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MANNHEIM CURVES IN n–DIMENSIONAL EUCLIDEAN SPACE
HANDAN ÖZTEK·IN AND ESRA MET·IN
Abstract. In this study, we de…ne the generalized Mannheim Curves in n-dimensional Euclidean Space and obtain the caracterizations of the generalized Mannheim curves.
1. Introduction
The curves are a fundamental structure of di¤erential geometry. In di¤erential geometry, to study the corresponding relations between the curves is very important problem. Especially, Mannheim curves are one of them. Space curves of which prin-cipal normals are the binormal of another curFve are called Mannheim curves. the notion of Mannheim curve was discovered by A. Mannheim in 1878. These curves have beeen studied by many mathematicians (see [1] and [3-11]). For instance, In [5], Liu and Wang had obtained the necessary and su¢ cient conditions between the curvature and the torsion for a curve to be the Mannheim partner curves. In [3], Önder, Kazaz and U¼gurlu had studied some characterizations of Mannheim part-ner curves in Minkowski 3-space. In [4], K¬z¬ltu¼g and Yayl¬ had given a study on the quaternionic Mannheim curves of AW(k)-type in Euclidean space. In [6] and [9], the authors had studied the generalized Mannheim curves in Euclidean 4-space and Minkowski space-time. In [10], Önder and K¬z¬ltu¼g had studied Bertrand and Mannheim Partner D-curves on Parallel surfaces in Minkowski 3-Space.
On the other hand, the articles concerning Mannheim curves in n-dimensional space are rather few. In [11], D.W. Yoon studied non-null Mannheim curve and null Mannheim curve in an n-dimensional Lorentzian manifold. To the best of our knowledge, Mannheim curves have not been presented in n-dimensional Euclidean space. Thus, the study is proposed to serve such a need.
The main goal of this paper is to carryout some results which were given in [6] to Mannheim curves in n-dimensional Euclidean space En:
Received by the editors: March 03, 2016, Accepted: July 28, 2016. 2010 Mathematics Subject Classi…cation. 53A35, 53B30. Key words and phrases. Mannheim curves, Euclidean space.
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2. Preliminaries
Let En be an n-dimensional Euclidean space with cartesian coordinates (x1; x2;
::: ; xn): By a parametrized curve of class C1, we mean a mapping of a certain
interval I into En given by
(t) = x1(t) ; x2(t) ; ; xn(t) ; 8t 2 I.
If d (t)dt = Dd (t)dt ;d (t)dt E1=2 6= 0 for t 2 I, then is called a regular curve in
En. Here h:; :i denotes the Euclidean inner product on En. A regular curve is
parametrized by the arc-length parameter s. Then the tangent vector …elddds along has unit length, that is, d (s)ds = 1 for all s 2 I.
During this paper, curves considered are regular C1-curves in Enparametrized by the arc-length parameter. The Frenet equations for such a curve are given by as follow: de1(s) ds = k1(s) e2(s) de2(s) ds = k1(s) e1(s) + k2(s) e3(s) ::: den 1(s) ds = kn 2(s) en 2(s) + kn 1(s) en(s) den(s) ds = kn 1(s) en 1(s) ;
for all s 2 I. The unit vector …eld ej+1; j = 1; 2; :::; n 1, along is called the
Frenet j-normal vector …eld along . A straight line is called the Frenet j-normal line of at (s), if it passes throught the point (s) and is collinear to the j-normal vector ej+1; j = 1; 2; :::; n 1, of at (s), [2]:
3. Mannheim Curves in En
De…nition 3.1. Let be a special Frenet curve in En. The curve is called a
generalized Mannheim curve if there exists a distinct special Frenet curve e in En
such that the …rst normal line at each point of is included in the space spanned by the second normal line, the third normal line ,...,nth normal line of e at
corre-sponding point under . Here is a bijection from to e. The curve e is called the generalized Mannheim mate curve of .
Now, we give following theorems for generalized Mannheim curves in En:
Theorem 3.1. Let be a special Frenet curve in En. If the curve is a generalized
Mannheim curve, then the …rst curvature function k1and second curvature function
k2 of satisfy the equality
k1(s) =
n
(k1(s))2+ (k2(s))2
o ; s 2 I, where is a positive constant number.
