Jaime Cruz Sampedro - Newton's Eq

8/2/2019 Jaime Cruz Sampedro - Newton's Eq

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Exotic solutions to Newtons equationwith angular potentials

Jaime Cruz Sampedro

UAMAemail: [email protected]

II EIM, 2009

Jaime Cruz Sampedro (UAMA) Exotic solutions II EIM, 2009 1 / 25
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Abstract

In this talk we describe some results about the asymptotic behavior ofsolutions to Newtons equation with angular potentials. First we sketch

some classical results and then we present, in more detail, some recentcontributions of the author.

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I. Newtons equation

Equation number one in the study of the motion of a classical particle isNewtons equation

r(t) = F(r(t)).

F is a vector field on Rn that represents the force acting on the particle of

mass m = 1, whose position at time t is given by r(t).

Definition

We say thatFis conservative ifF = V , where V is a real-valuedfunction on Rn called the potential.

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Theorem

IfF is conservative, then

E(t) = |r(t)

|2

2 + V(r(t))

is constant in time.

QuestionLet r(t) be a unboundedsolution to Newtons equation

r(t) = V(r(t))

and define

(t) =r(t)

|r(t)| .

What is the asymptotic behavior of (t) as t ?



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II. Motion in a radial potential

DefinitionA force fieldF is called radial if

F(r) = f(r), = r/r, r = |r|.

Fact: Every radial field is conservative.

Theorem

The orbit of a particle in a radial fieldRn is a plane orbit. In R3 this factis due to the conservation of the angular momentum L = r(t) r(t).



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Law of gravitation

The force of attraction between two bodies of masses M and m = 1 is

given by F = V(r),where V(r) = /r and = GM.

In this case the orbit of m is a conic given by

r =|L|2/

1 + ecos ,

which is determined by the energy E, since its eccentricity is given by

e =

1 +

2|L|2

G2M2

E.

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Let us consider unbounded orbits in potentials of the form

V(r) = /r, > 0, > 0,

Theorem

Let r(t) be an unbounded solution to r = r/r+2 with E > 0. Then,there exists the limit

= limt(t).

Theorem

If E = 0, 0 < < 2, then there exists orbits of the form

r(t)2 =2

1 + cos ( (2 )(t)) .

These orbits are known as Cote spirals.



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We keep V(r) = /r and E = 0.

For Cote spirals we have

limt

(t) = 2/(2 ) y = limt

(t) exists.

If = 2 and c = 2 |L|2 > 0, we have a spiral of the form

r(t)2 = 2

ct+ , (t) = (

|L

|/2

c) log(2

ct+ ) + .

In this case, = limt (t) does not exist ifL = 0.

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Theorem

Let V be a radial potential such that |V(r)| C/r, with > 2, then:1 IfL = 0, the orbit is a straight line.

2 IfL = 0, the orbits are bounded.

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Theorem

Let V be a radial potential such that |V(r)| C/r, with > 2, then:1 IfL = 0, the orbit is a straight line.

2 IfL = 0, the orbits are bounded.This last is not valid anymore if the potential is not radial.

Theorem (JC-Gutierrez, AIMS09)

There exists a non-radial potential V(r, ), with > 2, that satisfiesV(r, ) = O(r) and possesses an orbit which goes to infinity in spirals.

Proof. If S(r, ) = r(2)/2

sin(log r ), thenV(r, ) = |S(r, )|2/2

possesses the desired property.



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We remind our question:What is the behavior of (t) = r(t)/

|r(t)

|as t goes to infinity?

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We remind our question:What is the behavior of (t) = r(t)/

|r(t)

|as t goes to infinity?

Theorem (Herbst, 91)

If V is an angular potential and r(t) is a solution to

r(t) = V(r(t)),then

limt

p(t) = 0 and limt

V((t)) = 0.

Here, p(t) = r(t) and p(t) = p(t) ((t) p(t)) (t).



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Let V be an an angular potential and let

Cr= { Sn1 : V() = 0}.

If r(t) is a solution to r(t) = V(r(t)), then:(i) For n = 2,

limt(t) = Cr.

(ii) For n 3 y Cr finite,

limt

(t) =

Cr.


( )


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Let V be an an angular potential and let

Cr= { Sn1 : V() = 0}.If r(t) is a solution to r(t) = V(r(t)), then:

(i) For n = 2,

limt(t) = Cr.

(ii) For n 3 y Cr finite,

limt

(t) =

Cr.

Conjecture (1991)

The limit in (ii) always exists.


Sk h f h f f 2


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Sketch of the proof for n = 2

In polar coordinates Newtons equations take the form

r = r2,

r2

= (),

where () = V(cos , sin ).


