Download 7. INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN En Author: E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi

Int. J. of Mathematical Sciences and Applications,
Vol. 1, No. 1, January 2011
Copyright c Mind Reader Publications
www.ijmsa.yolasite.com
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
E. B. Koc Ozturk, U. Ozturk1 , Y. Yayli2 , S. Ozkaldi3
Ankara University, Department of Mathematics,
Faculty of Sciences, 06100
Tandogan-Ankara, Turkey
e-mail: [email protected]
K¬r¬kkale University, Hac¬lar Hüseyin Aytemiz1
Vocational School,
Hac¬lar-K¬r¬kkale, Turkey
e-mail: [email protected]
Ankara University, Department of Mathematics2 ,
Faculty of Sciences, 06100
Tandogan-Ankara, Turkey
e-mail: [email protected]
and
K¬r¬kkale University, Department of Mathematics3 ,
Faculty of Sciences, 71450
Yahşihan-K¬r¬kkale, Turkey
Abstract
In this study we de…ned the spiral vector …elds and found the integral
curves of this spiral vector …elds in E 3 . Also we gave the relation with the
instantenous motion and homothetic motion of this integral curves.
In the special case we obtain the study of Karger and Novak [1] and
Taleshian[2]. We generalized all the results to E n :
Keywords: Curves in Euclidean space, Kinematics, Free motion of a rigid
body.
AMS Mathematics Classi…cation: 53A04, 53A17, 70E15
1
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
1
Inroduction
The de…nitions and applications of a linear vector …eld have been given by
Karger and Novak [1]. Karger and Novak [1] gave that the integral curves of
W c
a linear vector …eld X in E 3 described by a matrix of form
, where
0 0
W is skew-symmetric matrix, can be:i) Helices with common axes and same
parameter, if rank W; c = 3, ii) Circles which lie in planes paralel to
each other and which have centers on the axis perpendicular to those planes,
if rank W; c = 2, iii) Paralel straight lines, if rank W; c = 1:
In additon to Karger and Novak’s results [1] Taleshian [2] showed that all the
results are also valid in the general case. Taleshian [1] examined the integral
curves of a linear vector …eld for Euclidian space. They are circles or helices
in the cases when the matrix of the linear vector …eld has respectively, even
or odd rank.
In this paper, we examined the relation with the linear vector …elds and
instantenous motions. An application for homothetic motions of the linear
vector …elds is given Spiral vector …elds are assumed as linear vector …elds and
linear vector …leds are examined with the aid of the instantaneous homothetic
motions. As a result of it we obtain that the path of a point is a spiral in an
instantaneous homothetic motion.
Furthermore, the type of the integral curves of one parameter homothetic
motion is determined with the aid of linear vector …eld in Euclidian space by
I3 + W v
can be spiral curves, if 6= 0 and
a matrix form
0
0
I3 + W; v = 3 , where W is skew-symmetric. We generalized
rank
the integral curves of a spiral vector …elds to E n .
2
Preliminaries
Let
t
:
I R ! En
! (t) = ( 1 (t);
2 (t); : : : ;
n (t))
be a parameterized curve and let X be a vector …eld in E n ([3]). If
2
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
d
dt
= X( (t)) ; 8t 2 I holds true, then the curve is called an integral curve
of the vector …eld X ([3]). Let V be a vector space over R of dimension n. A
vector …eld X on V is called linear if Xv = A(v), 8v 2 V , where A is a linear
mapping from V into V [3]. Let A be a linear mapping given skew-symmetric
matrix in E 3 . Then we can chose an orthonormal basis in E 3 , such that the
matrix A reduces to the form
2
3
0
0
0 0 5
A=4
([3]).
0 0 0
The orthogonal matrices in R33 de…nes with the set of
O(3) = A 2 R33 : AT A = AAT = I :
This set is a group with the matrix multiplication and is called an orthogonal
group ([3]). It is called a special subgroup of O(3) and de…nes with the set
of
SO(3) = A 2 R33 : AT A = AAT = I; det A = 1
([3]).
