Singular points and Lie rotated vector fields
Internat. J. Math. & Math. Sci.
Vol.
SINGULAR POINTS AND LIE ROTATED VECTOR FIELDS
JIE WANG
CHEN CHEN
This paper gives the definition of Lie rotated vector fields in the plane and the conditions of movement of singular points on Lie rotated vector fields with variable parameters. 1. Introduction. Many engineering problems are usually run into a class of nonlinear equations that contain variable parameters. In order to study whole orbits or whole phase diagrams of vector fields that contain parameters, it is a complicated and interesting problem how the whole orbit or whole phase diagram change as parameter is changed. It is extremely complicated for general containing parameter vector fields to change in the plane, but for some special containing parameter rotated vector fields, their change has regular rule as parameter is changed. These are many results in this respects [3, 4, 5, 6, 7]. In Section 2, we present the basic definitions of Lie rotated vector fields. We define Lie rotated vector fields using one parameter group approach. In accordance with the strict definition of rotated vector field, the singular points of X(?) must be kept fixed, but in this paper, the singular points of X(?) can be moved as parameter ? is changed. In Section 3, we discuss the motion of singular points on Lie rotated vector fields. In the section, we require the singular points of X(?) to be strictly moved as parameter ? is changed, and permit the moved singular points to disappear or decompose, which do not coincide with the singular points of original vector field. We give some conditions and properties corresponding to the vector field Y . In this paper, we give some examples to illustrate the concept and notion of Lie rotated vector fields.
and phrases; Lie rotated vector fields; Lie bracket; one parameter group; singular points

Y = Y1(x), Y2(x) .
For the vector fields (2.1), we define
X ? Y = X1Y2 ? X2Y1,
X, Y = X1Y1 + X2Y2.
If X and Y are vector fields, then [X, Y ] is a vector field which is operated by Lie
(2.1)
(2.2)
[X, Y ] = Z1, Z2 ,
(2.3)
(2.4)
(2.5)
(2.6)
(2.7)
? I, is
(2.8)
(2.9)
where Z1 and Z2 are expressed as,
Z1 = X, ?Y1 ? Y , ?X1 ,
Z2 = X, ?Y2 ? Y , ?X2 ,
respectively, where ? is gradient operator.
Let the plane vector fields X(?) = (X1(x, ?), X 2(x, ?)) be defined by the following
differential equations:
dx1
dt = X1(x, ?),
dx2
dt = X2(x, ?),
where X1 and X2 are functions of x and parameter ? ? I ? R, and the singular points
are isolated.
Definition 2.1. Let the plane vector field X(?) be determined by (2.5), where
X1, X2 ? C3(R2 ? I, R), I = {?  ?  < ?} is a real interval, ? is a given positive number.
If vector field Y exists which is defined by the following differential equations:
dx1
dt = Y1(x),
dx2
dt = Y2(x),
where Y1 and Y2 ? C3(R2, R). At all ordinary points of X(0), such that the following
relation holds
L(0) d=ef X(0) ? X? (0) + [X(0), Y ] > 0 (< 0),
where X? (0) is the derivative of the vector field X(?) at ? = 0, then X(?), ?
called Lie rotated vector fields.
Remark 2.2. If the vector field X(?) is defined on D ? I, where D ? R2, such that
X(0) satisfies relation (2.7) at all ordinary points of X(0) on D, then X(?), ? ? I, is
called Lie rotated vector fields on D.
Lemma 2.3. Let ?s be a one parameter transform group which is produced by C1
vector field Y , s ? R, and let X be C1 vector field. If s is fixed, and ?p(t) is an integral
curve of X through the point p, ?p(0) = p, then ?s ??p(t) is an integral curve of ??sX
through the point ?s (p). If Xp = 0, then (??sX)?s(p) = 0.
Proof. The proof follows from [1] and [2]. In fact, if ?p(t) is an integral curve of
X through the point p, then
and
?s ? ?p(t) t=0 = ?s (p)
d d
?s ? ?p(t) ?t ? dt t = ??s?p(t) ? ?p(t)?t ? dt t
= ??s?p(t) ? X?p(t) = ??sX ?s??p(t).
It follows that ?s ? ?p(t) is an integral curve of ??sX through the point ?s(p).