Sk h f h f f 2
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Sketch of the proof for n = 2

In polar coordinates Newtons equations take the form

r = r2,

r2

= (),

where () = V(cos , sin ).

from this equations and conservation of energy,

1

2(r2 + (r)2) + V(cos , sin ) = C,

we find that there exists 0 such thatr(t) .


If i ( ) ( ) d ( ( )) ( ) h
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If we write 1() = (t) and ((t)) = r(t), where

(t) =

t

1

ds

r(s),

then

d

d=

d1d

2,

d

d

1

2

d1d

2+ (1)

=

d1d

2.


If it ( ) (t) d ( (t)) (t) h


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If we write 1() = (t) and ((t)) = r(t), where

(t) =

t

1

ds

r(s),

then

d

d=

d1d

2,

d

d

1

2

d1d

2+ (1)

=

d1d

2.

Note that (t) , as t , and ((t)) . Then

lim

d1

d = 0

and since is of constant sign for large , then the limit

lim

12d1

d

2

+ (1) = E exists.Jaime Cruz Sampedro (UAMA) Exotic solutions II EIM, 2009 16 / 25

Suppose that


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pp

= lim inf

1() < = lim sup

1().

Given (, ), by continuity there is a sequence k going to infinitysuch that 1(k) = .



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Since d1/d 0 we have

V(cos , sin ) = limk

12

d1d

2

+ (1(k))

= E, < < .



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Since d1/d 0 we have

V(cos , sin ) = limk

12

d1d

2

+ (1(k))

= E, < < .

Thus, V is constant in the angular sector < < and therefore

r(t) = 0,

on that sector.

Finally, since p(t) 0, then 1() has a limit. Impossible!


IV A contribution of the author


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IV. A contribution of the author

Theorem(JC, QTDS08)

Let V be an angular potential in Rn

, n 3, and suppose thatCr = { Sn1 : V() = 0}

is totally disconnected. If r(t) is a solution to

r(t) = V(r(t)),then

limt

(t) = Cr.

The proof of this result is based on the following facts:1 lim

tV((t)) = 0.

2 If K is a compact totally disconnected subset of a separable metricspace X, then for each x, y in K, x = y, there exist disjoint sets Uand V open in X such that x

U, y

V, and K

U

V.


This result is optimal in the sense that if Cr possesses a non trivial


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connected component, then the limit may not exist.

Theorem (JC-Gutierrez, AIMS09)

There exists an angular potential V, smooth onR

3

\ {0}, such that Crcontains the equator ofS2 and such that Newtons equation possesses anorbit r(t) for which (t) = r(t)/|r(t)| does not have a limit as t .

Proof. Let T be a homogeneous function of degree zero, smooth on

R3 \ {0}, and considerS(x, y, z) = rT(x, y, z),

where r = x2 + y2 + z2. Then S is a homogeneous function of degreeone on R3 \ {0} and thus

V(r) = 12|S(r)|2

is a smooth angular potential on R3

\ {0}

.


If r(t) is an orbit of
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If r(t) is an orbit ofr = S(r),

one can show that r(t) is a solution to Newtons equation and that

(t) = r(t)/|r(t)| satisfies = T(). (1)

To construct the example we define T in cylindrical coordinates by

T(,, z) = er2/z2 sin

r

r

, z = 0;

= 0, z = 0.

Our construction is inspired in an example, due to Palis and de Melo, of atwo dimensional gradient vector field whose limit is a circle.


2
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Note that the equator ofS2 is a subset of Cr. We show that (1) has anorbit that goes to the equator ofS2 in spirals.


L t U b th b t f S2 d fi d b


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Let U be the open subset ofS2 defined by

U =

(,, z) S2 : > 0, z > 0, and T(,, z) < 0

.

U does not contain critical points of Tand if

[a, b] = {(, 0, z) S2 : 1 1

1 12

},

then U is the regionS2

limited by [a, b] and by the spirals

A =

(,, z) S2 : = 1 1

+ , 0

T1{0}

and

B =

(,, z) S2 : = 1 1

+ 2, 0

T1{0}.


Since T (A B) and points out towards the exterior of U, then


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J = { [a, b] : the positive orbit of (, 0, z) goes out of U transv. to B }is non empty and bounded below for some 1 > a. If

0 = infJ y P = (0, 0, z),then the positive orbit of P stays inside U and thus has in spirals towardsthe equator ofS2.



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Thanks a lot for your attention.

Email: [email protected]

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Jaime Cruz Sampedro - Newton's Eq

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