198991015840

1119894(119904) ∓

V1198941198991015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198671

120597V119894

= 11989910158401015840

1119894(119904) ∓

V11989411989910158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(21)

Therefore the 2-jet of (1205971198671120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198671

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198671

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198671

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(22)

It is enough to show that the rank of the matrix 119860 is 2 where

119860 =(

1198991015840

11(119904) ∓

V11198991015840

13(119904)

radic1 minus V21minus V22

1198991015840

12(119904) ∓

V21198991015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

11(119904) ∓

V111989910158401015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

12(119904) ∓

V211989910158401015840

13(119904)

radic1 minus V21minus V22

)

(23)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119860 = (11989910158401015840121198991015840

11minus 1198991015840

1211989910158401015840

11) minus

V1

V3

(11989910158401015840

121198991015840

13minus 1198991015840

1211989910158401015840

13)

minusV2

V3

(11989910158401015840

131198991015840

11minus 1198991015840

1311989910158401015840

11)

=1

V3

⟨kN10158401(119904) and N10158401015840

1(119904)⟩

=1

V3

⟨k minus1198961 (119904)T (119904) and minus119896

1015840

1(119904)T (119904)

minus1198962

1(119904)N1 (119904) minus 1198961 (119904) 1198962 (119904)N2 (119904)⟩

=1

V3

⟨k 11989631(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

=1

V3

⟨∓N2 (119904) 119896

3

1(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

= ∓1

V3

1198963

1(119904) = 0

(24)

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

functions to get some families of functions We show thatthese singularities are versally unfolded by the families ofBishop normal indicatrix height functions If the singularityof ℎ119894V = 119867

119894(119904 k) and ℎ

119895V = 119867119895(119904 k) is 119860

119896-type (119896 =

1 2) and the corresponding 2-parameter unfolding is versalthen applying the theory of singularity [14 15] we knowthat discriminant set of the 2-parameter unfolding is locallydiffeomorphic to a line or a cusp thus we finished theclassification of singularities of the Bishop spherical imagesand type-2 Bishop spherical images because the discriminantsets of the unfolding are precisely the Bishop spherical imagesand type-2 Bishop spherical images Moreover we see thatthe 119860

119896-singularity (119896 = 1 2) of ℎ

119894V and ℎ119895V is closely

related to Bishop curvatures and type-2 Bishop curvaturesThe singular points of the Bishop spherical images and type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero Thus we get themain results in this paper which are stated in Theorem 1 Itis important that the properties of those functions relatedwith space curve 120574(119904) needed to be generic [15] Once weproved that the properties of those functions were genericwe could deduce that all singularities were stable under smallperturbations for our family of functions By consideringtransversality we prove that these properties given by us aregeneric As applications of our main results we give twoexamples

The rest of this paper is organized as follows Firstlywe introduce some basic concepts and the main results inthe next two sections Then we introduce four differentfamilies of functions on 120574 that will be useful to the study ofBishop spherical images and type-2 Bishop spherical imagesAfterwards some general results on the singularity theory areused for families of function germs and the main results areproved Finally we give two examples to illustrate the mainresults and the conclusion of the work is drawn

2 Preliminaries and Notations

In this section wewill introduce the notions of Bishop FrameldquoType-2 Bishop Framerdquo Bishop spherical images and type-2Bishop spherical images of unit speed regular curve Let 120574 =120574(119904) be a regular unit speed Frenet curve in E3 We know thatthere exist accompanying three frames called Frenet frame forFrenet curve Denote by (T(119904)N(119904)B(119904)) the moving Frenetframe along the unit speed Frenet curve 120574(119904)Then the Frenetformulas are given by

(

T1015840 (119904)N1015840 (119904)B1015840 (119904)

) = (

0 119896 (119904) 0

minus119896 (119904) 0 120591 (119904)

0 minus120591 (119904) 0

)(

T (119904)N (119904)B (119904)

) (1)

Here 119896(119904) and 120591(119904) are called curvature and torsion respec-tively [16] The Bishop Frame of the 120574(119904) is expressed by thealternative frame equations

(

T1015840 (119904)N10158401(119904)

N10158402(119904)

) = (

0 1198961 (119904) 1198962 (119904)

minus1198961 (119904) 0 0

minus1198962 (119904) 0 0

)(

T (119904)N1 (119904)

N2 (119904)

) (2)

Here we will call the set (T(119904)N1(119904)N2(119904)) as Bishop Frame

and 1198961(119904) = ⟨T1015840(119904)N

1(119904)⟩ and 119896

2(119904) = ⟨T1015840(119904)N

2(119904)⟩ as

Bishop curvatures The relation matrix can be expressed as

(

T (119904)N (119904)B (119904)

) = (

1 0 0

0 cos 120579 (119904) sin 120579 (119904)0 minus sin 120579 (119904) cos 120579 (119904)

)(

T (119904)N1 (119904)

N2 (119904)

) (3)

One can show that

119896 (119904) = radic1198962

1(119904) + 119896

2

2(119904)

120579 (119904) = arctan(1198962 (119904)1198961 (119904)

) where 1198961 (119904) = 0 120591 (119904) =

119889120579 (119904)

119889119904

(4)

so that 1198961(119904) and 119896

2(119904) effectively correspond to a Cartesian

coordinate system for the polar coordinates 119896(119904) and 120579(119904)with120579 = int 120591(119904)119889119904 Here Bishop curvatures are also defined by

1198961 (119904) = 119896 (119904) cos 120579 (119904)

1198962 (119904) = 119896 (119904) sin 120579 (119904)

(5)

The orientation of the parallel transport frame includes thearbitrary choice of integration constant 120579

0 which disappears

from 120591 (and hence from the Frenet frame) due to thedifferentiation [1]

The ldquoType-2 Bishop Framerdquo of the 120574(119904) is defined by thealternative frame equations see [7]

(

1205771015840

1(119904)

1205771015840

2(119904)

B1015840 (119904)) = (

0 0 minus1205981 (119904)

0 0 minus1205982 (119904)

1205981 (119904) 1205982 (119904) 0

)(

1205771 (119904)

1205772 (119904)

B (119904)) (6)

The relation matrix between Frenet-Serret and ldquoType-2Bishop Framerdquo can be expressed as

(

T (119904)N (119904)B (119904)

) = (

sin 120579 (119904) minus cos 120579 (119904) 0cos 120579 (119904) sin 120579 (119904) 0

0 0 1

)(

1205771 (119904)

1205772 (119904)

B (119904)) (7)

Here the type-2 Bishop curvatures are defined by

1205981 (119904) = minus120591 (119904) cos 120579 (119904)

1205982 (119904) = minus120591 (119904) sin 120579 (119904)

(8)

We will call the set (1205771(119904) 1205772(119904)B(119904)) ldquoType-2 Bishop Framerdquo

which is properly oriented and 1205981(119904) = ⟨B1015840(119904) 120577

1(119904)⟩ and

1205982(119904) = ⟨B1015840(119904) 120577

2(119904)⟩ type-2 Bishop curvatures One also can

show that

120581 (119904) =119889120579 (119904)

119889119904 (9)

so that 1205981(119904) and 120598

2(119904) also effectively correspond to a

Cartesian coordinate system for the polar coordinates 120591(119904)120579(119904) with 120579 = int 120581(119904)119889119904

The following notions are the main objects in this paperThe unit sphere with center in the origin in the space E3 isdefined by

1198782= x isin E

3| ⟨x x⟩ = 1 (10)

Mathematical Problems in Engineering 3

Translating framersquos vector fields to the center of unit spherewe obtain Bishop spherical images or Bishop spherical indi-catrixes The first Bishop spherical indicatrix and the secondBishop spherical indicatrix are denoted byFBN(119904) = N

1(119904)

and SBN(119904) = N2(119904) separately The first type-2 Bishop

spherical indicatrix and the second type-2 Bishop sphericalindicatrix in [7] is denoted by FNBN(119904) = 120577

1(119904) and

SNBN(119904) = 1205772(119904) separately

3 Cusps of Bishop Spherical Indicatrixes andTheir Visualizations

Themain results of this paper are in the following theorem

Theorem 1 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(1) Suppose that 1198962(119904) = 0 Then one have the following

claims

(a) The first Bishop spherical image N1(119904) is locally

diffeomorphic to a line 0 timesR at 1199040if 1198961(1199040) = 0

(b) The first Bishop spherical image N1(119904) is locally

diffeomorphic to the cusp119862 at 1199040if 1198961(1199040) = 0 and

1198961015840

1(1199040) = 0

(2) Suppose that 1198961(119904) = 0 Then one has the following

claims

(a) The second Bishop spherical image N2(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1198962(1199040) = 0

(b) The second Bishop spherical image N2(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1198962(1199040) =

0 and 11989610158402(1199040) = 0

(3) Suppose that 1205982(119904) = 0 Then one has the following

claims

(a) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1205981(1199040) = 0

(b) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205981(1199040) =

0 and 12059810158401(1199040) = 0

(4) Suppose that 1205981(119904) = 0 Then one has the following

claims

(a) The second type-2 Bishop spherical image 1205772(119904)

is locally diffeomorphic to a line 0 times R at 1199040if

1205982(1199040) = 0

(b) The second type-2 Bishop spherical image 1205772(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205982(1199040) =