Next, due to
??sXq = D?s ??s(q) ? X ??s(q) , q ? R2.
Set q = ?s(p), note that we already suppose Xp = 0, again note that ?s is a one
parameter transform which is produced by Y , then
??sX?s(p) = D?s(p) ? X(p) = D?s(p) ? Xp = 0,
i.e., ?s(p) is a singular point of ??sX.
Lemma 2.4. Let ?s be a one parameter transform group which is produced by C1
vector field Y , s ? R, fix s, then the index of isolated singularity of C1 vector field X is
not changed under the ?s transform.
Proof. In fact, by the condition of the lemma, it is known that ?s is a differentiable
homeomorphism, then the lemma follows from [8, Theorem 4.2].
Next, if X(?) is a Lie rotated vector field, then Y is a corresponding vector field
which satisfies (2.7), and ?s is a one parameter transform group which is produced
by Y , s ? R.
Lemma 2.5. Let X(?) be a Lie rotated vector field, for all ? > 0, there exist ? = ?(?),
such that when ?  < ?, ??sX(?) constitutes a rotated vector field.
Proof. Let the singular points of ??sX(?), ? = 0, on the plane R2 be p? 1 , . . . , p? k
and the singular points of X(0) on the plane R2 are p1, . . . , pm, ?? > 0, 0 < ? 1, let
S?(p? i ) or S?(pj) (1 ? i ? k, 1 ? j ? m) be open neighborhood p? i (1 ? i ? k) and
pj (1 ? j ? m), and radius ?, such that S?(p)?S?(q) = ?, where p and q ? {p? i }?{pj}
(1 ? i ? k, 1 ? j ? m), p = q. Let ?s be a one parameter transform group which is
produced by C1 vector field Y , s ? R. By the limit definition of Lie bracket, we have
s d s2 d2
??sX(?) = ??0X(?) + 1! dt s=0??sX(?) + 2! dt2 s=0??sX(?) + ? ? ?
s s2
= X(?) + 1! [X(?), Y ] + 2! [[X(?), Y ], Y ] + ? ? ? .
Next, we notice that X(?) can be unfolded as
since
? 2
X(?) = X(0) + 1?! X? (0) + 2! X? (0) + ? ? ? ,
? ? 2
[X(?), Y ] = [X(0), Y ] + 1! X? (0), Y + 2! X? (0), Y + ? ? ? .
Let s = ? , it follows from (2.12), (2.13), and (2.14) that
???X(?) = X(0) + ? X? (0) + [X(0), Y ]
+ 12 ? 2 X? (0) + 2 X? (0), Y + [[X(0), Y ], Y ] + ? ? ? .
(2.10)
(2.11)
(2.12)
(2.13)
(2.14)
(2.15)
At the ordinary points of R2\ ik=1 S?(p? i ) jm=1 S?(pj) , for given ? > 0, we
sooner or later can find ?1 = ?1(?) > 0, such that when ?  < ?1, we have
and let ?(?) be the crossing angle of ???X(?) and the x1 axis, for given ? > 0, we
sooner or later can find ?2 = ?2(?), such that when ?  < ?2, at the ordinary points of
R2 \ ik=1 S?(p? i ) jm=1 S?(pj) (???X(?) is X(0) when ? = 0, ?(0) is the crossing
angle of X(0) and the x1 axis), so
0 < ?(?) ? ?(0) < ? .
Take ? = min{?1, ?2}, then when ?  < ?, ???X(?) constitutes a rotated vector field.
Remark 2.6. In accordance with the strict definition of rotated vector field, the
? X(?) in Lemma 2.5
singular points must be kept fixed, but the singular points of ??
can be moved as parameter ? is changed. In the unmistakable circumstance, when
?  < ?, we call ???X(?) a rotated vector.
In the above lemma, ? needs not be a quite small positive number, i.e., 0 < ? 1
need not be set up. For the sake of distinctness, we cite an example to illustrate this
equation.
Example 2.7. Let X(?) = (x2, ?x1 + ?x 2), if we take Y = (?x2/2, 0), then at all the
ordinary points of X(0), we have
1
X(0) = 2 x12 + x22 > 0,
that is, X(?) is a Lie rotated vector field.