Proof. Let be a generalized Mannheim curve andsbe the generalized Mannheim mate curve of .
By the de…nition, a generalized Mannheim mates is given by
s
(s) = (s) + (s) e2(s) ; s 2 I; (3.1)
where is a smooth function on I. Generally, the parameter s isn’t arc-length of
s
. Letss be the arc-parameter ofs de…ned by
s s = s Z 0 ds(s) ds ds:
We can consider a smooth function f : I !sI given by f (s) =ss. Then we have f0(s) =
q
[1 (s) k1(s)]2+ [ 0 (s)]2+ [ (s) k2(s)]2 (3.2)
Thus , we can write the reparametrization ofsby
s
(f (s)) = (s) + (s) e2(s) (3.3)
here is a bijection from tos de…ned by
( (s)) =s(f (s)) :
By di¤erentiating both sides of (3:3) with respect to s, we obtain
f0(s)es1(f (s)) = (1 (s) k1(s)) e1(s) + 0 (s) e2(s) + (s) k2(s) e3(s) (3.4)
From De…nition 3.1. and since the …rst normal line at the each point of is lying in the plane generated by the second normal line, the third normal line ,...,nth normal line s at the correspending points under bijection , we have e2(s) =
x3(s)es3(f (s)) + x4(s)es4(f (s)) + ::: + xn(s)esn(f (s)), where x3; x4; :::; xn are
some smooth functions on I. Thus we obtain 0 (s) = 0, that is, the function is constant. Then we rewrite the equation (3:4) by
s e1(f (s)) = (1 k1(s)) f0(s) e1(s) + k2(s) f0(s) e3(s) ; (3.5) where f0(s) = q [1 k1(s)]2+ [ k2(s)]2; s 2 I:
By taking di¤erentiation both sides of the equation (3:5) with respect to s, we obtain f0(s) ~k1(f (s)) ~e2(f (s)) = 1 k1(s) f0(s) 0 e1(s) + (1 k1(s)) k1(s) (k2(s)) 2 f0(s) ! e2(s) + k2(s) f0(s) 0 e3(s) + k2(s) k3(s) f0(s) e4(s) ; s 2 I: By the fact D ~ e2(f (s)) ; x3(s)es3(f (s)) + x4(s)es4(f (s)) + ::: + xn(s)esn(f (s)) E = 0; s 2 I; we have that coe¢ cient of e2in the above equation is zero, that is,
(1 k1(s)) k1(s) (k2(s))2= 0; s 2 I: Thus we obtain k1(s) = h (k1(s))2+ (k2(s))2 i ; s 2 I which completes the proof.
Following theorem gives a parametric representation for Mannheim curves in n-dimensional Euclidean space En:
Theorem 3.2. Let be a curve de…ned by
(u) = Z
f (u) sin (u) du; Z
f (u) cos (u) du; Z
f (u) h1(u) du;
; Z
where is a positive constant, hi : U ! IR; i 2 f1; 2; :::; n 2g are any smooth
functions, and the smooth function f : U ! IR+; i 6= j is given by
f = 1 + n 2X i=1 h2i ! 3=2 2 6 6 41 + n 2X i=1 h2i + _h2i + n 2X i;j=1 j<i hi_hj _hihj 2 3 7 7 5 5=2 8 > > < > > : 2 6 6 41 + n 2 X i=1 h2i + _h2i + n 2 X i;j=1 j<i hi_hj _hihj 2 3 7 7 5 3 + 1 + n 2X i=1 h2i !3 9 > > = > > ; 8 > > < > > : n 2 X i=1 hi+ •hi 2 + n 2X i;j=1 j<i hi_hj _hihj _hi•hj •hi_hj 2 + hi•hj •hihj 2 9 > > = > > ; for u 2 U. Then the curvature functions k1 and k2 of satisfy
k1(u) =
n
[k1(u)]2+ [k2(u)]2
o at the each point (u) of .