0 and 12059810158402(1199040) = 0 where the ordinary cusp is 119862 =

(1199091 1199092) | 1199092

1= 1199093

2 The picture of cusp will be

seen in Figure 1

minus05 minus04 minus03 minus02 minus01 0 01 02 03 04 050

01

02

03

04

05

06

07

Figure 1 Cusp

4 Bishop Normal Indicatrix Height Functions

In this section we will introduce four different families offunctions on 120574 that will be useful to study the singular pointsof the Bishop spherical images and type-2 Bishop sphericalimages of unit speed regular curve Let 120574 119868 rarr E3 bea regular unit speed curve Now we define two families ofsmooth functions on 119868 as follows

119867119894 119868 times 119878

2997888rarr R by 119867

119894 (119904 k) = ⟨N119894 (119904) k⟩ (11)

where 119894 = 1 2 We call it the first (second) Bishop normalindicatrix height function for the case 119894 = 1 (119894 = 2) Forany k isin 1198782 we denote ℎ

119894V(119904) = 119867119894(119904 k) We also define two

families of smooth functions on 119868 as follows

119867119895 119868 times 119878

2997888rarr R by 119867

119895 (119904 k) = ⟨120577119895 (119904) k⟩ (12)

where 119895 = 3 4 We call it the first (second) type-2 Bishopnormal indicatrix height function for the case 119895 = 3 (119895 = 4)For any k isin 1198782 we denote ℎ

119895V(119904) = 119867119895(119904 k) Then we have thefollowing proposition

Proposition 2 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(A) Suppose that 1198961(119904) = 0 Then one has the following

claims

(1) ℎ1V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that v = 120582T(119904)+120583N2(119904) and 1205822+1205832 =

1(2) ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)(3) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0 if and only if k =∓N2(119904) and 119896

2(119904) = 0

(4) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = ℎ(3)

1V (119904) = 0 if and onlyif k = ∓N

2(119904) and 119896

2(119904) = 119896

1015840

2(119904) = 0

4 Mathematical Problems in Engineering

(B) Suppose that 1198962(119904) = 0 Then one has the following

claims

(1) ℎ2V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 120582T(119904)+120583N1(119904) and 1205822+1205832 =

1(2) ℎ2V(119904) = ℎ

1015840

2V(119904) = 0 if and only if k = ∓N1(119904)(3) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = 0 if and only if k =∓N1(119904) and 119896

1(119904) = 0

(4) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = ℎ(3)

2V (119904) = 0 if and onlyif k = ∓N

1(119904) and 119896

1(119904) = 119896

1015840

1(119904) = 0

(C) Suppose that 1205981(119904) = 0 Then one has the following

claims

(1) ℎ3V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205772(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ3V(119904) = ℎ

1015840

3V(119904) = 0 if and only if k = ∓1205772(119904)(3) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = 0 if and only if k =∓1205772(119904) and 120598

2(119904) = 0

(4) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = ℎ(3)

3V (119904) = 0 if and onlyif k = ∓120577

2(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0

(D) Suppose that 1205982(119904) = 0 Then one has the following

claims

(1) ℎ4V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205771(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ4V(119904) = ℎ

1015840

4V(119904) = 0 if and only if k = ∓1205771(119904)(3) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = 0 if and only if k =∓1205771(119904) and 120598

1(119904) = 0

(4) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = ℎ(3)

4V (119904) = 0 if and onlyif k = ∓120577

1(119904) and 120598

1(119904) = 120598

1015840

1(119904) = 0

Proof (A) (1) If ℎ1V(119904) = ⟨N

1(119904) k⟩ = 0 then we have that

there are real numbers 120582 and 120583 such that k = 120582T(119904) + 120583N2(119904)

Moreover in combinationwith k isin 1198782 whichmeans 1205822+1205832 =1 it follows that ℎ

1V(119904) = 0 if and only if k = 120582T(119904) + 120583N2(119904)

and 1205822 + 1205832 = 1(2) When ℎ

1V(119904) = 0 the assertion (2) follows from thefact that ℎ1015840

1V(119904) = ⟨1198961(119904)T(119904) k⟩ = 1198961(119904)120582 and 1198961(119904) = 0 Thus

we get that ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)

(3) When ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 the assertion (3) followsfrom the fact that ℎ10158401015840

1V(119904) = ⟨1198961015840

1(119904)T(119904) + 119896

2

1(119904)N1(119904) +

1198961(119904)1198962(119904)N2(119904) ∓N

2(119904)⟩ = ∓119896

1(119904)1198962(119904) and 119896

1(119904) = 0

(4) Under the condition that ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0this derivative is computed as follows ℎ(3)

1V (119904) = ⟨(11989610158401015840

1(119904) +

1198963

1(119904) + 119896

1(119904)1198962

2(119904))T(119904) + 3119896

1(119904)1198961015840

1(119904)N1(119904) + (2119896

1015840

1(119904)1198962(119904) +

1198961(119904)1198961015840

2(119904))N2(119904) ∓N

2(s)⟩ = 2119896

1015840

1(119904)1198962(119904) + 119896

1(119904)1198961015840

2(119904) =

1198961(119904)1198961015840

2(119904) Since 119896

1(119904) = 0 we get that ℎ(3)

1V (119904) = 0 is equivalentto condition 1198961015840

2(119904) = 0 The assertion (4) follows

(B) Using the same computation as the proof of (A) wecan get (B)

(C) (1) Since ℎ3V(119904) = ⟨120577

1(119904) k⟩ = 0 we have that there

are real numbers 120582 and 120583 such that k = 120582B(119904) + 1205831205772(119904) By

the condition that k isin 1198782 we get 1205822 + 1205832 = 1 The conversedirection also holds

(2) Since ℎ10158403V(119904) = ⟨minus1205981(119904)B(119904) k⟩ = 0 by the conditions

of assertion (2) and 1205981(119904) = 0 we have that 120598

1(119904)120582 = 0 It

follows from the fact that 1205981(119904)120582 = 0 and 120598

1(119904) = 0 that this is

equivalent to condition 120582 = 0 Therefore we have k = plusmn1205772(119904)

The converse direction also holds(3) Since ℎ

10158401015840

3V(119904) = ⟨minus1205981015840

1(119904)B(119904) + 120598

2

1(119904)1205771(119904) +

12059811205982(119904)1205772(119904) k⟩ = 0 we have 120598

1(119904)1205982(119904) = 0 by the

conditions of assertion (3) and 1205981(119904) = 0 that this is equivalent

to condition 1205982(119904) = 0 Therefore we have k = plusmn120577

2(119904) and

1205982(119904) = 0 The converse direction also holds(4) Since ℎ(3)

3V (119904) = ⟨(1205981015840

1(119904)1205981(119904))1205771(119904) + 120598

1(119904)1205981015840

2(119904)1205772(119904) minus

(1205983

1(119904) + 120598

1(119904)1205982(119904)2minus 12059810158401015840

1(119904))B(119904) k⟩ = 0 by the conditions

of assertion (4) and 1205981(119904) = 0 we have 120598

1(119904)1205981015840

2(119904) = 0 This is

equivalent to condition 12059810158402(119904) = 0 Therefore we have k =

plusmn1205772(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0 The converse direction also

holds(D) Using the same computation as the proof of (C) we

can get (D)

Proposition 3 Let 120574 119868 rarr E3 be a regular unit speed curveOne has the following claims

(1) Suppose 1198961(119904) = 0 Then 119896

2(119904) = 0 if and only if each

of k = ∓N2(119904) is a constant vector

(2) Suppose 1198962(119904) = 0 Then 119896

1(119904) = 0 if and only if each

of k = ∓N1(119904) is a constant vector

(3) Suppose 1205981(119904) = 0 Then 120598

2(119904) = 0 if and only if each

of k = ∓1205772(119904) is a constant vector

(4) Suppose 1205982(119904) = 0 Then 120598

1(119904) = 0 if and only if each

of k = ∓1205771(119904) is a constant vector

Proof (1) Suppose 1198961(119904) = 0 By straightforward calculations

we have

k1015840 (119904) = plusmn1198962 (119904)T (119904) (13)

Thus k1015840(119904) equiv 0 if and only if 1198962(119904) = 0

(2) Using the same computation as the proof of (1) we canget (2)

(3) Suppose 1205981(119904) = 0 By straightforward calculations we

have

k1015840 (119904) = plusmn1205982 (119904)B (119904) (14)

Thus k1015840(119904) equiv 0 if and only if 1205982(119904) = 0

(4) Using the same computation as the proof of (3) wecan get (4)

5 Versal Unfolding and Proofof the Main Result

In this section we use some general results on the singularitytheory for families of function germs Detailed descriptionscan be found in the book [16] Let 119865 (R times R119903 (119904

0 1199090)) rarr

Mathematical Problems in Engineering 5

R be a function germ We call 119865 an 119903-parameter unfoldingof 119891(119904) where 119891(119904) = 119865(119904 119909

0) We say that 119891(119904) have 119860

119896-

singularity at 1199040if 119891(119901)(119904

0) = 0 for all 1 le 119901 le 119896 and

119891(119896+1)

(1199040) = 0 We also say that 119891(119904) has 119860

ge119896-singularity at 119904

0

if 119891(119901)(1199040) = 0 for all 1 le 119901 le 119896 Let 119865 be an unfolding

of 119891(119904) and let 119891(119904) has 119860119896-singularity (119896 ge 1) at 119904

0 We

denote the (119896 minus 1)-jet of the partial derivative 120597119865120597119909119894at 1199040by

119895(119896minus1)

((120597119865120597119909119894)(119909 1199040))(1199040) = sum

119896minus1

119895=1119886119895119894119904119895 for 119894 = 1 119903 Then

119865 is called a versal unfolding if the 119896times 119903matrix of coefficients(1198860119894 119886119895119894) has rank 119896 (119896 le 119903) where 119886

