CADERNOS DE MATEMÁTICA
ARTIGO NÚMERO
03, 315–323 October (2002)
SMA#154
Generalized homogeneous functions and the two-body problem
Carlos Biasi
and
S. M. S. Godoy
Departamento de Matemática, Instituto de Ciências Matemáticas e de Computação, Universidade de
São Paulo-Campus de São Carlos, Caixa Postal 668, 13560-970 São Carlos SP, Brazil
E-mail: [email protected]
E-mail: [email protected]
In this article we study a generalization of the homogeneous function concept. An application is done with a solution of the two-body problem. October,
2002 ICMC-USP
1. INTRODUCTION
In this paper we generalize the classic concept of homogeneous function of degree α and
we study the relation between the homogeneous function concept and the movement of a
body that satisfies the Kepler’s second law.
As an application of the involved techniques that were used, we present a solution of
the two-body problem, giving a way to obtain a time equation for the body that rotates
around the other, using the concept of homogeneous function.
We obtained a series like as that was presented in [1].
2. GENERALIZED HOMOGENEOUS FUNCTIONS
Let U be an open subset of Rn so that if x ∈ U and λ is a real number, 0 < λ < 1, then
λ.x ∈ U .
Let f : U → R be a C r function. We say that f is an homogeneous
function of degree α if f (λ.x) = λα .f (x), if λ > 0.
Definition 2.1.
We put bellow the well known concept of homogeneous function.
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C. BIASI AND S.M.S. GODOY
Let θ be a function of class C r such that θ : (0, ∞) × (0, ∞) → (0, ∞) and
(
θ(1, z) = z,
θ(λ1 .λ2 , z) = θ(λ1 , θ(λ2 , z)).
(1)
We observe that θ is an action of the multiplicative group (0, ∞) to (0, ∞).
∂θ(1, z)
Consider the function
α(z) =
, z ∈ (0, ∞).
∂λ
Let us generalize the concept of an homogeneous function.
Definition 2.2.
Let f : U → R be a function of class C r . We say that f is θ −
homogeneous if
i) f (λ. x) = θ (λ, f (x))
ii) α (f (x)) > 0.
Lemma 2.1. Let θ be an action of (0, ∞).
~ f (x), x i = α (f (x))
Then f is a θ − homogeneous function ⇐⇒ h ∇
Proof: ⇒) We note that:
∂θ
~
~ f (x), x i = ∂θ (1, f (x)) =
h ∇ f (λ x), x i =
(λ, f (x)). Then, for λ = 1, we have that, h ∇
∂λ
∂λ
α (f (x)).
⇐) We define for each value of x, the functions:
ϕ(λ) = f (λ x) and ϕ̃(λ) = θ(λ, f (x))
We note that ϕ(1) = f (x) = ϕ̃(1). We will prove that ϕ and ϕ̃ are solutions of an ordinary
differential equation with the same initial condition.
~ f (λ x), λ x i = λh ∇
~ f (λ x), x i = λ ϕ0 (λ)
We have that: α(f (λ x)) = h ∇
0
Then, α(ϕ(λ)) = λϕ (λ)
α
So ϕ is a solution of the equation ϕ0 = ϕ.
λ
For the function ϕ̃(λ) we have,
∂θ
λϕ̃0 (λ) = λ (tλ, f (x)).
∂λ
Consider the function h(t) = θ(t, θ(λ, f (x))) = θ(t λ, f (x)).
∂θ
∂θ
Then, h0 (t) = λ (t λ, f (x)). So, h0 (1) = λ (λ, f (x)).
∂λ
∂λ
∂θ
By other side, h0 (1) =
(1, θ(λ, f (x)) = α(θ(λ, f (x))). Then, λϕ̃0 (λ) = α(ϕ̃(λ)). So,
∂t
α
ϕ̃0 = ϕ̃, and then ϕ = ϕ̃.
λ
Remark 2. 1. Let f be a θ − homogeneous function with θ(λ, z) = λα z. Then f is
homogeneous of degree α, with α(z) = α z.
In fact, if f is a θ − homogeneous function, then f (λ x) = θ(λ, f (x)) = λα f (x). But
∂θ
(1, z) = αλα−1 z, for λ = 1. So, α(z) = α z.
α(z) =
∂λ
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Remark 2. 2. When P0 is any point, we say that f is a θ − homogeneousf unction
relative to P0 if f (P0 + λ(x − P0 )) = θ(λ, f (x)).
~ f (x), x − P0 i = α(f (x)).