Proof. Let be a curve de…ned by (u) =
Z
f (u) sin (u) du; Z
f (u) cos (u) du; Z
f (u) h1(u) du;
; Z
f (u) hn 2(u) du ;
where is a non-zero constant number, h1; h2; :::; hn 2 are any smooth functions,
f is a positive valued smooth function. Thus, we obtain
_ (u) = [ f (u) sin (u) ; f (u) cos (u) ; f (u) h1(u) ; ; f (u) hn 2(u)] ; u 2 U
where the subscript dot ( ) denotes the di¤erentiation with respect to u. The arc-length parameter s of is given by
s = (u) =
u
Z
u0
k _ (u)k du; and k _ (u)k = f (u) ( 1 + n 2X i=1 (hi(u))2 )1 2 : Let ' denotes the inverse function of : U ! I IR; then u = ' (s) and
'0(s) = d (u)
du ju='(s) ; s 2 I;
The unit tangent vector e1(s) of the curve at the each point (' (s)) is given by e1(s) = 2 4 ( 1 + n 2 X i=1 (hi(' (s)))2 ) 1=2 sin (' (s)) ; ( 1 + n 2 X i=1 (hi(' (s)))2 ) 1=2 cos (' (s)) ; ( 1 + n 2X i=1 (hi(' (s)))2 ) 1=2 h1(' (s)) ; ; ( 1 + n 2X i=1 (hi(' (s)))2 ) 1=2 hn 2(' (s)) 3 5 Now, we use the following abbreviations for the sake of brevity:
h1= h1(' (s)) ; h2= h2(' (s)) ; :::; hn 2= hn 2(' (s)) _h1(' (s)) = dh1(u) du ju=('(s)); _h2(' (s)) = dh2(u) du ju=('(s)); :::; _hn 2(' (s)) = dhn 2(u) du ju=('(s)); • h1(' (s)) = d2h 1(u) du2 ju=('(s)); •h2(' (s)) = d2h 2(u) du2 ju=('(s)); :::; •hn 2(' (s)) = d 2h n 2(u) du2 ju=('(s)); '0 = '0(s) = d' (s) du js; A = 1 + n 2X i=1 h2i; B = n 2X i=1 hi_hi; C = n 2 X i=1 _h2 i; D = n 2 X i=1 hih•i; E = n 2X i=1 _hi•hi; F = n 2X i=1 • h2i Then we have _ A = 2B; _B = C + D; _C = 2E; ' = 1f 1A 1=2 and e1(s) = h A 1=2sin (' (s)) ; A 1=2cos (' (s)) ; A 1=2h; A 1=2hn 2 i By di¤erentiation the last equation with respect to s, we …nd
By the fact that e2= (k1) 1e01, we have e2= 2 6 6 6 6 6 6 4
A 1=2B(A + AC B)2 1=2sin (' (s)) +A1=2(A + AC B2) 1=2cos('(s)) A 1=2B(A + AC B2) 1=2cos('(s)) A1=2 A + AC B2 1=2sin('(s))
A 1=2B A + AC B2 1=2h 1 A1=2 A + AC B2 1=2 _h1 .. . A 1=2B A + AC B2 1=2h n 2 A1=2 A + AC B2 1=2 _hn 2 3 7 7 7 7 7 7 5 .