0119894= (120597119865120597119909

119894)(1199090 1199040) We

now introduce an important set concerning the unfoldingThe discriminant set of 119865 is the set

D119865= 119909 isin R

119903| 119865 (119904 119909) =

120597119865

120597119904(119904 119909) = 0 (15)

Then we have the following well-known result (cf [16])

Theorem4 Let119865 (RtimesR119903 (1199040 1199090)) rarr R be an 119903-parameter

unfolding of 119891(119904) which has the 119860119896singularity at 119904

0

Suppose that 119865 is a (119901) versal unfolding Then one has thefollowing

(a) If 119896 = 1 thenD119865is locally diffeomorphic to 0 timesR119903minus1

(b) If 119896 = 2 thenD119865is locally diffeomorphic to 119862 timesR119903minus2

By Proposition 2 and the definition of discriminant setwe have the following proposition

Proposition 5 (1) The discriminant sets of 1198671and 119867

2are

respectively

D1198671

= k = ∓N2 (119904) | 119904 isin 119868

D1198672

= k = ∓N1 (119904) | 119904 isin 119868

(16)

(2) The discriminant sets of1198673and119867

4are respectively

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868

(17)

For the Bishop normal indicatrix height functions we canconsider the following propositions

Proposition 6 (1) If ℎ1V0

(119904) has 119860119896-singularity (119896 = 1 2) at

1199040 then119867

1(119904 k) is a (119901) versal unfolding of ℎ

1V0

(119904)(2) If ℎ

2V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198672(119904 k) is a (119901) versal unfolding of ℎ

2V0

(119904)(3) If ℎ

3V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198673(119904 k) is a (119901) versal unfolding of ℎ

3V0

(119904)(4) If ℎ

4V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198674(119904 k) is a (119901) versal unfolding of ℎ

4V0

(119904)

Proof (1) We denote that

N1 (119904) = (11989911 (119904) 11989912 (119904) 11989913 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(18)

Under this notation we have that

1198671 (119904 k) = 11989911 (119904) V1 + 11989912 (119904) V2 plusmn 11989913 (119904)radic1 minus V2

1minus V22 (19)

Thus we have that

1205971198671

120597V119894

= 1198991119894 (119904) ∓

V11989411989913 (119904)

radic1 minus V21minus V22

119894 = 1 2 (20)

We also have that

120597

120597119904

1205971198671

120597V119894

= 1198991015840

1119894(119904) ∓

V1198941198991015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198671

120597V119894

= 11989910158401015840

1119894(119904) ∓

V11989411989910158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(21)

Therefore the 2-jet of (1205971198671120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198671

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198671

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198671

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(22)

It is enough to show that the rank of the matrix 119860 is 2 where

119860 =(

1198991015840

11(119904) ∓

V11198991015840

13(119904)

radic1 minus V21minus V22

1198991015840

12(119904) ∓

V21198991015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

11(119904) ∓

V111989910158401015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

12(119904) ∓

V211989910158401015840

13(119904)

radic1 minus V21minus V22

)

(23)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119860 = (11989910158401015840121198991015840

11minus 1198991015840

1211989910158401015840

11) minus

V1

V3

(11989910158401015840

121198991015840

13minus 1198991015840

1211989910158401015840

13)

minusV2

V3

(11989910158401015840

131198991015840

11minus 1198991015840

1311989910158401015840

11)

=1

V3

⟨kN10158401(119904) and N10158401015840

1(119904)⟩

=1

V3

⟨k minus1198961 (119904)T (119904) and minus119896

1015840

1(119904)T (119904)

minus1198962

1(119904)N1 (119904) minus 1198961 (119904) 1198962 (119904)N2 (119904)⟩

=1

V3

⟨k 11989631(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

=1

V3

⟨∓N2 (119904) 119896

3

1(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

= ∓1

V3

1198963

1(119904) = 0

(24)

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

Translating framersquos vector fields to the center of unit spherewe obtain Bishop spherical images or Bishop spherical indi-catrixes The first Bishop spherical indicatrix and the secondBishop spherical indicatrix are denoted byFBN(119904) = N

1(119904)

and SBN(119904) = N2(119904) separately The first type-2 Bishop

spherical indicatrix and the second type-2 Bishop sphericalindicatrix in [7] is denoted by FNBN(119904) = 120577

1(119904) and

SNBN(119904) = 1205772(119904) separately

3 Cusps of Bishop Spherical Indicatrixes andTheir Visualizations

Themain results of this paper are in the following theorem

Theorem 1 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(1) Suppose that 1198962(119904) = 0 Then one have the following

claims

(a) The first Bishop spherical image N1(119904) is locally

diffeomorphic to a line 0 timesR at 1199040if 1198961(1199040) = 0

(b) The first Bishop spherical image N1(119904) is locally

diffeomorphic to the cusp119862 at 1199040if 1198961(1199040) = 0 and

1198961015840

1(1199040) = 0

(2) Suppose that 1198961(119904) = 0 Then one has the following

claims

(a) The second Bishop spherical image N2(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1198962(1199040) = 0

(b) The second Bishop spherical image N2(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1198962(1199040) =

0 and 11989610158402(1199040) = 0

(3) Suppose that 1205982(119904) = 0 Then one has the following

claims

(a) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to a line 0 times R at 1199040if

1205981(1199040) = 0

(b) The first type-2 Bishop spherical image 1205771(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205981(1199040) =

0 and 12059810158401(1199040) = 0

(4) Suppose that 1205981(119904) = 0 Then one has the following

claims

(a) The second type-2 Bishop spherical image 1205772(119904)

is locally diffeomorphic to a line 0 times R at 1199040if

1205982(1199040) = 0

(b) The second type-2 Bishop spherical image 1205772(119904) is

locally diffeomorphic to the cusp119862 at 1199040if 1205982(1199040) =

0 and 12059810158402(1199040) = 0 where the ordinary cusp is 119862 =

(1199091 1199092) | 1199092

1= 1199093

2 The picture of cusp will be

seen in Figure 1

minus05 minus04 minus03 minus02 minus01 0 01 02 03 04 050

01

02

03

04

05

06

07

Figure 1 Cusp

4 Bishop Normal Indicatrix Height Functions

In this section we will introduce four different families offunctions on 120574 that will be useful to study the singular pointsof the Bishop spherical images and type-2 Bishop sphericalimages of unit speed regular curve Let 120574 119868 rarr E3 bea regular unit speed curve Now we define two families ofsmooth functions on 119868 as follows

119867119894 119868 times 119878

2997888rarr R by 119867

119894 (119904 k) = ⟨N119894 (119904) k⟩ (11)

where 119894 = 1 2 We call it the first (second) Bishop normalindicatrix height function for the case 119894 = 1 (119894 = 2) Forany k isin 1198782 we denote ℎ

119894V(119904) = 119867119894(119904 k) We also define two

families of smooth functions on 119868 as follows

119867119895 119868 times 119878

2997888rarr R by 119867

119895 (119904 k) = ⟨120577119895 (119904) k⟩ (12)

where 119895 = 3 4 We call it the first (second) type-2 Bishopnormal indicatrix height function for the case 119895 = 3 (119895 = 4)For any k isin 1198782 we denote ℎ

119895V(119904) = 119867119895(119904 k) Then we have thefollowing proposition

Proposition 2 Let 120574 119868 rarr E3 be a regular unit speed curveThen one has the following

(A) Suppose that 1198961(119904) = 0 Then one has the following

claims

(1) ℎ1V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that v = 120582T(119904)+120583N2(119904) and 1205822+1205832 =

1(2) ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)(3) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0 if and only if k =∓N2(119904) and 119896

2(119904) = 0

(4) ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = ℎ(3)

1V (119904) = 0 if and onlyif k = ∓N

2(119904) and 119896

2(119904) = 119896

1015840

2(119904) = 0

4 Mathematical Problems in Engineering

(B) Suppose that 1198962(119904) = 0 Then one has the following

claims

(1) ℎ2V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 120582T(119904)+120583N1(119904) and 1205822+1205832 =

1(2) ℎ2V(119904) = ℎ

1015840

2V(119904) = 0 if and only if k = ∓N1(119904)(3) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = 0 if and only if k =∓N1(119904) and 119896

1(119904) = 0

(4) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = ℎ(3)

2V (119904) = 0 if and onlyif k = ∓N

1(119904) and 119896

1(119904) = 119896

1015840

1(119904) = 0

(C) Suppose that 1205981(119904) = 0 Then one has the following

claims

(1) ℎ3V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205772(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ3V(119904) = ℎ

1015840

3V(119904) = 0 if and only if k = ∓1205772(119904)(3) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = 0 if and only if k =∓1205772(119904) and 120598

2(119904) = 0

(4) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = ℎ(3)

3V (119904) = 0 if and onlyif k = ∓120577

2(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0

(D) Suppose that 1205982(119904) = 0 Then one has the following

claims

(1) ℎ4V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205771(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ4V(119904) = ℎ

1015840

4V(119904) = 0 if and only if k = ∓1205771(119904)(3) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = 0 if and only if k =∓1205771(119904) and 120598

1(119904) = 0

(4) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = ℎ(3)

4V (119904) = 0 if and onlyif k = ∓120577

1(119904) and 120598

1(119904) = 120598

1015840

1(119904) = 0

Proof (A) (1) If ℎ1V(119904) = ⟨N

1(119904) k⟩ = 0 then we have that

there are real numbers 120582 and 120583 such that k = 120582T(119904) + 120583N2(119904)

Moreover in combinationwith k isin 1198782 whichmeans 1205822+1205832 =1 it follows that ℎ