Now we consider the range of ? , because
so
???X(?) =
1 1
2 ?x 1 + 1 ? 14 ? 2 x2, ?x1 + 2 ?x 2
yet go a step further calculating, we have
(2.22)
which is larger than zero at the ordinary points of X(0) and ???X(?) for all ? ? R,
but the range of ? that satisfies formula (2.17) is ?  < 4, thus we take ? = 4, when
?  < ? = 4, ???X(?) constitutes rotated vector field.
3. The motion of singular points. Let X(?) be a Lie rotated vector field, we require
the singular points of X(?) to be strictly moved as parameter ? is changed, and permit
the singular points that have been moved disappear or decompose, but require the
singular points that have been decomposed to be at most limited in number, which
do not coincide with the singular points of the original vector field.
If p is a singular point of X(?) , we name J? (p) for index of singular point p of X(?) ,
under the same circumstances, J0(p0) for index of singular point p0 of X(?) , J? ? (q)
for index of singular point q of ???X(?) (? = 0).
Theorem 3.1. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, and let Y p0 = 0. If
the singular point p0 of X(0) disappears or decomposes as pi (1 ? i ? k) in X(?) (? =
0), then J0(p0) = 0, and J? (pi) = 0 (? = 0, 1 ? i ? k).
Proof. First of all, we prove that J0(p0) = 0. In fact, because of X(?) p0 = 0 (? = 0),
utilize Lemma 2.3 and condition Y p0 = 0, we know that ???X(?) p0 = 0 (? = 0), it
follows from Lemma 2.5, for given ? > 0, when ?  < ?, ???X(?) constitutes a rotated
vector field. Take ? > 0 as quite small positive number, such that S?(p0) does not
contain the singular points of ???X(?) (? = 0), and only contains the isolate
singular point p0 of X(0). It is easy to know that J? ? (p0) = 0 about ?S?(p0). By (2.17)
of Lemma 2.5, it follows that J0(p0) = 0 when ?  < ?.
Using the same method, we prove J? ? (?? (pi)) = 0 (? = 0, 1 ? i ? k) and by
Lemma 2.4, we find J? (pi) = 0 (? = 0, 1 ? i ? k).
Corollary 3.2. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, if Y p0 = 0,
and moved the singular points pi = ??? (p0) (? = 0, 1 ? i ? k), then J0(p0) = 0 and
J? (pi) = 0 (? = 0, 1 ? i ? k).
Proof. Since X(0)p0 = 0, let the singular point of X(?) (? = 0) disappears or
decomposes into p1, . . . , pk points which do not coincide with singular point p0 of
X(0), i.e., X(?) pi = 0 (1 ? i ? k), yet because of X(?) p0 = 0 (? = 0) and Y p0 = 0.
? X(?) ?? (p0) = 0 and ???X(?) ?? (pi) = 0, but by condition
By Lemma 2.3, we have ??
?? (pi) = p0, we know that ???X(?) p0 = 0, as in the proof of Theorem 3.1, we can
prove that J0(p0) = 0 and J? (pi) = 0 (? = 0, 1 ? i ? k).
Corollary 3.3. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, if Y p0 = 0,
but for some i0 (1 ? i0 ? k), set up ?? (pi0 ) = p0 (? = 0), then J0(p0) = j? (pi0 ),
J? (pi) = 0 (? = 0, 1 ? i ? k and i = i0).
2
Example 3.4. Let X(?) = x2 , ?x1 + ? , and let
Y = 3x1 ? ?x2, 2x2
(3.1)
when ?  < ?, we take ? > 0 and ? 1, on the range of D = {(x1, x2)  x2 < ??1} ? R2,
at all ordinary points ? D of X(0), set up
L(0) = ?x12 + x2 ? ?x32 > 0,
2
(3.2)
that is, X(?) constitutes a Lie rotated vector field on D, the singular points of X(?) are
strictly moved as parameter ? is changed. We note that Y p0 = 0, p0 = (0, 0) is singular
point of X(0), by Theorem 3.1, we can find that J0(p0) = 0 and J? (pi) = 0 (? = 0),
where pi = (?, 0).
Theorem 3.5. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, p0 is elementary.
(1) If Y p0 = 0, then p0 cannot be moved as parameter ? is changed.
(2) If Y p0 = 0, then p0 can be moved as parameter ? is changed, and the moved point
is the singular point ??? (p0) of X(?) (? = 0).