Lie algebra of the SO(3) Lie groups is de…ned with the set of
9
8
2
3
0
w3 w2
=
<
0
w1 5 ; wT = w
so(3) = w 2 R33 : w = 4 w3
;
:
w2 w1
0
3
Helical Vector Field in E 3
One parameter motion is given by
x
:
E3 ! E3
! (x) = g(t)x + c(t)
where g(t) 2 SO(3); c(t) 2 R3 and matrix form of this mapping is;
g(t) c(t)
x
y(t)
=
0
1
1
1
{z
}| {z }
| {z } |
Y (t)
A(t)
3
X
([1]).
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
One parameter matrices such as A(t) create a Lie group according to the
matrix multiplication. We denote this group with
SE(3) =
A:A=
g c
0 1
; g 2 SO(3); c 2 R3 :
When SE(3) is considered with its topologic structure, it describes 6-dimensional
topologic manifold. Then, SE(3) is a matrix Lie group. Let Se(3) be a Lie
algebra of SE(3) and it can …nd with
A(t)A 1 (t) =
W c(t)
0
0
;
_
where g(t)g 1 (t) = W is skew-symmetric and c(t) = c(t)
Lie algebra Se(3) of SE(3) Lie group can be written as
Se(3) =
3.1
W c
0 0
: WT =
W (t)c(t). Then,
W; c 2 R3 :
Integral Curves of Helical Vector Fields in E 3
Elements of Lie algebra match with the helical vector …elds. Now let give
a relationship between helical vector …eld and one parameter motions. One
parameter motion is given by
y(t) = g(t)x + c(t):
The velocity y(t), if expressed in terms of y(t), has the form
T
y(t)
_ = v(t) = g(t)g
_
(t)y(t)
T
g(t)g
_
(t)c(t) + c(t):
_
Since g is orthonormal, the matrix gg
_ T = W is skew-symmetric and the
product W x can be written in the form w
~ ~x. Then we obtain velocity
vector of instantenous motion as follows,
v(t0 ) = w
~ ~y + c
v(t0 )
W c
=
0 0
0
4
y
1
:
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
W c
: W T = W; c 2 R3 denote a
0 0
linear vector …eld and we called this linear vector …eld as a helical vector …eld.
Let X be a helical vector …eld in E 3 and letf0; u1 ; u2 ; u3 g be an orthonormal
frame of E 3 , then the matrix in this frame can be written as
2
3
0
0 a
6
W c
0 0 b 7
7:
=6
4
0 0
0 0 0 c 5
0 0 0 0
Therefore elements of Se(3) =
Case 1. If c 6= 0, =
6 0 and rank
point P = (x; y; z) of E 3 is
2
X(P )
0
or X(P ) = ( y + a;
6
=6
4
W; c
0
= 3:Then the value of X at a
3 2
a
x
7
6
b 7 6 y
:
c 5 4 z
0
1
0
0 0
0 0
0 0
0
0
3
7
7
5
x + b; c) : On the other hand, if the curve
t
:
I R ! E3
! (t) = ( 1 (t);
2 (t);
3 (t))
is an integral curve of X, then we can write the di¤erential equation
d
= X( (t));
dt
8t 2 I
(1)
as the system of di¤erential equations
dx
= y+a
dt
dy
=
x+b
dt
dz
= c
dt
(2)
Then the solution of the last equation of this system is
dz
= c
dt
z = ct + d
5
(3)
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
For the solutions of the …rst two equations we derivative the second equation
and obtain that
d2 y
dx
=
2
dt
dt
d2 y
+y =
a
(4)
dt2
which is the …rst order linear di¤erential equation with constant coe¢ cent.
We know the solution of this equation is
y = A cos t + B sin t
(5)
a
On the other hand the derivation of (4) and the second of (2) give us that
x = A sin t
(6)
B cos t + b
Thus the integral curves of X can be written as
(t) = (A sin t
B cos t + b; A cos t + B sin t
a; ct + d)
(7)
This is family of inclined curves with common axes and same parameter,
since we have
1p 2
K1
A + B2;
(8)
=
H=
K2
2
where K1 and K2 are the curvatures of the curve and H is constant for each
one of the curves.
W; c
Case 2. If c 6= 0, 6= 0 and rank
equation of integral curves as
(t) = (A sin t
= 2:Hence (6) gives us the
B cos t + b; A cos t + B sin t
a; d)
(9)
The curves are the circles each one of which lies on the parallel planes and the
centers of these circles are located on an axis perpendicular to those parallel
planes.