1V(119904) = 0 if and only if k = 120582T(119904) + 120583N2(119904)

and 1205822 + 1205832 = 1(2) When ℎ

1V(119904) = 0 the assertion (2) follows from thefact that ℎ1015840

1V(119904) = ⟨1198961(119904)T(119904) k⟩ = 1198961(119904)120582 and 1198961(119904) = 0 Thus

we get that ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)

(3) When ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 the assertion (3) followsfrom the fact that ℎ10158401015840

1V(119904) = ⟨1198961015840

1(119904)T(119904) + 119896

2

1(119904)N1(119904) +

1198961(119904)1198962(119904)N2(119904) ∓N

2(119904)⟩ = ∓119896

1(119904)1198962(119904) and 119896

1(119904) = 0

(4) Under the condition that ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0this derivative is computed as follows ℎ(3)

1V (119904) = ⟨(11989610158401015840

1(119904) +

1198963

1(119904) + 119896

1(119904)1198962

2(119904))T(119904) + 3119896

1(119904)1198961015840

1(119904)N1(119904) + (2119896

1015840

1(119904)1198962(119904) +

1198961(119904)1198961015840

2(119904))N2(119904) ∓N

2(s)⟩ = 2119896

1015840

1(119904)1198962(119904) + 119896

1(119904)1198961015840

2(119904) =

1198961(119904)1198961015840

2(119904) Since 119896

1(119904) = 0 we get that ℎ(3)

1V (119904) = 0 is equivalentto condition 1198961015840

2(119904) = 0 The assertion (4) follows

(B) Using the same computation as the proof of (A) wecan get (B)

(C) (1) Since ℎ3V(119904) = ⟨120577

1(119904) k⟩ = 0 we have that there

are real numbers 120582 and 120583 such that k = 120582B(119904) + 1205831205772(119904) By

the condition that k isin 1198782 we get 1205822 + 1205832 = 1 The conversedirection also holds

(2) Since ℎ10158403V(119904) = ⟨minus1205981(119904)B(119904) k⟩ = 0 by the conditions

of assertion (2) and 1205981(119904) = 0 we have that 120598

1(119904)120582 = 0 It

follows from the fact that 1205981(119904)120582 = 0 and 120598

1(119904) = 0 that this is

equivalent to condition 120582 = 0 Therefore we have k = plusmn1205772(119904)

The converse direction also holds(3) Since ℎ

10158401015840

3V(119904) = ⟨minus1205981015840

1(119904)B(119904) + 120598

2

1(119904)1205771(119904) +

12059811205982(119904)1205772(119904) k⟩ = 0 we have 120598

1(119904)1205982(119904) = 0 by the

conditions of assertion (3) and 1205981(119904) = 0 that this is equivalent

to condition 1205982(119904) = 0 Therefore we have k = plusmn120577

2(119904) and

1205982(119904) = 0 The converse direction also holds(4) Since ℎ(3)

3V (119904) = ⟨(1205981015840

1(119904)1205981(119904))1205771(119904) + 120598

1(119904)1205981015840

2(119904)1205772(119904) minus

(1205983

1(119904) + 120598

1(119904)1205982(119904)2minus 12059810158401015840

1(119904))B(119904) k⟩ = 0 by the conditions

of assertion (4) and 1205981(119904) = 0 we have 120598

1(119904)1205981015840

2(119904) = 0 This is

equivalent to condition 12059810158402(119904) = 0 Therefore we have k =

plusmn1205772(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0 The converse direction also

holds(D) Using the same computation as the proof of (C) we

can get (D)

Proposition 3 Let 120574 119868 rarr E3 be a regular unit speed curveOne has the following claims

(1) Suppose 1198961(119904) = 0 Then 119896

2(119904) = 0 if and only if each

of k = ∓N2(119904) is a constant vector

(2) Suppose 1198962(119904) = 0 Then 119896

1(119904) = 0 if and only if each

of k = ∓N1(119904) is a constant vector

(3) Suppose 1205981(119904) = 0 Then 120598

2(119904) = 0 if and only if each

of k = ∓1205772(119904) is a constant vector

(4) Suppose 1205982(119904) = 0 Then 120598

1(119904) = 0 if and only if each

of k = ∓1205771(119904) is a constant vector

Proof (1) Suppose 1198961(119904) = 0 By straightforward calculations

we have

k1015840 (119904) = plusmn1198962 (119904)T (119904) (13)

Thus k1015840(119904) equiv 0 if and only if 1198962(119904) = 0

(2) Using the same computation as the proof of (1) we canget (2)

(3) Suppose 1205981(119904) = 0 By straightforward calculations we

have

k1015840 (119904) = plusmn1205982 (119904)B (119904) (14)

Thus k1015840(119904) equiv 0 if and only if 1205982(119904) = 0

(4) Using the same computation as the proof of (3) wecan get (4)

5 Versal Unfolding and Proofof the Main Result

In this section we use some general results on the singularitytheory for families of function germs Detailed descriptionscan be found in the book [16] Let 119865 (R times R119903 (119904

0 1199090)) rarr

Mathematical Problems in Engineering 5

R be a function germ We call 119865 an 119903-parameter unfoldingof 119891(119904) where 119891(119904) = 119865(119904 119909

0) We say that 119891(119904) have 119860

119896-

singularity at 1199040if 119891(119901)(119904

0) = 0 for all 1 le 119901 le 119896 and

119891(119896+1)

(1199040) = 0 We also say that 119891(119904) has 119860

ge119896-singularity at 119904

0

if 119891(119901)(1199040) = 0 for all 1 le 119901 le 119896 Let 119865 be an unfolding

of 119891(119904) and let 119891(119904) has 119860119896-singularity (119896 ge 1) at 119904

0 We

denote the (119896 minus 1)-jet of the partial derivative 120597119865120597119909119894at 1199040by

119895(119896minus1)

((120597119865120597119909119894)(119909 1199040))(1199040) = sum

119896minus1

119895=1119886119895119894119904119895 for 119894 = 1 119903 Then

119865 is called a versal unfolding if the 119896times 119903matrix of coefficients(1198860119894 119886119895119894) has rank 119896 (119896 le 119903) where 119886

0119894= (120597119865120597119909

119894)(1199090 1199040) We

now introduce an important set concerning the unfoldingThe discriminant set of 119865 is the set

D119865= 119909 isin R

119903| 119865 (119904 119909) =

120597119865

120597119904(119904 119909) = 0 (15)

Then we have the following well-known result (cf [16])

Theorem4 Let119865 (RtimesR119903 (1199040 1199090)) rarr R be an 119903-parameter

unfolding of 119891(119904) which has the 119860119896singularity at 119904

0

Suppose that 119865 is a (119901) versal unfolding Then one has thefollowing

(a) If 119896 = 1 thenD119865is locally diffeomorphic to 0 timesR119903minus1

(b) If 119896 = 2 thenD119865is locally diffeomorphic to 119862 timesR119903minus2

By Proposition 2 and the definition of discriminant setwe have the following proposition

Proposition 5 (1) The discriminant sets of 1198671and 119867

2are

respectively

D1198671

= k = ∓N2 (119904) | 119904 isin 119868

D1198672

= k = ∓N1 (119904) | 119904 isin 119868

(16)

(2) The discriminant sets of1198673and119867

4are respectively

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868

(17)

For the Bishop normal indicatrix height functions we canconsider the following propositions

Proposition 6 (1) If ℎ1V0

(119904) has 119860119896-singularity (119896 = 1 2) at

1199040 then119867

1(119904 k) is a (119901) versal unfolding of ℎ

1V0

(119904)(2) If ℎ

2V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198672(119904 k) is a (119901) versal unfolding of ℎ

2V0

(119904)(3) If ℎ

3V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198673(119904 k) is a (119901) versal unfolding of ℎ

3V0

(119904)(4) If ℎ

4V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198674(119904 k) is a (119901) versal unfolding of ℎ

4V0

(119904)

Proof (1) We denote that

N1 (119904) = (11989911 (119904) 11989912 (119904) 11989913 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(18)

Under this notation we have that

1198671 (119904 k) = 11989911 (119904) V1 + 11989912 (119904) V2 plusmn 11989913 (119904)radic1 minus V2

1minus V22 (19)

Thus we have that

1205971198671

120597V119894

= 1198991119894 (119904) ∓

V11989411989913 (119904)

radic1 minus V21minus V22

119894 = 1 2 (20)

We also have that

120597

120597119904

1205971198671

120597V119894

= 1198991015840

1119894(119904) ∓

V1198941198991015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198671

120597V119894

= 11989910158401015840

1119894(119904) ∓

V11989411989910158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(21)

Therefore the 2-jet of (1205971198671120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198671

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198671

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198671

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(22)

It is enough to show that the rank of the matrix 119860 is 2 where

119860 =(

1198991015840

11(119904) ∓

V11198991015840

13(119904)

radic1 minus V21minus V22

1198991015840

12(119904) ∓

V21198991015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

11(119904) ∓

V111989910158401015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

12(119904) ∓

V211989910158401015840

13(119904)

radic1 minus V21minus V22

)

(23)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119860 = (11989910158401015840121198991015840

11minus 1198991015840

1211989910158401015840

11) minus

V1

V3

(11989910158401015840

121198991015840

13minus 1198991015840

1211989910158401015840

13)

minusV2

V3

(11989910158401015840

131198991015840

11minus 1198991015840

1311989910158401015840

11)