As in the proof of Lemma 1 we easily demonstrate that: h ∇
Theorem 2.1. Let θ be an action as in (1), C a curve and P0 6∈ C. There exists a
θ − homogeneous function f relative to P0 , so that f (x) = 1, ∀x ∈ C.
Proof: Define ψ(λ) = θ(λ, 1). Suppose P0 = 0. For every x ∈ C, choose y so that
x
∈ C. Then we define f (x) = y. It is clear that f (x) = 1, ∀x ∈ C.
ψ −1 (y)
λx
λx
We have that −1
=
∈ C.
ψ (θ(λ, f (x)))
λψ −1 (f (x))
−1
−1
−1
So, θ(λψ (f (x)), 1) = ψ(λψ (f (x))) = ψ(ψ (θ(λ, f (x)))) = θ(λ, f (x)).
So, f (λ, x) = θ(λ, f (x)) and the function f is θ − homogeneous.
of a homogeneous function
Corollary 2.1. If C is any curve, then it is a level curve
y
of degree α > 0.
P2
We remember now the Kepler’s second law:
The Area’s Law
Let P1 and P2 be two successive positions of a
body in a interval of time δt. The element of
area in this interval of time is δA = r2 δφ/2,
∂A
r2 ∂φ
or,
=
.
is a constant, that is, the
∂t
2 ∂t
area is proportional to time.
φ
x
Consider a C r plane curve C, r ≥ 1, and a
C
point P = (P1 , P2 ) ∈
/ C. Suppose C is given by
x = x(t) = (x1 (t), x2 (t)), so that
x(t1 )
¯
¯
¯
¯
¯x(t) − P ¯
¯x1 (t) − P1 x2 (t) − P2 ¯
¯
¯
¯
¯>0
det ¯ 0
= det ¯
x (t) ¯
x01 (t)
x02 (t) ¯
P1
dφ
x(t)
x(t0 )
(2)
•
P = (P1 , P2 )
We know that the area swept out by a body that moves from Q0 = (x1 (t0 ), x2 (t0 )) to
Q1 = (x1 (t1 ), x2 (t1 )) is
¯
Z ¯
1 t1 ¯¯x(t) − P ¯¯
A=
dt
2 t0 ¯ x0 (t) ¯
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C. BIASI AND S.M.S. GODOY
So,
¯
¯
1 ¯¯x(t) − P ¯¯
A (t) = ¯ 0
=c
2 x (t) ¯
0
Definition 2.3. A curve C satisfies the Kepler’s second law relative to the point P if
A0 (t) = c, for some c > 0.
¯
Z t¯
¯x(u) − P ¯
1
¯
¯ 0
¯ x (u) ¯ du = 2 2c (t − t0 ) = c (t − t0 ).
t0
So, the area is proportional to the time to go from x(t0 ) to x(t), and then satisfies the
area’s Kepler’s law.
Then, A(t) =
1
2
Remark 2. 3. If x(t), t ∈ (a, b) is a parametric curve C that satisfies the Kepler’s second
law with constant c in relation to P , we have that:
A=
1
2
Z
b
a
¯
¯
¯x(t) − P ¯
¯ 0
¯
¯ x (t) ¯ dt = c(b − a) = cp
where p = b − a.
3. PARAMETRIC REPRESENTATION BY SURFACE MEASURE
It is often convenient to shift from one parameter representation of a curve, to another, to
achieve once a special parametric representation for the Kepler’s second law to be satisfied.
Let x̃(u), u ∈ (c, d), the parameter representation of a curve C. We have that:
Ã(u) =
So,
Ã0 (u) =
1
2
Z
u
u0
¯
¯
¯x̃(v) − P ¯
¯ 0
¯
¯ x̃ (v) ¯ dv
¯
¯
1 ¯¯x̃(u) − P ¯¯
> 0, ∀u
2 ¯ x̃0 (u) ¯
We make the follow change of parameter: t = Ã(u), t ∈ (a, b), and h(t) = u, and define
x(t) = x̃(h(t)).
With this choice of the parametric representation, the curve C satisfies the Kepler’s
second law in relation to P.
In fact, we have that:
¯
¯
¯
¯
¯
¯
¯
¯
¯x̃(u) − P ¯
1
1 ¯¯x(t) − P ¯¯ 1 ¯¯ x̃(u) − P ¯¯ 1 0 ¯¯x̃(u) − P ¯¯ 1
¯
¯=1
¯.