After long process of calculation, we have
e02+ k1e1= '0A1=2 A + AC B2 3=2 2 6 6 6 6 6 6 6 4 P Q sin (' (s)) R cos (' (s)) P Q cos (' (s)) + R sin (' (s)) P h1 R _h1+ Q•h1 .. . P hn 2 R _hn 2+ Q•hn 2 3 7 7 7 7 7 7 7 5 ; (3.7) where P = (1 + C + BE D CD) ; Q = A + AC B2 ; R = (B + AE BD) Consequently, from the (3:7) and (3:8) we …nd
ke02+ k1e1k2= ('0)2A A + AC B2 3 fP2 2P Q + Q2+ R2 + P2 h21+ h22+ ::: + h2n 2 + R2 _h21+ _h22+ ::: + _h2n 2 + Q2 •h21+ •h22+ ::: + •h2n 2 2P R h1_h1+ h2_h2+ ::: + hn 2_hn 2 2RQ _h1•h1+ _h2•h2+ ::: + _hn 2•hn 2 + 2P Q h1•h1+ h2•h2+ ::: + hn 2•hn 2 g: Thus we obtain (k2)2= ('0)2A A + AC B2 2 A + AC B2 (1 + F ) + 2 (D 1) (1 + C + BE D CD) 2E (B + AE BD) + 1 + C 2D 2CD + D2 +CD2+ 2BE 2BDE + AE2 = ('0)2A A + AC B2 2 f A + AC B2 (1 + F ) 1 C + 2D +2CD 2BE AE2 D2 CD2+ 2BDEg:
Moreover, from the equation (3:6) we have
(k1)2= '0A 2 A + AC B2 :
The last two equation gives us
(k1)2+ (k2)2= 2f 2A 3 A + AC B2 2 f A + AC B2 3 + A3(A + AC B2+ AF + ACF B2F 1 C + 2D + 2CD 2BE AE2 D2 CD2+ 2BDE)g and k1= 1f 1A 3=2 A + AC B2 1=2: Thus, by setting f = 1 + n 2 X i=1 h2i ! 3=2 2 6 6 41 + n 2X i=1 h2i + _h2i + n 2X i;j=1 j<i hi_hj _hihj 2 3 7 7 5 5=2 8 > > < > > : 2 6 6 41 + n 2 X i=1 h2i + _h2i + n 2 X i;j=1 j<i hi_hj _hihj 2 3 7 7 5 3 + 1 + n 2X i=1 h2i !3 9 > > = > > ; (n 2 X i=1 hi+ •hi 2 + n 2X i;j=1 j<i hi_hj _hihj _hi•hj •hi_hj 2 + hi•hj •hihj 2 9 > > = > > ; we obtain k1= (k1)2+ (k2)2 :
Theorem 3.3. Let f ; eg , be a generalized Mannheim mate in En. Let M; fM
be the curvature centers at two corresponding point of and e , respectively. Then the ratio ke(es)Mk
k (s)Mk : k
e(es) fMk
Proof. If M is the curvature centers of , then we can write M = (s) + 1 k1 e2 and k (s) Mk = kM (s)k = 1 k1: Similarly, we have e (es) fM = 1 ~ k1 , (s) fM = v u u t1 + ~k2 1: n X i=3 2 i ~ k1 ; ke (es) M k = (1 + k1) k1 v u u tXn i=3 2 i: Therefore, we obtain ke (es) M k k (s) Mk : e (es) fM (s) fM = (1 + k1) v u u tXn i=3 2 i v u u t1 + ~k2 1: n X i=3 2 i
which completes the proof.
From the above Theorem 3.3, we have following Corollory:
Corollary 3.4. The Mannheim theorem for the generalized Mannheim curves in En is not valid.
Theorem 3.5. Let ande be two curves parametrized by the arc-length parameter s. If f ; eg is a generalized Mannheim mate in En , then there exists following
relation
n
X
i=3
i i 1eki 1 i+1eki = 0,
where i; i; i 2 f1; 2; :::; ng are arbitrary constants and ek1; ek2; :::; ekn 1 are
curva-tures of e.
Proof. Since the curve is Mannheim curve, then we have (es) = e (es)+
n
X
i=3
i(es) eei(es) and e1
ds des=ee1 3ek2ee2+ n X i=3 i 1eki 1 i+1eki eei. (3.9) By taking inner product with e2=
n
X
i=3
ieei second equation of the equation (3:9),
we have ,
n
X
i=3
i i 1eki 1 i+1eki = 0:Thus the proof is completed.
Theorem 3.6. The distance between corresponding points of a generalized Mannheim curve and of its generalized Mannheim partner curve in En is a constant.
References
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[8] Tigano O., Sulla determinazione delle curve di Mannheim, Matematiche Catania, 3(1948), 25-29.
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[11] Yoon, DW., Mannheim Curves in an n-dimensional Lorentz Manifold, International journal of Pure and Applied Mathematics, Vol. 96 No. 2 (2014), 165-174.
Current address : Handan ÖZTEK·IN: Department of Mathematics, Firat University, 23119 Elaz¬¼g TURKEY
E-mail address : handanoztekin@gmail.com
Current address : Esra MET·IN: Department of Mathematics, Firat University, 23119 Elaz¬¼g TURKEY