=1

V3

⟨kN10158401(119904) and N10158401015840

1(119904)⟩

=1

V3

⟨k minus1198961 (119904)T (119904) and minus119896

1015840

1(119904)T (119904)

minus1198962

1(119904)N1 (119904) minus 1198961 (119904) 1198962 (119904)N2 (119904)⟩

=1

V3

⟨k 11989631(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

=1

V3

⟨∓N2 (119904) 119896

3

1(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

= ∓1

V3

1198963

1(119904) = 0

(24)

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

(B) Suppose that 1198962(119904) = 0 Then one has the following

claims

(1) ℎ2V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 120582T(119904)+120583N1(119904) and 1205822+1205832 =

1(2) ℎ2V(119904) = ℎ

1015840

2V(119904) = 0 if and only if k = ∓N1(119904)(3) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = 0 if and only if k =∓N1(119904) and 119896

1(119904) = 0

(4) ℎ2V(119904) = ℎ

1015840

2V(119904) = ℎ10158401015840

2V(119904) = ℎ(3)

2V (119904) = 0 if and onlyif k = ∓N

1(119904) and 119896

1(119904) = 119896

1015840

1(119904) = 0

(C) Suppose that 1205981(119904) = 0 Then one has the following

claims

(1) ℎ3V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205772(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ3V(119904) = ℎ

1015840

3V(119904) = 0 if and only if k = ∓1205772(119904)(3) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = 0 if and only if k =∓1205772(119904) and 120598

2(119904) = 0

(4) ℎ3V(119904) = ℎ

1015840

3V(119904) = ℎ10158401015840

3V(119904) = ℎ(3)

3V (119904) = 0 if and onlyif k = ∓120577

2(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0

(D) Suppose that 1205982(119904) = 0 Then one has the following

claims

(1) ℎ4V(119904) = 0 if and only if there are real numbers 120582

and 120583 such that k = 1205821205771(119904)+120583B(119904) and 1205822 +1205832 =

1(2) ℎ4V(119904) = ℎ

1015840

4V(119904) = 0 if and only if k = ∓1205771(119904)(3) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = 0 if and only if k =∓1205771(119904) and 120598

1(119904) = 0

(4) ℎ4V(119904) = ℎ

1015840

4V(119904) = ℎ10158401015840

4V(119904) = ℎ(3)

4V (119904) = 0 if and onlyif k = ∓120577

1(119904) and 120598

1(119904) = 120598

1015840

1(119904) = 0

Proof (A) (1) If ℎ1V(119904) = ⟨N

1(119904) k⟩ = 0 then we have that

there are real numbers 120582 and 120583 such that k = 120582T(119904) + 120583N2(119904)

Moreover in combinationwith k isin 1198782 whichmeans 1205822+1205832 =1 it follows that ℎ

1V(119904) = 0 if and only if k = 120582T(119904) + 120583N2(119904)

and 1205822 + 1205832 = 1(2) When ℎ

1V(119904) = 0 the assertion (2) follows from thefact that ℎ1015840

1V(119904) = ⟨1198961(119904)T(119904) k⟩ = 1198961(119904)120582 and 1198961(119904) = 0 Thus

we get that ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 if and only if k = ∓N2(119904)

(3) When ℎ1V(119904) = ℎ

1015840

1V(119904) = 0 the assertion (3) followsfrom the fact that ℎ10158401015840

1V(119904) = ⟨1198961015840

1(119904)T(119904) + 119896

2

1(119904)N1(119904) +

1198961(119904)1198962(119904)N2(119904) ∓N

2(119904)⟩ = ∓119896

1(119904)1198962(119904) and 119896

1(119904) = 0

(4) Under the condition that ℎ1V(119904) = ℎ

1015840

1V(119904) = ℎ10158401015840

1V(119904) = 0this derivative is computed as follows ℎ(3)

1V (119904) = ⟨(11989610158401015840

1(119904) +

1198963

1(119904) + 119896

1(119904)1198962

2(119904))T(119904) + 3119896

1(119904)1198961015840

1(119904)N1(119904) + (2119896

1015840

1(119904)1198962(119904) +

1198961(119904)1198961015840

2(119904))N2(119904) ∓N

2(s)⟩ = 2119896

1015840

1(119904)1198962(119904) + 119896

1(119904)1198961015840

2(119904) =

1198961(119904)1198961015840

2(119904) Since 119896

1(119904) = 0 we get that ℎ(3)

1V (119904) = 0 is equivalentto condition 1198961015840

2(119904) = 0 The assertion (4) follows

(B) Using the same computation as the proof of (A) wecan get (B)

(C) (1) Since ℎ3V(119904) = ⟨120577

1(119904) k⟩ = 0 we have that there

are real numbers 120582 and 120583 such that k = 120582B(119904) + 1205831205772(119904) By

the condition that k isin 1198782 we get 1205822 + 1205832 = 1 The conversedirection also holds

(2) Since ℎ10158403V(119904) = ⟨minus1205981(119904)B(119904) k⟩ = 0 by the conditions

of assertion (2) and 1205981(119904) = 0 we have that 120598

1(119904)120582 = 0 It

follows from the fact that 1205981(119904)120582 = 0 and 120598

1(119904) = 0 that this is

equivalent to condition 120582 = 0 Therefore we have k = plusmn1205772(119904)

The converse direction also holds(3) Since ℎ

10158401015840

3V(119904) = ⟨minus1205981015840

1(119904)B(119904) + 120598

2

1(119904)1205771(119904) +

12059811205982(119904)1205772(119904) k⟩ = 0 we have 120598

1(119904)1205982(119904) = 0 by the

conditions of assertion (3) and 1205981(119904) = 0 that this is equivalent

to condition 1205982(119904) = 0 Therefore we have k = plusmn120577

2(119904) and

1205982(119904) = 0 The converse direction also holds(4) Since ℎ(3)

3V (119904) = ⟨(1205981015840

1(119904)1205981(119904))1205771(119904) + 120598

1(119904)1205981015840

2(119904)1205772(119904) minus

(1205983

1(119904) + 120598

1(119904)1205982(119904)2minus 12059810158401015840

1(119904))B(119904) k⟩ = 0 by the conditions

of assertion (4) and 1205981(119904) = 0 we have 120598

1(119904)1205981015840

2(119904) = 0 This is

equivalent to condition 12059810158402(119904) = 0 Therefore we have k =

plusmn1205772(119904) and 120598

2(119904) = 120598

1015840

2(119904) = 0 The converse direction also

holds(D) Using the same computation as the proof of (C) we

can get (D)

Proposition 3 Let 120574 119868 rarr E3 be a regular unit speed curveOne has the following claims

(1) Suppose 1198961(119904) = 0 Then 119896

2(119904) = 0 if and only if each

of k = ∓N2(119904) is a constant vector

(2) Suppose 1198962(119904) = 0 Then 119896

1(119904) = 0 if and only if each

of k = ∓N1(119904) is a constant vector

(3) Suppose 1205981(119904) = 0 Then 120598

2(119904) = 0 if and only if each

of k = ∓1205772(119904) is a constant vector

(4) Suppose 1205982(119904) = 0 Then 120598

1(119904) = 0 if and only if each

of k = ∓1205771(119904) is a constant vector

Proof (1) Suppose 1198961(119904) = 0 By straightforward calculations

we have

k1015840 (119904) = plusmn1198962 (119904)T (119904) (13)

Thus k1015840(119904) equiv 0 if and only if 1198962(119904) = 0

(2) Using the same computation as the proof of (1) we canget (2)

(3) Suppose 1205981(119904) = 0 By straightforward calculations we

have

k1015840 (119904) = plusmn1205982 (119904)B (119904) (14)

Thus k1015840(119904) equiv 0 if and only if 1205982(119904) = 0

(4) Using the same computation as the proof of (3) wecan get (4)

5 Versal Unfolding and Proofof the Main Result

In this section we use some general results on the singularitytheory for families of function germs Detailed descriptionscan be found in the book [16] Let 119865 (R times R119903 (119904

0 1199090)) rarr

Mathematical Problems in Engineering 5

R be a function germ We call 119865 an 119903-parameter unfoldingof 119891(119904) where 119891(119904) = 119865(119904 119909

0) We say that 119891(119904) have 119860

119896-

singularity at 1199040if 119891(119901)(119904

0) = 0 for all 1 le 119901 le 119896 and

119891(119896+1)

(1199040) = 0 We also say that 119891(119904) has 119860

ge119896-singularity at 119904

0

if 119891(119901)(1199040) = 0 for all 1 le 119901 le 119896 Let 119865 be an unfolding

of 119891(119904) and let 119891(119904) has 119860119896-singularity (119896 ge 1) at 119904

0 We

denote the (119896 minus 1)-jet of the partial derivative 120597119865120597119909119894at 1199040by

119895(119896minus1)

((120597119865120597119909119894)(119909 1199040))(1199040) = sum

119896minus1

119895=1119886119895119894119904119895 for 119894 = 1 119903 Then

119865 is called a versal unfolding if the 119896times 119903matrix of coefficients(1198860119894 119886119895119894) has rank 119896 (119896 le 119903) where 119886

0119894= (120597119865120597119909

119894)(1199090 1199040) We

now introduce an important set concerning the unfoldingThe discriminant set of 119865 is the set

D119865= 119909 isin R

119903| 119865 (119904 119909) =

120597119865

120597119904(119904 119909) = 0 (15)

Then we have the following well-known result (cf [16])