Proof. (1) We note J0(p0) = ?1 = 0, it is proved immediately from Theorem 3.1.
(2) First of all, we prove that p0 is indeed moved as ? is changed, suppose that
it is not real, i.e., p0 is not moved as ? is changed, then that X(?) p0 = 0 (? = 0),
by Lemma 2.3, we know that ???X(?) ?? (p0) = 0. Because p0 is isolate singular point
of X(0), we take ?? > 0 and ample small ? > 0, it follows that ?? (p0) ? S?(p0). When
0 < ?  < ?? < ?, then for ?S?(p0), we have J? ? (p0) = 0 (since ???X(?) p0 = 0), where
? = 0. But J0(p0) = ?1 = 0, this is a contradiction from Lemma 2.5. Thus we have
proved p0 is indeed moved as ? is changed, and by Corollaries 3.2 and 3.3, it follows
that p0 is moved as the singular point ??? (p0) of X(?) (? = 0) when ? is changed.
Lemma 3.6. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, and there is an elliptic
region at the singular point p0.
(1) If Y p0 = 0, then the singular point p0 cannot be moved when parameter ? = 0.
(2) If Y p0 = 0, then when parameter ? = 0, singular point p0 is moved, and p0 be
moved as singular point ??? (p0) of X(?) .
Proof. (1) We already know that Y p0 , suppose the original equation is not real,
then when ? = 0, singular point p0 is moved, thus we let p0 moved as the
singular point p? of X(?) , X(?) p? = 0, p? = p0, ? = 0. From Lemma 2.3, we know that
???X(?) ?? (p? ) = 0, and by Y p0 = 0, we know that ?? (p? ) = p0 (? = 0). Let ? be an
elliptic region at the singular point p0 of X(0), for arbitrary fixed ? ( 0 < ?  < ?), it
is sure to have some elliptic trajectory r of X(0), which does not contain the point
of ?? (p? ) on r and in r . Since r has single direction, and there is no singular point
of ???X(?) on r and in r . By Lemma 2.5, we can know that positive half trajectory or
negative half trajectory of ???X(?) which pass through the point p will wander about
without a home to go to, where p is any point which passes the inner region of r , this
is a contradiction.
(2) Now we know Y p0 = 0, yet use reduction to absurdity. Suppose, when ? = 0,
singular point p0 is not moved, i.e., establish X(?) p0 = 0, namely we have ???X(?) ?? (p? )=
0, and ?? (p0) = p0. The method of the proof is completely alike as part (1), we can
prove it is a contradiction. Thus let ? = 0, singular point p0 is moved as singular
point p? (p0 = p? ) of X(?) , i.e., ???X(?) ?? (p? ) = 0 (? = 0). If ?? (p? ) = p0, the
method of the proof is alike as in part (1), yet it is a contradiction, thus only establish
?? (p? ) = p0, or p? = ??? (p0).
Lemma 3.7. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, and when ? = 0, p0
is moved as the singular point p? (p? = p0) of X(?) as ? is changed. If Y p0 = 0, then
for singular point p? (or p0), at least there are a positive half trajectory and a negative
half trajectory of X(?) (or X(0)) to get into it.
Proof. We only prove the circumstance of point p? (the proof is completely alike
as the circumstance of point p0).
From Lemma 3.6, we know that there is no elliptic region which links with the
singular point p0 of X(0), the same do the singular point p? of X(?) , and from Theorem 3.1,
we know that the index of p? of X(?) is zero. Take p? as circular center, make the
circumference of a circle l with radius rather small, and let that hyperbolic region of
point p? which intersects with the circumference of a circle l has h. By the Bendixson?s
formula in ?6of Chapter 3 of [8], we can immediately find h = 2.
From Lemmas 3.6and 3.7, we have the following theorem.
Theorem 3.8. Let X(?) be a Lie rotated vector field, X(0)p0 = 0, and let Y p0 = 0,
then some singulars while can be moved as parameter ? is changed in X(?) only contain
two hyperbolic regions and their index is zero.
Acknowledgement. The authors express grateful thanks to Professor Yaoxian
Wang for his help and direction in the work.
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Jie Wang and Chen Chen : School of Electric Power, Shanghai Jiaotong University, Shanghai, 200030, China
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