Hirsch M., Smale S., Devaney R., Differential Equations, Dynamical Systems, and an Introduction to Chaos

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286 Chapter 13 Applications in Mechanics

We visualize phase space as the collection of all tangent vectors at each point

X ∈

C. For a given X ∈ R

−{0}, let T

={(X , V ) |V ∈ R

}. T

is the

tangent plane to the conﬁguration space at X . Then

P =



X∈C

is the tangent space to the conﬁguration space, which we may naturally identify

with a subset of

The dimension of phase space is four. However, we can cut this dimension

in half by making use of the two known ﬁrst integrals, total energy and angu-

lar momentum. Recall that energy is constant along solutions and is given by

E(X , V ) = K (V ) + U (X ) =

|V |

−

|X|

Let 

denote the subset of P consisting of all points (X, V ) with E(X, V ) = h.

The set 

is called an energy surface with total energy h.Ifh ≥ 0, then 

meets each T

in a circle of tangent vectors satisfying

|V |

= 2



h +

|X|



The radius of these circles in the tangent planes at X tends to ∞as X → 0 and

decreases to 2h as |X | tends to ∞.

When h<0, the structure of the energy surface 

is different. If |X | >

−1/h, then there are no vectors in T

∩ 

. When |X |=−1/h, only the zero

vector in T

lies in 

. The circle r =−1/h in conﬁguration space is therefore

known as the zero velocity curve.IfX lies inside the zero velocity curve, then

meets the energy surface in a circle of tangent vectors as before. Figure 13.1

gives a caricature of 

for the case where h<0.

We now introduce polar coordinates in conﬁguration space and new

variables (v

, v

) in the tangent planes via

V = v



cos θ

sin θ



+ v



−sin θ

cos θ



We have

V = X



= r





cos θ

sin θ



+ rθ





−sin θ

cos θ



so that r



= v

and θ



= v

/r. Differentiating once more, we ﬁnd

−1



cosθ

sinθ



=−

|X|







−



cosθ

sinθ











−sinθ

cosθ



13.4 The Newtonian Central Force System 287

zero velocity

curve

Figure 13.1 Over each nonzero point

inside the zero velocity curve, T

meets the energy surface 

in a circle

of tangent vectors.

Therefore in the new coordinates (r,θ ,v

), the system becomes



=−



=−

In these coordinates, total energy is given by





−

and angular momentum is given by  =rv

. Let 

h,

consist of all points in

phase space with total energy h and angular momentum . For simplicity, we

will restrict attention to the case where h<0.

If  =0, we must have v

=0. So if X lies inside the zero velocity curve, the

tangent space at X meets 

h,0

in precisely two vectors of the form

±v



cosθ

sinθ



both of which lie on the line connecting 0 and X, one pointing toward 0, the

other pointing away. On the zero velocity curve, only the zero vector lies in



h,0

. Hence we see immediately that each solution in 

h,0

lies on a straight line

288 Chapter 13 Applications in Mechanics

through the origin. The solution leaves the origin and travels along a straight

line until reaching the zero velocity curve, after which time it recedes back to

the origin. In fact, since the vectors in 

h,0

have magnitude tending to ∞ as

X →0, these solutions reach the singularity in ﬁnite time in both directions.

Solutions of this type are called collision-ejection orbits.

When  =0, a different picture emerges. Given X inside the zero velocity

curve, we have v

=/r, so that, from the total energy formula,

=2hr

+2r −

. (A)

The quadratic polynomial in r on the right in Eq. (A) must therefore be

nonnegative, so this puts restrictions on which r-values can occur for X ∈

h,

The graph of this quadratic polynomial is concave down since h<0. It has no

real roots if 

> −1/2h. Therefore the space 

h,

is empty in this case. If



=−1/2h, we have a single root that occurs at r =−1/2h. Hence this is the

only allowable r-value in 

h,

in this case. In the tangent plane at (r, θ), we

have v

=0, v

=−2h, so this represents a circular closed orbit (traversed

clockwise if  < 0, counterclockwise if  > 0).

If 

< −1/2h, then this polynomial has a pair of distinct roots at α,β with

α < −1/2h<β. Note that α > 0. Let A

α,β

be the annular region α ≤r ≤β in

conﬁguration space. We therefore have that motion in conﬁguration space is

conﬁned to A

α,β

Proposition. Suppose h <0 and 

< −1/2h. Then 

h,

⊂P is a two-

dimensional torus.