Theorem4 Let119865 (RtimesR119903 (1199040 1199090)) rarr R be an 119903-parameter

unfolding of 119891(119904) which has the 119860119896singularity at 119904

0

Suppose that 119865 is a (119901) versal unfolding Then one has thefollowing

(a) If 119896 = 1 thenD119865is locally diffeomorphic to 0 timesR119903minus1

(b) If 119896 = 2 thenD119865is locally diffeomorphic to 119862 timesR119903minus2

By Proposition 2 and the definition of discriminant setwe have the following proposition

Proposition 5 (1) The discriminant sets of 1198671and 119867

2are

respectively

D1198671

= k = ∓N2 (119904) | 119904 isin 119868

D1198672

= k = ∓N1 (119904) | 119904 isin 119868

(16)

(2) The discriminant sets of1198673and119867

4are respectively

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868

(17)

For the Bishop normal indicatrix height functions we canconsider the following propositions

Proposition 6 (1) If ℎ1V0

(119904) has 119860119896-singularity (119896 = 1 2) at

1199040 then119867

1(119904 k) is a (119901) versal unfolding of ℎ

1V0

(119904)(2) If ℎ

2V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198672(119904 k) is a (119901) versal unfolding of ℎ

2V0

(119904)(3) If ℎ

3V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198673(119904 k) is a (119901) versal unfolding of ℎ

3V0

(119904)(4) If ℎ

4V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198674(119904 k) is a (119901) versal unfolding of ℎ

4V0

(119904)

Proof (1) We denote that

N1 (119904) = (11989911 (119904) 11989912 (119904) 11989913 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(18)

Under this notation we have that

1198671 (119904 k) = 11989911 (119904) V1 + 11989912 (119904) V2 plusmn 11989913 (119904)radic1 minus V2

1minus V22 (19)

Thus we have that

1205971198671

120597V119894

= 1198991119894 (119904) ∓

V11989411989913 (119904)

radic1 minus V21minus V22

119894 = 1 2 (20)

We also have that

120597

120597119904

1205971198671

120597V119894

= 1198991015840

1119894(119904) ∓

V1198941198991015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198671

120597V119894

= 11989910158401015840

1119894(119904) ∓

V11989411989910158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(21)

Therefore the 2-jet of (1205971198671120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198671

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198671

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198671

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(22)

It is enough to show that the rank of the matrix 119860 is 2 where

119860 =(

1198991015840

11(119904) ∓

V11198991015840

13(119904)

radic1 minus V21minus V22

1198991015840

12(119904) ∓

V21198991015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

11(119904) ∓

V111989910158401015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

12(119904) ∓

V211989910158401015840

13(119904)

radic1 minus V21minus V22

)

(23)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119860 = (11989910158401015840121198991015840

11minus 1198991015840

1211989910158401015840

11) minus

V1

V3

(11989910158401015840

121198991015840

13minus 1198991015840

1211989910158401015840

13)

minusV2

V3

(11989910158401015840

131198991015840

11minus 1198991015840

1311989910158401015840

11)

=1

V3

⟨kN10158401(119904) and N10158401015840

1(119904)⟩

=1

V3

⟨k minus1198961 (119904)T (119904) and minus119896

1015840

1(119904)T (119904)

minus1198962

1(119904)N1 (119904) minus 1198961 (119904) 1198962 (119904)N2 (119904)⟩

=1

V3

⟨k 11989631(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

=1

V3

⟨∓N2 (119904) 119896

3

1(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

= ∓1

V3

1198963

1(119904) = 0

(24)

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

R be a function germ We call 119865 an 119903-parameter unfoldingof 119891(119904) where 119891(119904) = 119865(119904 119909

0) We say that 119891(119904) have 119860

119896-

singularity at 1199040if 119891(119901)(119904

0) = 0 for all 1 le 119901 le 119896 and

119891(119896+1)

(1199040) = 0 We also say that 119891(119904) has 119860

ge119896-singularity at 119904

0

if 119891(119901)(1199040) = 0 for all 1 le 119901 le 119896 Let 119865 be an unfolding

of 119891(119904) and let 119891(119904) has 119860119896-singularity (119896 ge 1) at 119904

0 We

denote the (119896 minus 1)-jet of the partial derivative 120597119865120597119909119894at 1199040by

119895(119896minus1)

((120597119865120597119909119894)(119909 1199040))(1199040) = sum

119896minus1

119895=1119886119895119894119904119895 for 119894 = 1 119903 Then

119865 is called a versal unfolding if the 119896times 119903matrix of coefficients(1198860119894 119886119895119894) has rank 119896 (119896 le 119903) where 119886

0119894= (120597119865120597119909

119894)(1199090 1199040) We

now introduce an important set concerning the unfoldingThe discriminant set of 119865 is the set

D119865= 119909 isin R

119903| 119865 (119904 119909) =

120597119865

120597119904(119904 119909) = 0 (15)

Then we have the following well-known result (cf [16])

Theorem4 Let119865 (RtimesR119903 (1199040 1199090)) rarr R be an 119903-parameter

unfolding of 119891(119904) which has the 119860119896singularity at 119904

0

Suppose that 119865 is a (119901) versal unfolding Then one has thefollowing

(a) If 119896 = 1 thenD119865is locally diffeomorphic to 0 timesR119903minus1

(b) If 119896 = 2 thenD119865is locally diffeomorphic to 119862 timesR119903minus2

By Proposition 2 and the definition of discriminant setwe have the following proposition

Proposition 5 (1) The discriminant sets of 1198671and 119867

2are

respectively

D1198671

= k = ∓N2 (119904) | 119904 isin 119868

D1198672

= k = ∓N1 (119904) | 119904 isin 119868

(16)

(2) The discriminant sets of1198673and119867

4are respectively

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868

(17)

For the Bishop normal indicatrix height functions we canconsider the following propositions

Proposition 6 (1) If ℎ1V0

(119904) has 119860119896-singularity (119896 = 1 2) at

1199040 then119867

1(119904 k) is a (119901) versal unfolding of ℎ

1V0

(119904)(2) If ℎ

2V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198672(119904 k) is a (119901) versal unfolding of ℎ

2V0

(119904)(3) If ℎ

3V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198673(119904 k) is a (119901) versal unfolding of ℎ

3V0

(119904)(4) If ℎ

4V0

(119904) has 119860119896-singularity (119896 = 1 2) at 119904

0 then

1198674(119904 k) is a (119901) versal unfolding of ℎ

4V0

(119904)

Proof (1) We denote that

N1 (119904) = (11989911 (119904) 11989912 (119904) 11989913 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(18)

Under this notation we have that

1198671 (119904 k) = 11989911 (119904) V1 + 11989912 (119904) V2 plusmn 11989913 (119904)radic1 minus V2

1minus V22 (19)

Thus we have that

1205971198671

120597V119894

= 1198991119894 (119904) ∓

V11989411989913 (119904)

radic1 minus V21minus V22

119894 = 1 2 (20)

We also have that

120597

120597119904

1205971198671

120597V119894

= 1198991015840

1119894(119904) ∓

V1198941198991015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198671

120597V119894

= 11989910158401015840

1119894(119904) ∓

V11989411989910158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(21)

Therefore the 2-jet of (1205971198671120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198671

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198671

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198671

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(22)

It is enough to show that the rank of the matrix 119860 is 2 where

119860 =(

1198991015840

11(119904) ∓

V11198991015840

13(119904)

radic1 minus V21minus V22

1198991015840

12(119904) ∓

V21198991015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

11(119904) ∓

V111989910158401015840

13(119904)

radic1 minus V21minus V22

11989910158401015840

12(119904) ∓

V211989910158401015840

13(119904)

radic1 minus V21minus V22

)

(23)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119860 = (11989910158401015840121198991015840

11minus 1198991015840

1211989910158401015840

11) minus

V1

V3

(11989910158401015840

121198991015840

13minus 1198991015840

1211989910158401015840

13)

minusV2

V3

(11989910158401015840

131198991015840

11minus 1198991015840

1311989910158401015840

11)

=1

V3

⟨kN10158401(119904) and N10158401015840

1(119904)⟩

=1

V3

⟨k minus1198961 (119904)T (119904) and minus119896

1015840

1(119904)T (119904)

minus1198962

1(119904)N1 (119904) minus 1198961 (119904) 1198962 (119904)N2 (119904)⟩

=1

V3

⟨k 11989631(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

=1

V3

⟨∓N2 (119904) 119896

3

1(119904)N2 (119904) minus 119896

2

1(119904) 1198962 (119904)N1 (119904)⟩

= ∓1

V3

1198963

1(119904) = 0

(24)

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Note that k isin D1198671

is a singular point where

k = ∓N2 (119904) (25)

This completes the proof(2) Using the same computation as the proof of (1) we can

get (2)(3) We denote that

1205771 (119904) = (12057711 (119904) 12057712 (119904) 12057713 (119904))

k = (V1 V2 plusmnradic1 minus V2

1minus V22)

(26)

Under this notation we have

1198673 (119904 k) = 12057711 (119904) V1 + 12057712 (119904) V2 plusmn 12057713 (119904)radic1 minus V2

1minus V22 (27)

Thus we have

1205971198673

120597V119894

= 1205771119894 (119904) ∓

V11989412057713 (119904)

radic1 minus V21minus V22

119894 = 1 2 (28)

We also have

120597

120597119904

1205971198673

120597V119894

= 1205771015840

1119894(119904) ∓

V1198941205771015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

1205972

1205971199042

1205971198673

120597V119894

= 12057710158401015840

1119894(119904) ∓

V11989412057710158401015840

13(119904)

radic1 minus V21minus V22

119894 = 1 2

(29)

Therefore the 2-jet of (1205971198673120597V119894)(119904 V) (119894 = 1 2) at 119904