= .2 ¯
=
= h (t) ¯ 0
x̃ (u) ¯ 2 ¯¯x̃(u) − P ¯¯ ¯ x̃0 (u) ¯
2 ¯ x0 (t) ¯ 2 ¯x̃0 (u).h0 (t)¯ 2
¯ x̃0 (u) ¯
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So
¯
¯
¯x(t) − P ¯
¯ 0
¯
¯ x (t) ¯ = 2 and the Kepler’s second law is satisfied.
Definition 3.1.
If a curve C satisfies the Kepler’s second law with constant c = 1 in
relation to P we say that the curve C has a parametric representation by surface measure.
Then, above we prove that any curve can have a parametric representation by surface
measure, what is analogous that we know by the parametric representation of one curve
by arc length. [2]
So we prove that if x(t) is a parametric representation of a curve by surface measure and
the time to go from a point Q1 to Q2 is T , then the area swept out is T.
4. GENERALIZED HOMOGENEITY AND THE KEPLER’S SECOND
LAW
Let U be an open subset of R2 . Let f : U → R be a C 1 function whose derivative at a
point x is denoted by f 0 (x). There exists a unique vector g(x) ∈ R2 such that f 0 (x) · v =
hg(x), vi, for all v ∈ R2 , where h , i denotes the inner product in R2 . Let u(x) be the
hamiltonian field obtained by rotating g(x) by an angle of π2 radians. Observe that the
vectors u(x) are tangent to the level curves of f , so the vectors g(x) are orthogonal to the
level curves of f .
Theorem 4.1. Let f be a θ − homogeneous function and x(t) a solution of the initial
value problem
(
ẋ = u(x)
x(t0 ) = x0
(3)
where u is defined above. Then x(t) satisfies the Kepler’s second law in relation to the
origin.
Proof We note that because x(t) is a solution of (3), then it is a parametric function of a
part of the level curve f −1 (f (x0 )). This fact and lemma 1 imply that
¯
¯
¯ x (t) x2 (t) ¯
¯ = h∇f (x), xi = α(f (x)) = αf (x0 ))
A0 (t) = ¯¯ 10
x1 (t) x2 0 (t)¯
which is constant.
Corollary 4.1. Let f be homogeneous of degree α > 0. Then the solution of the equaα
tion ẋ = u(x) satisfies the Kepler’s second law with constant c = .
2
Proof The function f is homogeneous of degree α, then by Remark 1, α(z) = αz. Let
x(t) be the solution of ẋ = u(x) so that f (x(t)) = 1. Then by Theorem 2,
α
2c = A0 (t) = α.f (x0 ) = α.1, and so c = .
2
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C. BIASI AND S.M.S. GODOY
Lemma 4.1. Let x(t), t ∈ (a, b) a parameter representation of a curve of a curve C
that satisfies the second Kepler’s law with constant c in relation to P. Then to obtain
another parametric representation x1 (s), s ∈ (c, d) that satisfies the Kepler’s second law
with constant c1 it is suficient to take x1 (s) = x(s cc1 ).
Proof Let h : (c, d) → (a, b) be a function so that t = h(s) and x(t) = x1 (h(s)). Then,
¯
¯
¯
¯ ¯
¯
¯x1 (h(s)) − P ¯
¯x(t) − P ¯ ¯ x1 (h(s)) − P ¯
0
0
¯=¯ 0
¯
¯
¯
=
h
(s)
2c = ¯¯ 0
¯ x1 0 (h(s)) ¯ = h (s)2c1
x (t) ¯ ¯h (s)x1 0 (h(s))¯
c
c
which implies that t = h(s) = s + t0 ,
c1
c1
c
So, x1 (s) = x(s + t0 ).
c1
Then h0 (s) =
Lemma 4.2. Let x(t), t ∈ J and x̃(s), s ∈ J1 , parametric representations of a curve
C that satisfies the Kepler’s second law with the same constant c in relation to P. Then
t = s + d, with d constant.
Proof Because x(t) and x̃(s) parameterize the same curve C we have that x̃(s) = x(t) =
x(h(s)), t = h(s), and then x̃0 (s) = h0 (s)x0 (h(s)) = h0 (s)x0 (t). So,
¯
¯
¯
¯
¯x̃(s) − P ¯
¯x(t) − P ¯
0
¯
¯
¯
¯ = h0 (s)2c
2c = ¯ 0
= h (s) ¯ 0
x̃ (s) ¯
x (t) ¯
Then, h0 (s) = 1 and so h(s) = s + d.