Proof: We compute the set of tangent vectors lying in T

∩

h,

for each

X ∈A

α,β

.IfX lies on the boundary of the annulus, the quadratic term on the

right of Eq. (A) vanishes, and so v

=0 while v

=/r. Hence there is a unique

tangent vector in T

∩

h,

when X lies on the boundary of the annulus. When

X is in the interior of A

α,β

, we have

=±



2hr

+2r −

, v

=/r

so that we have a pair of vectors in T

∩

h,

in this case. Note that these

vectors all point either clockwise or counterclockwise in A

α,β

, since v

has the

same sign for all X . See Figure 13.2. Thus we can think of 

h,

as being given

by a pair of graphs over A

α,β

: a positive graph given by v

and a negative

graph given by v

−

which are joined together along the boundary circles r =α

and r =β. (Of course, the “real” picture is a subset of

.) This yields the

torus.



It is tempting to think that the two curves in the torus given by r =α and

r =β are closed orbits for the system, but this is not the case. This follows

13.5 Kepler’s First Law 289

a,b

Figure 13.2 A selection of

vectors in 

,

since, when r =α, we have



=−

(−α+

However, since the right-hand side of Eq. (A) vanishes at α, we have

2hα

+2α −

=0,

so that

−α+

=(2hα +1)α.

Since α < −1/2h, it follows that r





> 0 when r =α, so the r coordinate of

solutions in 

h,

reaches a minimum when the curve meets r =α. Similarly,

along r =β, the r coordinate reaches a maximum.

Hence solutions in A

α,β

must behave as shown in Figure 13.3. As a remark,

one can easily show that these curves are preserved by rotations about the

origin, so all of these solutions behave symmetrically.

More, however, can be said. Each of these solutions actually lies on a closed

orbit that traces out an ellipse in conﬁguration space. To see this, we need to

turn to analysis.

13.5 Kepler’s First Law

For most nonlinear mechanical systems, the geometric analysis of the previous

section is just about all we can hope for. In the Newtonian central force system,

290 Chapter 13 Applications in Mechanics

Figure 13.3 Solutions in 

,

that meet r = α or r = β.

however, we get lucky: As has been known for centuries, we can write down

explicit solutions for this system.

Consider a particular solution curve of the differential equation. We have

two constants of the motion for this system, namely, the angular momentum

 and total energy E. The case  =0 yields collision-ejection solutions as we

saw above. Hence we assume  =0. We will show that in polar coordinates

in conﬁguration space, a solution with nonzero angular momentum lies on a

curve given by r(1+ cosθ) =κ where  and κ are constants. This equation

deﬁnes a conic section, as can be seen by rewriting this equation in Cartesian

coordinates. This fact is known as Kepler’s ﬁrst law.

To prove this, recall that r



is constant and nonzero. Hence the sign of



remains constant along each solution curve. Thus θ is always increasing or

always decreasing in time. Therefore we may also regard r as a function of θ

along the curve.

Let W (t ) =1/r(t ); then W is also a function of θ. Note that W =−U . The

following proposition gives a convenient formula for kinetic energy.

Proposition. The kinetic energy is given by

K =







dθ





Proof: In polar coordinates, we have

K =





)

+(rθ



)



Since r =1/W , we also have



−1

dθ



=−

dθ

13.5 Kepler’s First Law 291

Finally,

rθ





=W .

Substitution into the formula for K then completes the proof.



Now we ﬁnd a differential equation relating W and θ along the solution

curve. Observe that K =E −U =E +W . From the proposition we get



dθ





(E +W ). (B)

Differentiating both sides with respect to θ, dividing by 2 dW /dθ, and using

dE/dθ =0 (conservation of energy), we obtain

dθ

+W =



where 1/

is a constant.

Note that this equation is just the equation for a harmonic oscillator with

constant forcing 1/

. From elementary calculus, solutions of this second-

order equation can be written in the form

W (θ) =



+A cos θ +B sin θ

or, equivalently,

W (θ) =



+C cos

(

θ +θ

)

where the constants C and θ

are related to A and B.

If we substitute this expression into Eq. (B) and solve for C (at, say, θ +θ

π/2), we ﬁnd

C =±





1+2

Inserting this into the solution above, we ﬁnd

W (θ) =





1±



1+2E

cos(θ +θ

)



There is no need to consider both signs in front of the radical since

cos(θ +θ

+π ) =−cos(θ +θ

292 Chapter 13 Applications in Mechanics

Moreover, by changing the variable θ to θ −θ

we can put any particular

solution into the form







1+2E

cosθ



This looks pretty complicated. However, recall from analytic geometry (or

from Exercise 2 at the end of this chapter) that the equation of a conic in polar

coordinates is

(1+ cos θ).