0is given

by

1198952(1205971198673

120597V119894

(119904 V0)) (1199040) =

120597

120597119904

1205971198673

120597V119894

(119904 minus 1199040) +

1

2

1205972

1205971199042

1205971198673

120597V119894

(119904 minus 1199040)2

= 1198861119894(119904 minus 1199040) +

1

21198862119894(119904 minus 1199040)2

(30)

It is enough to show that the rank of the matrix 119861 is 2 where

119861 =(

1205771015840

11(119904) ∓

V11205771015840

13(119904)

radic1 minus V21minus V22

1205771015840

12(119904) ∓

V21205771015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

11(119904) ∓

V112057710158401015840

13(119904)

radic1 minus V21minus V22

12057710158401015840

12(119904) ∓

V212057710158401015840

13(119904)

radic1 minus V21minus V22

)

(31)

Denote that V3= plusmnradic1 minus V2

1minus V22 Then we have

det119861 = (12057710158401015840121205771015840

11minus 1205771015840

1212057710158401015840

11) minus

V1

V3

(12057710158401015840

121205771015840

13minus 1205771015840

1212057710158401015840

13)

minusV2

V3

(12057710158401015840

131205771015840

11minus 1205771015840

1312057710158401015840

11)

=1

V3

⟨k 12057710158401(119904) and 120577

10158401015840

1(119904)⟩

=1

V3

⟨k (minus1205981 (119904)B (119904))

and (minus1205981015840

1(119904)B (119904) minus 1205982

1(119904) 1205771 (119904)

minus (1205981 (119904) 1205982 (119904)) 1205772 (119904)) ⟩

=1

V3

⟨k 12059831(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

=1

V3

⟨∓1205772 (119904) 120598

3

1(119904) 1205772 (119904) minus 120598

3

1(119904) 1205982 (119904) 1205771 (119904)⟩

= ∓1

V3

1205983

1(119904) = 0

(32)

Note that k isin D1198673

is a singular point where

k = ∓1205772 (119904) (33)

This completes the proof

(4) Using the same computation as the proof of (3) wecan get (4)

Proof of Theorem 1 (1) By Proposition 5 the discriminant setD1198672

of1198672is

D1198672

= k = ∓N1 (119904) | 119904 isin 119868 (34)

The assertion (1) ofTheorem 1 follows fromProposition 6 andTheorem 4

(2) By Proposition 5 the discriminant setD1198671

of1198671is

D1198671

= k = ∓N2 (119904) | 119904 isin 119868 (35)

The assertion (2) of Theorem 1 follows from Proposition 6andTheorem 4

(3) By Proposition 5 the discriminant setD1198674

of1198674is

D1198674

= k = ∓1205771 (119904) | 119904 isin 119868 (36)

The assertion (3) of Theorem 1 follows from Proposition 6andTheorem 4

(4) By Proposition 5 the discriminant setD1198673

of1198673is

D1198673

= k = ∓1205772 (119904) | 119904 isin 119868 (37)

The assertion (4) of Theorem 1 follows from Proposition 6andTheorem 4

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

6 Generic Properties

In this section we consider generic properties of regularcurves in E3 The main tool is a kind of transversalitytheorem Let Emb

119861(119868E3) be the space of embeddings 120574

119868 rarr E3 with 119896119894(119904) = 0 or 120598

119895(119904) = 0 equipped with Whitney

119862infin-topology Here 119894 = 1 2 and 119895 = 3 4 We also consider the

function H119896 E3 times 1198782 rarr R defined by H

119896(u k) = ⟨u k⟩

Here 119896 = 1 2 3 4 We claim thatH119896k is a submersion for any

k isin 1198782 where ℎ119896V(u) = H

119896(u k) For any 120574 isin Emb

119861(119868E3)

we have119867119896=H119896∘(120574times119894119889

1198782) We also have the ℓ-jet extension

119895ℓ

1119867119896 119868times1198782rarr 119869ℓ(119868R) defined by 119895ℓ

1119867119896(119904 k) = 119895ℓℎ

119896V(119904) Weconsider the trivialization 119869ℓ(119868R) equiv 119868 timesR times 119869ℓ(1 1) For anysubmanifold119876 sub 119869ℓ(1 1) we denote that119876 = 119868times0times119876 It isevident that both 119895ℓ

1119867119896is a submersions and119876 is an immersed

submanifold of 119895ℓ(119868R) Then 119895ℓ1119867119896is transversal to 119876 We

have the following proposition as a corollary of Lemma 6 inWassermann [17]

Proposition 7 Let119876 be a submanifold of 119869ℓ(1 1)Then the set

119879119876= 120574 isin Emb

119861(119868E3) | 119895ℓ119867119896is transversal to 119876 (38)

is a residual subset of Emb(119868E3) If 119876 is a closed subset then119879119876is open

Let 119891 (R 0) rarr (R 0) be a function germ whichhas an 119860

119896-singularity at 0 It is well known that there exists

a diffeomorphism germ 120601 (R 0) rarr (R 0) such that119891 ∘ 120601 = plusmn119904

119896+1 This is the classification of 119860119896-singularities

For any 119911 = 119895119897119891(0) in 119869ℓ(1 1) we have the orbit 119871119897(119911) givenby the action of the Lie group of 119897-jet diffeomorphism germsIf 119891 has an 119860

119896-singularity then the codimension of the orbit

is 119896 There is another characterization of versal unfolding asfollows [15]

Proposition 8 Let 119865 (R times R119903 0) rarr (R 0) be an 119903-parameter unfolding of 119891 (R 0) rarr (R 0) which has an119860119896-singularity at 0 Then 119865 is a versal unfolding if and only

if 1198951198971119865 is transversal to the orbit 119871119897(119895119897119891(0)) for 119897 ge 119896 + 1 Here

119895119897

1(R times R119903 0) rarr 119869

ℓ(RR) is the 119897-jet extension of 119865 given by

119895119897

1119865(119904 x) = 119895119897119865

119909(119904)

The generic classification theorem is given as follows

Theorem 9 There exists an open and dense subset O sub

Emb119861(119868E3) such that for any 120574 isin O the Bishop spherical

images N119894(119904) 119894 = 1 2 and the type-2 Bishop spherical images

120577119895(119904) 119895 = 3 4 of 120574(119904) are locally diffeomorphic to the ordinary

cusp at any singular point

7 Examples

As applications and illustration of the main results(Theorem 1) we give two examples in this section

Figure 2 Curve

Example 1 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (3 cos 15119904 3 sin 1

51199044

5119904) (39)

with respect to a arclength parameter 119904 where 119904 isin [0 50120587]see Figure 2

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) =3

25

120591 (119904) =4

25

(40)

Using the Bishop curvature equations (5) and (8) we obtainthe following

1198961 (119904) =

3

25cos( 4

25119904)

1198962 (119904) =

3

25sin( 4

25119904)

1205981 (119904) = minus

4

25cos( 3

25119904)

1205982 (119904) = minus

4

25sin( 3

25119904)

(41)

Using (3) and (7) we obtain the vector parametric equationsof the Bishop spherical images N

1(119904)N2(119904) and the type-2

Bishop spherical images 1205771(119904) 1205772(119904) as follows

N1 (119904) = (minus cos(

4

25119904) cos (15119904)

minus4

5sin( 4

25119904) sin(1

5119904)

minus cos( 425119904) sin (15119904)

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

+4

5sin( 4

25119904) cos(1

5119904) minus

3

5sin( 4

25119904))

N2 (119904) = (minus sin( 4

25119904) cos (15119904)

+4

5cos( 4

25119904) sin(1

5119904)

minus sin( 425119904) sin (15119904)

minus4

5cos( 4

25119904) cos(1

5119904)

3

5cos( 4

25119904))

1205771 (119904) = (minus

3

5sin( 3

25119904) sin (15119904) minus cos( 3

25119904) cos(1

5119904)

3

5sin( 3

25119904) cos (15119904)

minus cos( 325119904) sin(1

5119904)

4

5sin( 3

25119904))

1205772 (119904) = (

3

5cos( 3

25119904) sin (15119904) minus sin( 3

25119904) cos(1

5119904)

minus3

5cos( 3

25119904) cos (15119904)

minus sin( 325119904) sin(1

5119904) minus

4

5cos( 3

25119904))

(42)

We see that 1198961(119904) = 0 gives some real roots 119904 = (254)119896120587 +

(258)120587 and 11989610158401(119904) = 0 gives some real roots 119904 = (254)119896120587

On the other hand equation 1198962(119904) = 0 gives some real

roots 119904 = (254)119896120587 and 11989610158402(119904) = 0 gives some real roots

119904 = (254)119896120587 + (258)120587 where 119896 = 0 1 7 For the casethat 119904 = (254)119896120587 + (258)120587 we have that 119896

1(119904) = 0 and

1198961015840

1(119904) = 0 Hence we have that the first Bishop spherical image

N1(119904) is locally diffeomorphic to cusp and it has eight cusps

see Figure 3We can also get that the second Bishop sphericalimage N

2(119904) is locally diffeomorphic to cusp at 119904 = (254)119896120587

and it also has eight cusps see Figure 4 For the case that119904 = (253)119896120587+ (256)120587 where 119896 = 0 5 we have 120598

1(119904) = 0

and 12059810158401(119904) = 0 Hence we have that the first type-2 Bishop

spherical image 1205771(119904) is locally diffeomorphic to cusp and it

has six cusps see Figure 5 We also can get that the secondtype-2 Bishop spherical image 120577