Case 3. If
the system
6= 0 and rank
W; c
= 1:Then the system (9) reduces to
dx
= a
dt
dy
= b
dt
dz
= c
dt
6
(10)
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
Then the solution of this system is
(11)
(t) = (at + d1 ; bt + d2 ; ct + d3 )
These integral curves are parallel straight lines [2].
4
Spiral Vector Field in E 3
One parameter homothetic motion is given by
:
E3 ! E3
! (x) = h(t)g(t)x + v(t);
x
where g(t) 2 SO(3); v(t) 2 R3 and matrix form of this mapping is,
y(t)
h(t)g(t) v(t)
x
=
:
1
0
1
1
| {z } |
{z
} | {z }
Y (t)
X
B(t)
If h(t) = 1, then the homothetic motion is an Euclidian motion. One parameter matrices such as B(t) create a Lie group according to the matrix
multiplication. We denote this group with
SD(3) =
hg v
0 1
B:B=
; h 6= 0; g 2 SO(3); v 2 R3 :
When SD(3) is considered with its topologic structure, it describes 7-dimensional
topologic manifold. Then, SD(3) is a matrix Lie group. Let Sd(3) be a Lie
algebra of SD(3) and it can …nd with
1
_
B(t)B
(t) =
h(t)I3 + W v(t)
0
0
v
0 0
=
=
v
;
_
1
where g(t)g
_
(t) = W is skew-symmetric and h(t)
= (t) and
h(t)
v(t) = v(t)
_
W (t)v(t) . Then, Lie algebra Sd(3) of SD(3) Lie group can
be written as
Sd(3) =
v
0 0
:
= h(t)I3 + W; W T =
7
W; v 2 R3 :
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
Elements of Lie algebra match with the linear vector …elds. Now let give a
relationship between the linear vector …eld and the instantaneous homothetic
motions. Let y(t) be a one-parameter homothetic motion y(t) = h(t)g(t)x +
v(t). The velocity y(t),
_
if expressed in terms of y(t), has the form
T
y(t)
_ = g(t)g
_
(t)y(t) +
_
h(t)
y(t)
h(t)
T
g(t)g
_
(t)v(t)
_
h(t)
v(t) + v(t):
_
h(t)
Since g is orthonormal, the matrix gg
_ T = W is skew-symmetric and the
product W x can be written in the form w
~ ~x. Then we obtain velocity
vector of instantenous motion as follows,
y(t)
_ =w
y + y + v:
In the case of t = 0; the velocity of vector
y(0)
_
= [ (0)I3 + W (0)] y(0) + v(0):
Matrix form of this vector is
y(0)
_
0
4.1
=
(0)I3 + W (0) v(0)
0
0
y(0)
1
(12)
:
Integral Curves of Spiral Vector Fields in E 3
De…nition 4.1.1 Spiral vector …eld is de…ned by
Y
M
:
R3 ! R3
! Y (M ) = W M + V;
linear mapping, where I3 + W =
2 R33 ; W skew-symmetric and v 2 R31 :
Spiral vector …elds can match with the elements of Lie algebra of oneparameter homothetic motions. Then, we arrive the path of one-parameter
instantaneous homothetic motions. Let Y be a spiral vector …eld. Then it
can be written as
Y (M )
0
=
I3 + W v
0
0
M
1
8
=
v
0 0
M
1
:
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
Y (t) = x + c is an ordinary linear di¤erential equation with constant coe¤cients. Its solution to the boundary condition y(0) = x can be given explicitly
with matrix exponentials,
y(t) = S(t) + e t x
S(t) = e t v
with
1
e tv
1
e
t
:
(0)I3 + W (0) v(0)
linear vector
0
0
…elds, we need to …nd the path of y(0) as the same of the velocity vector of the
form (12) in an instantaneous homothetic motion in the case t = 0. Let show
this instantaneous homothetic motion with y1 (t), and …nd the homothetic
motion with y1 (0) = y(0) = M boundary condition. If we solve y_ 1 (t) =
y1 (t)+v(t) di¤erential equation to the boundary condition y1 (0) = y(0) = x
is,
y1 (t) = e t y1 (0) e t v 1 e t v 1 e t :
If we want to …nd integral curves of
It describes a homothetic motion of the form y(t) = h(t)g(t)x + v(t); which
has a time independent velocity vector …eld. It can also be viewed as oneparameter subgroup of the group of similarities in R3 . If 6= 0 and W 6= 0
then, this homothetic motion is a spiral motion. We show this homothetic
motion as
v
M
y1
;
= exp
1
0 0
1
where y1 (t) curve is an integral curve of Y =
I3 + W (0) v
0
0
spiral vector
…eld.