Here κ is the latus rectum and  ≥0istheeccentricity of the conic. The origin

is a focus and the three cases  > 1,  =1,  < 1 correspond, respectively, to

a hyperbola, parabola, and ellipse. The case  =0 is a circle. In our case we

have

 =



1+2E

so the three different cases occur when E>0, E =0, or E<0. We have proved:

Theorem. (Kepler’s First Law) The path of a particle moving under the

inﬂuence of Newton’s law of gravitation is a conic of eccentricity



1+2E

This path lies along a hyperbola, parabola, or ellipse according to whether E > 0,

E =0,orE<0.



13.6 The Two-Body Problem

We now turn our attention brieﬂy to what at ﬁrst appears to be a more difﬁcult

problem, the two-body problem. In this system we assume that we have two

masses that move in space according to their mutual gravitational attraction.

Let X

denote the positions of particles of mass m

in R

.SoX

) and X

=(x

). From Newton’s law of gravitation, we ﬁnd

the equations of motion



=gm

−X



=gm

−X

13.7 Blowing Up the Singularity 293

Let’s examine these equations from the perspective of a viewer living on the

ﬁrst mass. Let X =X

−X

. We then have



−X



=gm

−X

−gm

−X

=−g (m

)

|X|

But this is just the Newtonian central force problem, with a different choice of

constants.

So, to solve the two-body problem, we ﬁrst determine the solution of X(t )

of this central force problem. This then determines the right-hand side of the

differential equations for both X

and X

as functions of t, and so we can

simply integrate twice to ﬁnd X

(t) and X

(t).

Another way to reduce the two-body problem to the Newtonian central

force is as follows. The center of mass of the two-body system is the vector

A computation shows that X



=0. Therefore we must have X

=At +B where

A and B are ﬁxed vectors in

. This says that the center of mass of the system

moves along a straight line with constant velocity.

We now change coordinates so that the origin of the system is located at X

That is, we set Y

−X

for j =1,2. Therefore m

(t)+m

(t) =0 for all

t. Rewriting the differential equations in terms of the Y

,weﬁnd



=−

)



=−

)

which yields a pair of central force problems. However, since we know that

(t)+m

(t) =0, we need only solve one of them.

13.7 Blowing Up the Singularity

The singularity at the origin in the Newtonian central force problem is the ﬁrst

time we have encountered such a situation. Usually our vector ﬁelds have been

294 Chapter 13 Applications in Mechanics

well deﬁned on all of R

. In mechanics, such singularities can sometimes be

removed by a combination of judicious changes of variables and time scalings.

In the Newtonian central force system, this may be achieved using a change of

variables introduced by McGehee [32].

We ﬁrst introduce scaled variables

1/2

In these variables the system becomes



−1/2



−3/2



−3/2



−1





−3/2



−



We still have a singularity at the origin, but note that the last three equations

are all multiplied by r

−3/2

. We can remove these terms by simply multiplying

the vector ﬁeld by r

3/2

. In doing so, solution curves of the system remain the

same but are parameterized differently.

More precisely, we introduce a new time variable τ via the rule

dτ

3/2

By the chain rule we have

dτ

and similarly for the other variables. In this new timescale the system

becomes

˙r =ru

θ =u

˙u

−1

˙u

=−

13.7 Blowing Up the Singularity 295

where the dot now indicates differentiation with respect to τ . Note that, when

r is small, dt /dτ is close to zero, so “time” τ moves much more slowly than

time t near the origin.

This system no longer has a singularity at the origin. We have “blown up”

the singularity and replaced it with a new set given by r =0 with θ,u

, and u

being arbitrary. On this set the system is now perfectly well deﬁned. Indeed,

the set r =0 is an invariant set for the ﬂow since ˙r =0 when r =0. We have thus

introduced a ﬁctitious ﬂow on r =0. While solutions on r =0 mean nothing

in terms of the real system, by continuity of solutions, they can tell us a lot

about how solutions behave near the singularity.

We need not concern ourselves with all of r =0 since the total energy relation

in the new variables becomes

hr =





−1.

On the set r =0, only the subset  deﬁned by

=2, θ arbitrary

matters. The set  is called the collision surface for the system; how solu-

tions behave on  dictates how solutions move near the singularity since any

solution that approaches r =0 necessarily comes close to  in our new coor-

dinates. Note that  is a two-dimensional torus: It is formed by a circle in the

θ direction and a circle in the u

–plane.

On  the system reduces to

θ =u

˙u

=−

where we have used the energy relation to simplify ˙u

. This system is easy

to analyze. We have ˙u

> 0 provided u

=0. Hence the u

coordinate must

increase along any solution in  with u

=0.

On the other hand, when u

=0, the system has equilibrium points. There

are two circles of equilibria, one given by u

=0,u

√

2, and θ arbitrary, the

other by u

=0,u

=−

√

2, and θ arbitrary. Let C

denote these two circles

with u

=±

√

2onC

. All other solutions must travel from C

−

to C

since

increases along solutions.