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Bound on the number of isolated rest points of a vector field
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Let $P(x,y,z)$ be a homogeneous polynomial of degree $n$. Consider the vector field
$$F_P = left[ beginarray*20c
fracpartial Ppartial x - nxP\
fracpartial Ppartial y - nyP\
fracpartial Ppartial z - nzP
endarray right].$$
I want to prove that if $n$ is fixed, the number of isolated rest points of this vector field $F_P$ is uniformly bounded over all homogeneous polynomials $P$ of degree $n$. What is the bound for $n=3$?
Thank you, in advance, for any suggestion, comment, or response!
real-analysis functional-analysis differential-equations differential-geometry algebraic-geometry
asked Jul 16 at 0:41
Arthur
19812
add a comment |Â
up vote
0
down vote
favorite
Let $P(x,y,z)$ be a homogeneous polynomial of degree $n$. Consider the vector field
$$F_P = left[ beginarray*20c
fracpartial Ppartial x - nxP\
fracpartial Ppartial y - nyP\
fracpartial Ppartial z - nzP
endarray right].$$
I want to prove that if $n$ is fixed, the number of isolated rest points of this vector field $F_P$ is uniformly bounded over all homogeneous polynomials $P$ of degree $n$. What is the bound for $n=3$?
Thank you, in advance, for any suggestion, comment, or response!
real-analysis functional-analysis differential-equations differential-geometry algebraic-geometry
asked Jul 16 at 0:41
Arthur
19812
I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $P(x,y,z)$ be a homogeneous polynomial of degree $n$. Consider the vector field
$$F_P = left[ beginarray*20c
fracpartial Ppartial x - nxP\
fracpartial Ppartial y - nyP\
fracpartial Ppartial z - nzP
endarray right].$$
I want to prove that if $n$ is fixed, the number of isolated rest points of this vector field $F_P$ is uniformly bounded over all homogeneous polynomials $P$ of degree $n$. What is the bound for $n=3$?
Thank you, in advance, for any suggestion, comment, or response!
real-analysis functional-analysis differential-equations differential-geometry algebraic-geometry
asked Jul 16 at 0:41
Arthur
19812
Let $P(x,y,z)$ be a homogeneous polynomial of degree $n$. Consider the vector field
$$F_P = left[ beginarray*20c
fracpartial Ppartial x - nxP\
fracpartial Ppartial y - nyP\
fracpartial Ppartial z - nzP
endarray right].$$
I want to prove that if $n$ is fixed, the number of isolated rest points of this vector field $F_P$ is uniformly bounded over all homogeneous polynomials $P$ of degree $n$. What is the bound for $n=3$?
Thank you, in advance, for any suggestion, comment, or response!
real-analysis functional-analysis differential-equations differential-geometry algebraic-geometry
asked Jul 16 at 0:41
Arthur
19812
edited Jul 16 at 22:33
asked Jul 16 at 0:41
Arthur
19812
asked Jul 16 at 0:41
Arthur
19812
asked Jul 16 at 0:41
Arthur
19812
19812
I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42
add a comment |Â
I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42
I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42
add a comment |Â
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I do not understand what uniformly bounded over all homogeneous polynomials $P$ of degree $n$ means.
– Arthur
Jul 16 at 0:43
This question is related to structurally stable vector fields.
– Arthur
Jul 16 at 0:58
I suppose it means: For all $nin mathbbN$ there is a constant $C>0$ such that for all homogeneous polynomials $P$ of degree $n$ we have $# ain mathbbR^3vert DF_P(a) = 0 le C$.
– Jan Bohr
Jul 16 at 7:41
@JanBohr: Thanks! I guess you mean $a in mathbbR^3 $, right?
– Arthur
Jul 16 at 15:38
You're right. That makes more sense.
– Jan Bohr
Jul 16 at 15:42