Theorem 4.2. Let x(t) be a parametric representation of a curve C that satisfies the
Kepler’s second law with a constant α = c in relation to origin. Then there exists a
homogeneous function of degree α = 2c so that x(t) is a solution of (3).
Proof By Corollary 1 we have that C = {x(t), t ∈ J} is a level curve of a homogeneous
function f of degree α = 2c, that is, f (x(t)) = 1, x(t0 ) = x0 , f (x0 ) = 1.
Consider the equation (3) and let x̃(s) be a solution so that x̃(t0 ) = x0 . Then, f (x̃(s)) = 1,
and because x̃(s) satisfies the Kepler’s second law with the same constant c and x̃(t0 ) = x0 ,
then x̃(t) = x(t).
We observe that if we change the point P and consider the same parameter, the relation
between the areas swept out by a point that moves from point P to a point P1 is given by:
Z
b
A=
a
Z
=
a
b
¯
¯
¯
Z b ¯
1 ¯¯x(t) − P ¯¯
1 ¯¯x(t) − P1 + P1 − P ¯¯
dt =
¯
¯ dt
x0 (t)
2 ¯ x0 (t) ¯
a 2
¯
¯
¯
¯
¯
Z b ¯
1 ¯¯ P1 − P ¯¯
1 ¯¯ P1 − P ¯¯
1 ¯¯x(t) − P1 ¯¯
dt +
¯R b x0 (t)¯ dt = A1 + 2 ¯x(t) − x(a)¯
2 ¯ x0 (t) ¯
a 2
a
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GENERALIZED HOMOGENEOUS FUNCTIONS AND THE TWO-BODY PROBLEM
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and x(t) is a solution of (3) for t ∈ (t0 , t1 ), we have that
¯ If f is ¯ θ homogeneous
¯
¯
¯x(t) − P ¯ ¯x(t) − P ¯
¯ 0
¯ ¯
¯
¯ x (t) ¯ = ¯ u(x) ¯ = α(f (x))
Z
Z
1 t1
1 t1
1
Then, A =
α(f (x(t))) dt =
α(z0 ) dt = α(z0 )(t1 − t0 ).
2 t0
2 t0
2
5. AN APPLICATION: THE TWO-BODY PROBLEM
Consider the classical problem in which an object of
mass m orbits another object of a much larger mass
M. Let F be the center of mass between m and M,
and suppose that the movement is elliptical, that is,
the object of mass m describes an ellipse whose focus
is F.
The orbital period P is knew in terms of the masses
4π 2 a3
m and M, that is: P =
, where G is the
G(m + M )
gravitational constant.
y
b
F
•
x
c
a
We observe that this is a movement that satisfies the Kepler’s second law and then we
know¯ that: ¯
1 ¯¯x(t) − F ¯¯
=c
and
A = πab
2 ¯ x0 (t) ¯
¯
Z t1 ¯
1 ¯¯x(t) − F ¯¯
Then, πab =
¯ x0 (t) ¯ dt = (t1 − t0 )c = P c.
t0 2
πab
.
So, c =
P
Our objective is to obtain the equation of the movement x(t) of the body of mass m. (with
Kepler’s constant c relatively to F ).
x1 2
x2 2
Given f = f (x1 , x2 ) = 2 + 2 − 1, the movement’s orbit is given by the level curve
a
b
x1 2
x2 2
one of f, that is,
+ 2 = 1.
a2
b
¡
¢
~ (x) = 2x1 , 2x2 , and
We observe that f is homogeneous of degree 2, ∇f
2
2
a
b
¡ −2x2 2x1 ¢
, 2 .
u(x) =
b2
a
¡ −2x2 2x1 ¢
, 2 .
The ordinary differential equation is ẋ =
b2
a
¡
2s ¢
2s
. For this solution the constant
The solution for this equation is x̃(s) = a cos , b sin
ab
ab
in relation to P = 0 is:
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C. BIASI AND S.M.S. GODOY
¯
¯
¯
¯
¯ a cos 2t̃ b sin 2t̃ ¯
ab
ab
¯
¯
¯
¯
1 ¯x(t̃) − P ¯ 1 ¯
¯=1
= ¯
0
¯
¯
¯
2 x (t̃)
2 ¯ −2
2t̃ 2
2t̃ ¯
sin
cos
b
ab a
ab
Let x̄(t̃) be the parametric representation with constant c = 1 in relation to F. We know
that any curve can
representation
the
¯ a parametric
¯
¯ for toZsatisfy
¯
¯ Kepler’s second law.