As a result of this we can give the following theorem.
Theorem 4.1.1 Let Y be a spiral vector …eld in E 3 determined in the orv
I3 + W (0) v
=
, where W is
thonormal frame by the matrix
0
0
0 0
_
skew-symmetric, v 2 R33 and h(t)
= (t): The Integral curves of this vector
h(t)
…eld Y are the following:
1) If = 0 then it is a rigid body motion and it
v = 3, Helices with common axes and same
i) If rank
parameter,
v = 2, Circles which lie in planes parallel to
ii) If rank
each other and which have centers on the axis perpendicular
to those planes,
9
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
v = 1, Parallel straight lines, [1]
iii) If rank
These cases coincide to the results of (1).
v = 3; Integral curves are spiral curves.
2) If 6= 0, rank
2
3
2 3
0 1 0
0
Example 4.1.1 Let W = 4 1 0 0 5 ; v = 4 0 5 ; h = et ; = 1 and X =
0 0 0
1
2
3
1 1 0 0
6 1 1 0 0 7
I3 + W (0) v
7
=6
4 0 0 1 1 5 be a spiral vector …eld then we arrive
0
0
0 0 0 0
3
2 t
3 2
t
cos t
sin t 0
0
e cos t
e sin t 0
0
6 et sin t et cos t 0
6 et sin t cos t 0
0 7
0 7
7:
7
6
etx = 6
=
t
4
0
0
1 e
1 5
0
0
et et 1 5 4
0
0
0
1
0
0
0
1
tx
One-parameter homothetic motion can be de…ned with e . The path of a
point in this motion is spiral curves.
5
Linear Vector Fields in E 2n+1
One parameter motion is given by
x
:
E 2n+1 ! E 2n+1
! (x) = g(t)x + c(t);
where g(t) 2 SO(2n + 1); c(t) 2 R2n+1 and matrix form of this mapping is
y(t)
g(t) c(t)
x
=
:
1
0
1
1
| {z } |
{z
}| {z }
Y (t)
A(t)
X
One parameter matrices such as A(t) create a Lie group according to the
matrix multiplication. We denote this group with
SE(2n + 1) =
A:A=
g c
0 1
10
; g 2 SO(2n + 1); c 2 R2n+1 :
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
When SE(2n + 1) is a matrix Lie group. Let Se(2n + 1) be a Lie algebra of
SE(2n + 1). Then, Lie algebra Se(2n + 1) of SE(2n + 1) Lie group can be
written as
W c
0 0
Se(2n + 1) =
: WT =
W; c 2 R2n+1 ;
1
where g(t)g
_
(t) = W is skew-symmetric and c(t) = c(t)
_
W (t)c(t):
De…nition 5.1 Helical vector …eld is de…ned by
:
E 2n+1 ! R2n+1
~ + c;
! X(M ) = W M
X
M
linear mapping, where W skew-symmetric and c 2 R2n+1 :
Elements of Lie algebra match with the helical vector …elds. Now let give
a relationship between helical vector …eld and one parameter motions. One
parameter motion is given by
y(t) = g(t)x + c(t):
The velocity y(t),
_
if expressed in terms of y(t), has the form
T
y(t)
_ = v(t) = g(t)g
_
(t)y(t)
| {z }
W (t)
T
g(t)g
_
(t)c(t) + c(t)
_ :
|
{z
}
c()
1
Since g is orthonormal, the matrix g(t)g
_
(t) = W is skew-symmetric.