2(119904) is locally diffeomorphic

to cusp at 119904 = (253)119896120587 where 119896 = 0 5 and it also has sixcusps see Figure 6

Example 2 Let 120574(119904) be a unit speed curve of E3 defined by

120574 (119904) = (2

5sin 2119904 minus 1

40sin 8119904

minus2

5cos 2119904 + 1

40cos 8119904 4

15sin 3119904)

(43)

with respect to a arclength parameter 119904 where 119904 isin

[034120587 066120587] see Figure 7 For more details about this kindof curve see [18]

Figure 3 The first Bishop spherical image which has eight cusps

Figure 4The second Bishop spherical image which has eight cusps

The curvature and torsion of this curve are respectivelyas follows

119896 (119904) = minus4 sin 3119904

120591 (119904) = 4 cos 3119904(44)

Using the Bishop curvature equations (5) and (8) we obtain

1198961 (119904) = minus4 sin (3119904) cos(

4

3sin 3119904)

1198962 (119904) = minus4 sin (3119904) sin(

4

3sin 3119904)

1205981 (119904) = minus4 cos (3119904) cos(

4

3cos 3119904)

1205982 (119904) = minus4 cos (3119904) sin(

4

3cos 3119904)

(45)

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

Figure 5 The first type-2 Bishop spherical image which has sixcusps

Figure 6 The second type-2 Bishop spherical image which has sixcusps

Using (3) and (7) we obtain the Bishop sphericalimages N

1(119904)N2(119904) and the type-2 Bishop spherical images

1205771(119904) 1205772(119904) as follows

N1 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times ( (sin 8119904 minus sin 2119904)

times (2

5cos(4

3sin 3119904)minus 3

25sin(4

3sin 3119904)sin 3119904)

+24

125sin(4

3sin 3119904) cos 3119904 (cos 8119904minuscos 2119904))

minus2

minus2

minus2

0

00

2

2

z

y

x

Figure 7 Curve

Figure 8 (a)Thefirst Bishop spherical image (b)The secondBishopspherical image

2

5cos(4

3sin 3119904) (cos 2119904 minus cos 8119904) minus sin(4

3sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5cos(4

3sin 3119904) sin 3119904 minus sin(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

+ (8

25sin 2119904 minus 2

25sin 8119904) (sin 2119904 minus sin 8119904)))

N2 (119904) =

1

radicsin2119904(4cos2119904 minus 1)2

times (2

5(sin 8119904 minus sin 2119904) sin(4

3sin 3119904)

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Figure 9 (a)The first type-2 Bishop spherical image (b)The secondtype-2 Bishop spherical image

+ cos(43sin 3119904) ( 3

25sin 8119904minus 12

25sin 2119904)sin3119904

minus8

25cos 3119904 (cos 2119904 minus cos 8119904)

2

5(cos 2119904 minus cos 8119904) sin(4

3sin 3119904)

+ cos(43sin 3119904)

times (8

25cos 3119904 (sin 8119904 minus sin 2119904)

+ (12

25cos 2119904 minus 3

25cos 8119904) sin 3119904)

minus3

5sin(4

3sin 3119904) sin 3119904 + cos(4

3sin 3119904)

times ((8

25cos 2119904 minus 2

25cos 8119904) (cos 2119904 minus cos 8119904)

minus (8

25sin2119904 minus 2

25sin8119904) (sin8119904minussin2119904)))

1205771 (119904) = (sin(4

3cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timescos((43) cos(3119904)) (minus (85) sin2119904+(85) sin8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

sin(43cos (3119904)) (4

5sin (2119904)minus 1

5sin (8119904)) + 1

4

timescos ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

4

5sin(4

3cos 3119904) cos 3119904

minus3

5

cos ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

1205772 (119904) = (minus cos(43cos (3119904)) (4

5cos (2119904) minus 1

5cos (8119904)) + 1

4

timessin ((43) cos (3119904)) (minus (85) sin 2119904+(85) sin 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus cos(43cos (3119904)) (4

5sin (2119904) minus 1

5sin (8119904))+ 1

4

timessin ((43) cos (3119904)) ((85) cos 2119904minus(85) cos 8119904)

radicsin2 (119904) (4cos2 (119904) minus 1)2

minus4

5cos(4

3cos 3119904) cos 3119904

minus3

5

sin ((43) cos 3119904) sin 3119904

radicsin2 (119904) (4cos2 (119904) minus 1)2)

(46)

We see that 1198961(119904) = 0 gives some real roots 119904 = (13)119896120587 and

1198961015840

1(119904) = 0 gives some real roots 119904 = (13)119896120587 + (1205876) where 119896

is an integer Since 119904 isin [034120587 066120587] we have that 1198961(119904) = 0

and 1198962(119904) = 0 Hence we have thatN

1(119904) andN

2(119904) are locally

diffeomorphic to a line The pictures of Bishop sphericalimages are visualized in Figure 8 For the case that 119904 = (12)120587we have that 120598

1(119904) = 0 and 1205981015840

1(119904) = 0 Hence the first type-2

Bishop spherical image 1205771(119904) is locally diffeomorphic to cusp

see Figure 9 We also can get that the second type-2 Bishopspherical image 120577

2(119904) is not locally diffeomorphic to cusp see

Figure 9

8 Conclusions

In this paper we introduce the notions of Bishop normalindicatrix height functions on a space curve embedded inEuclidean 3-space We use the Bishop Frame the type-2Bishop Frame and those functions to study Bishop sphericalimages from the singularity viewpoint We find that thesingular points of the Bishop spherical images and Type-2 Bishop spherical images correspond to the point whereBishop curvatures and type-2 Bishop curvatures vanishedand their derivatives are not equal to zero As applicationsof the main results we give two examples and make themvisualized by computer

Acknowledgments

The authors would like to thank the referee for hishervaluable suggestions which improved the first version ofthe paper The first author was partially supported by theFundamental Research Funds for the Central Universities(Graduate Innovation Fund of Northeast Normal Universityno 12SSXT140) Preparatory Studies of Provincial InnovationProject of Mudanjiang Normal University no SY201225 andno SY201320 The second author (Corresponding author)was partially supported by NSF of China no 11271063

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

References

[1] R L Bishop ldquoThere is more than one way to frame a curverdquoTheAmericanMathematicalMonthly vol 82 no 3 pp 246ndash2511975

[2] N Clauvelin W K Olson and I Tobias ldquoCharacterization ofthe geometry and topology of DNA pictured as a discrete col-lection of atomsrdquo Journal of ChemicalTheory and Computationvol 8 no 3 pp 1092ndash1107 2012

[3] C Y Han ldquoNonexistence of rational rotation-minimizingframes on cubic curvesrdquoComputer Aided Geometric Design vol25 no 4-5 pp 298ndash304 2008

[4] K Shoeemake ldquoAnimating rotation with quaternion curvesrdquo inProceedings of the 12thAnnual Conference onComputerGraphicsand Interactive Techniques pp 245ndash254 1985

[5] A J Hanson and H Ma ldquoParallel transport approach to curveframingrdquo Indiana University 425 1995 vol 11

[6] H Liu and D Pei ldquoSingularities of a space curve accordingto the relatively parallel adapted frame and its visualizationrdquoMathematical Problems in Engineering vol 2013 Article ID512020 12 pages 2013

[7] S Yılmaz and M Turgut ldquoA new version of Bishop frame andan application to spherical imagesrdquo Journal of MathematicalAnalysis and Applications vol 371 no 2 pp 764ndash776 2010

[8] B Bukcu and M K Karacan ldquoSpecial Bishop motion andBishop Darboux rotation axis of the space curverdquo Journal ofDynamical Systems and GeometricTheories vol 6 no 1 pp 27ndash34 2008

[9] B Bukcu and M K Karacan ldquoThe slant helices accordingto Bishop framerdquo International Journal of Computational andMathematical Sciences vol 3 no 2 pp 67ndash70 2009

[10] L Kula and Y Yayli ldquoOn slant helix and its spherical indicatrixrdquoApplied Mathematics and Computation vol 169 no 1 pp 600ndash607 2005

[11] S Yılmaz E Ozyılmaz and M Turgut ldquoNew spherical indica-trices and their characterizationsrdquo Analele stiintifice ale Univer-sitatii Ovidius Constanta vol 18 no 2 pp 337ndash353 2010

[12] L Kula N Ekmekci Y Yaylı andK Ilarslan ldquoCharacterizationsof slant helices in Euclidean 3-spacerdquo Turkish Journal of Mathe-matics vol 34 no 2 pp 261ndash273 2010

[13] M K Karacan and B Bukcu ldquoAn alternative moving frame fortubular surfaces around timelike curves in the Minkowski 3-spacerdquo Balkan Journal of Geometry and its Applications vol 12no 2 pp 73ndash80 2007

[14] D Pei and T Sano ldquoThe focal developable and the binormalindicatrix of a nonlightlike curve in Minkowski 3-spacerdquo TokyoJournal of Mathematics vol 23 no 1 pp 211ndash225 2000

[15] Z Wang D Pei L Chen L Kong and Q Han ldquoSingularitiesof focal surfaces of null Cartan curves in Minkowski 3-spacerdquoAbstract and Applied Analysis vol 2012 Article ID 823809 20pages 2012

[16] J W Bruce and P J Giblin Curves and Singularities CambridgeUniversity Press Cambridge UK 2nd edition 1992

[17] GWassermann ldquoStability of causticsrdquoMathematische Annalenvol 216 pp 43ndash50 1975

[18] P D Scofield ldquoCurves of constant precessionrdquo The AmericanMathematical Monthly vol 102 no 6 pp 531ndash537 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of