Z have
Z ¯
1 s ¯¯x̃(v) − F ¯¯
1 s ¯¯ x̃(v) ¯¯
1 s ¯¯ F ¯¯
So, t̃ = Ã(s) =
¯ x̃0 (v) ¯ dv = 2
¯ 0 ¯ dv − 2
¯ 0 ¯ dv =
s0 x̃ (v)
s0 x̃ (v)
¯ 2 s0
¯
¯
1¯
F
¯
s − s0 − ¯¯
2 √x̃(s) − x̃(s0 )¯
Since F = − a2 − b2 and
that:
¯ √taking s0 =¯ 0, it follows
¯ √
¯
1 ¯¯(− a2 − b2 , 0)¯¯
1 ¯¯ − a2 − b2
0 ¯¯
=
s
+
t̃ = Ã(s) = s − s0 + ¯
2s
2s ¯ .
− 1) b sin ab
2 x̃(s) − x̃(0) ¯
2 ¯a(cos ab
£ 2s
¤
1 √
2s
1 √
2s 1
2s 1
Then, t̃ = s − b a2 − b2 sin
= s − b a2 − b2
− ( )3 + ( )5 + ...
2
ab
2
ab
ab 3!
ab 5!
But t̃ = Ã(s), s = h(t̃), then x̄(t̃) = x̃(h(t̃)), and we remember that in this manner x̄(t̃)
satisfies the Kepler’s second law with constant c = 1.
Taking x(t) = x̄(ct), where t̃ = ct, the Kepler’s second law is satisfied with constant
πab
c=
.
P
This is the parametric representation of the planetary movement of two-body problem
whose Kepler’s constant is given as a function of the period.
1 2
Let us now consider the case when the orbit is a parabola, y =
(x − d2 ).
2d
y
6
F
x
d/2
6
?
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GENERALIZED HOMOGENEOUS FUNCTIONS AND THE TWO-BODY PROBLEM
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
x = u
The parametric representation
1 2
y =
(u − d2 )
2d
cannot satisfies the Kepler’s second law, and then we make a parametric representation by
surface
by ¯making:
¯ measure
1
2
2 ¯
¯
1 ¯u 2d (u − d )¯
1 u2
1
1 u2 +d2
1
2
2
2
2
u
2 ¯1
¯ = 2 ( d − 2d (u − d )) = 2 ( 2d ) = 4d (u + d )
d
µ
¶
R u 2 +d2
1 3
1
Then, t̃ = 12 0 v 2d
dv = 21 6d
u + d2 u = 12d
u3 + d4 u = θ(u) and so, u = θ−1 (t̃)
By making a change of parameters, we obtain a new parametric representation that
satisfies the Kepler’s second law with constant 1 in relation to F. Let (x̃(t̃), ỹ(t̃)) this
parametric representation. So, (x̃(ct), ỹ(ct)) will satisfies the Kepler’s second law with
constant c determined by the velocity v0 at t = 0.
1
1
But, (x̃(ct), ỹ(ct)) = ((u, 2d
(u2 − d2 )) = (θ−1 (ct), 2d
(θ−1 (ct))2 − d2 ) and
³
³ 3u2 + 3d2 ´
³
´
´
1
2
1
= θ0 (u)
, where θ0 (u) =
x̃0 (ct), ỹ 0 (ct)
, 2d
= 14 d and we
θ0 (u) u
12d
t=0
u=0
u=0
observe that in this example the constant c was not determined in function of the period
because we are treating of the parabolic case; it is done in function of the initial velocity.
Put now x(t) = x̃(ct) and y(t) = ỹ(ct). Then, (x0 (0), y 0 (0)) d4 = (1, 0) and then v~0 =
dv0
(c d4 , 0) and v0 = |v~0 | = c d4 and then, c =
. So, we have that (x(t), y(t)) is a parametric
4
representation that satisfies the Kepler’s second law with constant c in relation to F, where
c is given above.
We observe that in this case we obtain u in function of t̃ by resolving a cubic equation.
In the elliptic case the equation is transcendent.
In a analogous way we can describe the movement in the case that the orbit is a hyperbole.
In this case the parametric representation involves hyperbolic functions.
REFERENCES
1. Herrick, C., On the computation of nearly parabolic two-body orbits, Astronom.J., vol.65, number 6,
386-388(1960)
2. Stoker, J. J., Differential Geometry, Pure and Applied Mathematics,Wiley-Interscience, (1969)
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