y(t)
_ = v(t) = W (t)y(t) + c(t)
in the case of t = 0, the velocity vector is
y(0)
_
= W (0)y(0) + c(0):
Matrix form of this vector is
y(0)
_
0
=
W (0) c(0)
0
0
11
y(0)
1
:
(13)
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
W (0) c(0)
linear vector …elds, we
0
0
need to …nd the path of y(0) as the same of the velocity vector of the form
(13) in an instantaneous homothetic motion in the case t = 0. Let show
this instantaneous homothetic motion with y1 (t), and …nd the one parameter
motion with y1 (0) = y(0) = M boundary condition. If we solve y(t)
_
=
W (t)y(t)+c(t) di¤erential equation to the boundary condition y1 (0) = y(0) =
M is
If we want to …nd integral curves of
y1 (t)
1
= exp t
=
W c
0 0
g1 (t) c1 (t)
0
1
M
1
M
1
;
where g1 (t) 2 SO(2n + 1); c1 (t) 2 R2n+1 : This y1 (t) curve is an integral curve
W c
of X =
: This result supports Taleshian’s study.
0 0
Theorem 5.1 Let X be a linear vector …eld in E 2n+1 determined by the
matrix
W c
0 0
where, with respect to an orthonormal frame, W is a skew-symmetric and c
is a column matrix. Then the integral curves of this vector …eld X are as
follows:
i) The integral curves of X are the inclined curves whose axes
coincide.
ii) Each of the integral curves of X lie on a rankA-dimensional right
hypercylinder.
Proof
Let W 2 M2n+1;2n+1 (R) be a skew-symmetric matrix and c 2 M1;2n+1 (R) be
a column matrix such that
12
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
2
6
6
6
6
6
6
6
W =6
6
6
6
6
6
4
0
3
1
1
0
..
.
0
p
p
0
..
.
0
2n 1
2n 1
0
0
0
7
7
7
7
7
7
7
7
7
7
7
0 7
7
0 5
0
and c = (a1 ; a2 ; a3 ; : : :; a2n ; a2n+1 ). For a linear vector …eld X and all the
points M = (x1 ; x2 ; x3 ; : : :; x2n ; x2n+1 ) 2 E 2n+1 , we have
X(M )
0
=
W c
0 0
M
1
;
where rank W c = 2n + 1 and X(M ) = ( 1 X2 + a1 ;
1 X1 + 2 X3 +
a2 ; : : :;
X
+
a
;
a
):
2n 1 2n 1
2n
2n+1
In addition, if the curve : I
R ! E 2n+1 is an integral curve of the
linear vector …eld X, then it satis…es the di¤erential equation
d
= X( (t));
dt
8t 2 I
(14)
and (3) provides the system of di¤erential equations
dx1
=
dt
dx2
=
dt
1 X2
+ a1
1 X1
+
2 X3
+ a2
..
.
(15)
dx2n
=
2n 1 X2n 1 + a2n
dt
dx2n+1
= a2n+1
dt
If we rewrite the matrix W by renumbering non-zero elements
the following
2
3
0 A
X(M )
M
=4 0 0 B 5
;
0
1
0 0 0
13
i,
we obtain
(16)
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
where
2
= 4
0
0
1
0
A0
B
c0 =
0
1
2
6
6
=6
4
m
0
m
a1
a2
..
.
a2n+1
3
5 2 Mm+1;m+1 (R)
3
7
7
7
5
and A0 2 Mm+1;m+1 (R); B 2 M1;2n m (R):
0
If we use the notation =
; then (14) and (16) give us that
0 0
dx
dt
P = c0
(17)
which is a …rst order linear di¤erential equation with constant coe¢ cient.
The solutions of this equation are given by
Z t
t
(t) = e
c0 e u du + D
(18)
0
since we have that
et
0
(t) =
0
I2n
m
(18) can be rewritten as
(t) =
Rt
et
0
c0 e u A0 du
Bt
+D
(19)
or, since the matrix is skew-symmetric, the rank is even, so m is odd and
det 6= 0 and 1 exists. Then
1
(t) =
an we have
I. The last 2n
A0 + 1 et A0
Bt
+D
(20)
m components of the curve are of the form bi t + di :
14
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
2
II. k 0 (t)k2 = et A0 + kBk2 = kA0 k2 + kBk2 ; which means that
k (t)k2 = constant.
Thus we can say that, inE 2n+1 ; by choosing an orthogonal frame
f0; u1 ; u2 ; u3 ; : : :; u2n ; u2n+1 g ;
we have
h 0 (t); ur i = k 0 (t)k : kur k : cos r ;
q
=
kA0 k2 + kBk2 : cos
m+2
r
2n + 1
r
and hence the angle in between each of the curves of the family (t) and each
of the base vectors ur is constant. Therefore each of the curves (t) makes a
constant angle with the space. Sp fum+2 ; : : :; u2n+1 g : On the other hand,
since the …rst m + 1 components of each curve (t) can be represented by
the vector
1
A0 +
1 t
e A0 + D1 ;
D1 2 M1;m+1 (R);
D=
D1
D2
;
using the curves
(t) =
1
A0 +
1 t
e A0 + D
and the point
1
A0 + D1 = Q = (q1 ; q2 ; : : :; qm+1 )
we obtain that
d (Q;
(t)) = constant.
Therefore we can say that all the curves (t) are the inclined curves and
they lie on the right hypercylinders whose bases are S m having as centers the
1 0 2
points (q1 ; q2 ; : : :; qm+1 ; 0; : : :; 0), and the radii
A
[2].
6
Spiral Vector Field in E n
One parameter motion is given by
x
:
En ! En
! (x) = h(t)g(t)x + v(t);
15
E. B. Koc Ozturk, U. Ozturk, Y. Yayli, S. Ozkaldi
where g(t) 2 SO(n); v(t) 2 Rn and matrix form of this mapping is
y(t)
1
=
h(t)g(t) v(t)
0
1
|
{z
}
x
1
:
B(t)
One parameter matrices such as B(t) create a Lie group according to the
matrix multiplication. We denote this group with
G=
B:B=
hg v
0 1
; g 2 SO(n); v 2 Rn :
then G is a matrix Lie group. Let g be a Lie algebra of G and it can be
found with
g=
v
0 0
: In + W =
2 Rnn ; W T =
W; v 2 Rn1 :
De…nition 6.1 Spiral vector …eld is de…ned by
Y
M
:
Rn ! R n
! Y (M ) = M + v;
linear mapping with In + W =
2 Rnn ; W skew-symmetric and v 2 Rn1 :
De…nition 6.2 Let
t
:
I R ! E3
! (t) = ( 1 (t);
2 (t);
: : :;
n (t))
be a parameterized curve and let X be a spiral vector …eld in E n and if
d
= X( (t)), 8t 2 I holds true, then the curve is called an integral curve of
dt
the spiral vector …eld X.
A path of the point x in one-parameter homothetic motion is y(t) = h(t)g(t)x+
v(t) and denoted by dot derivative and the equation velocity vector at t = 0
can be written as
y(0)
_
= [ (0)In + W (0)] y(0) + v(0):
16
INTEGRAL CURVES OF A SPIRAL VECTOR FIELD IN E n
If we solve y_ 1 (t) = y(t)+v(t) di¤erential equation to the boundary condition
y1 (0) = y(0) = M with the equation
y_ 1
0
=
v
0 0
M
1
;
we arrive
y1
0
= exp t
=
v
0 0
h(t)g1 (t) v1 (t)
0
0
M
1
M
1
;
g1 (t) 2 SO(n); v1 (t) 2 Rn :
v
spiral vector …eld. As
0 0
a result of this, we can give the following theorem.
Theorem 6.1 Let Y be a spiral vector …eld in E n and f0; u1 ; u2 ; u3 ; : : :; un g
be an ortonormal frame of E n , then the matrix in this frame can be written
v
as
, where In + W =
2 Rnn ; W skew-symmetric and v 2 Rn1 :
0 0
v = n, the integral curves of Y are spiral curves
Therefore, if rank
and if
= 0 then this linear vector …eld is a helical vector …eld [1].
This y1 (t) curve is an integral curve of Y =
References
[1] A. Karger and J. Novak, "Kinematics and Lie Groups", Gordon and
Breach Science Publishers, 1985.
[2] A. Taleshian, "Integral curves of a linear vector …eld", Balkan Society of
Geometers, Vol.6, pp. 37-42, 2004.
[3] H. H. Hacisaliho¼
glu, "Di¤erential Geometry", Inonu University Faculty
of Arts and Sciences Publications, Malatya, Turkiye, 1980.
17