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Plan for PURE Math 2012 Seminar
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\noindent
{\it Week 3}
\bigskip
\item{} Monday:  Distance coordinates and configuration spaces
\item{} Tuesday:  Central configurations in
the $3$- and $n$-body problems
\item{} Wednesday: 
\item{} Thursday: 
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\centerline{PURE Math 2012 - Seminar}
\centerline{Week 3 Labs - Understanding central configurations in the 3-body problem}
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\noindent
{\it Background}
\bigskip
The Newtonian $n$-body problem is the study of the equations of motion
of $n$ point masses $m_i$ at positions $q_i \in {\bf R}^d$, acted
on by mutual gravitational forces following Newton's inverse square law
of gravitation.  The usual ``$F = ma$'' equation 
for the $i$th point mass reduces to the differential equation
$${d^2 q_i\over dt^2} = A_i = \sum_{j\ne i} {m_j (q_j - q_i)\over r_{ij}^3}$$
where $r_{ik} = \|q_{i} - q_k\|$ is the mutual distance between the locations of 
the $i$th and $k$th
masses.  (The notation $A_i$ is used since this is just the acceleration 
vector for the $i$th mass.)  As we are seeing in class and in the discussions 
this week, {\it central configurations} in the Newtonian $n$-body problem are 
special configurations of masses that satisfy the algebraic equations
$$A_i + \lambda(q_i - c) = 0\leqno(1)$$
for some scalar $\lambda$, where $c = {\sum_i m_i q_i\over \sum_i m_i}$ is
the center of mass of the system.

If masses in a central configuration are released from rest, the acceleration
is proportional to the vector from the initial position to the center of mass, 
and the whole configuration will
collapse to a total collision at the center of mass (in which all the masses
will reach the collision point at the same time).  The intermediate
positions will all be scaled (``homothetic'') versions of the original configuration.

If the masses in a central configuration all lie
in one plane, and they are released with suitable
initial velocities, they can also move in periodic orbits.  

We usually want to think of two central configurations as {\it equivalent} if
there is a rigid motion of ${\bf R}^d$ that carries one into the other (together with 
a possible relabeling of the masses).  This convention suggests that we should
try to recast the equations (1) entirely in terms of the mutual distances $r_{ij}$, 
since they are invariants under the group of rigid motions of ${\bf R}^d$.  As
we have seen in class, the (asymmetric) Albouy-Chenciner (``AC'') equations give one 
way to do this.  A collection of mutual distances $(r_{ij})$ for $1 \le i < j \le n$
along with masses $m_i$ defines a central configuration if and only if the following
equations are satisfied for real $r_{ij} > 0$.  We have effectively picked a particular 
distance scale for which $\lambda = 1$ in the following form of the AC equations.
The AC system consists of the equations 
$$G_{ij} = \sum_{k=1}^n m_k S_{ki}(r_{jk}^2 - r_{ik}^2 - r_{ij}^2) = 0,\leqno(2)$$
where 
$$S_{ki} = \cases{r_{ki}^{-3} - 1 & if $k \ne i$\cr
                          0 & if $k = i$\cr}$$
(Note: the mutual distances are symmetric in the 
indices, so $r_{\ell m} = r_{m \ell}$ for all $\ell,m$.  This means that the $S_{ki}$
with $k \ne i$ have the same symmetry:  $S_{ki} = S_{ik}$.)  We have one nontrivial
equation for each pair $1 \le i,j \le n$ with $i \ne j$ (everything cancels out if $i = j$).  

In the labs this week, we want to understand the solutions of the equations (2)
in the case $n = 3$ for all choices of $m_1,m_2,m_3$ real and positive.  

\bigskip
\noindent
{\it Historical Note:}  This determination of the central configurations in the 
3-body problem goes all the way back to work of Euler and Lagrange in late 18th 
century, but they approached the problem in other ways!  Interestingly enough, 
Lagrange's work on this problem contains one of the first occurences of the method
of Lagrange multipliers that you studied in multivariable calculus.

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\noindent
{\it Step 1 -- Entering the Equations}
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Unfortunately, there is no really good way to have Sage deal with doubly 
subscripted variables directly.  However we can have variables named 
{\tt r12, r13, r23, S12, S13, S23, m1, m2, m3}, etc.  
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\item{$\bullet$}  Set up a ``big'' polynomial ring $R$ with all of the above
as variables, and a graded reverse lex order with the variables ordered
as above.
\item{$\bullet$}  Define polynomials {\tt G12, G13, G23}  using the AC formulas
in (2)
\item{$\bullet$}  Note that {\tt S12, S13, S23} depend {\it rationally} on 
the {\tt r12, r13, r23}.  We want to get a system of polynomial equations
out of this, so what we want is to substitute in the expressions for
the {\tt S12, S13, S23}, then take the {\it numerators} of the resulting
rational functions (do you see why?)  To do this simply in Sage, we
can just apply the {\tt .substitute()} and {\tt .numerator()} methods
in sequence.  For instance:
\bigskip
\centerline{\tt G12S = G12.substitute(S12 = 1/r12\^{}3-1, S13 = 1/r13\^{}3-1).numerator()}
\bigskip
\item{} will produce the correct polynomial we want from $G_{12}$.
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\noindent
{\it Step 2 -- Studying the AC Ideal}
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\noindent
Continuing from the above, 
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\item{$\bullet$} Make an ideal using {\tt G12S} from above, and 
similar polynomials {\tt G13S} and {\tt G23S} produced in corresponding
and appropriate fashion. 
\item{$\bullet$}  Now compute a Gr\"obner basis for your ideal and 
examine {\it factored forms} the Gr\"obner basis polynomials one by 
one.  You should see some very interesting factorizations like
this in our output (possibly with the factors ordered differently):
\bigskip
\centerline{\tt 
m3*m2*(r23-1)*r12*(r23\^{}2+r23+1)*(r12-r13-r23)*(r12+r13-r23)$\backslash\backslash$}
\centerline{\tt *(r12-r13+r23)*(r12+r13+r23)}
\bigskip
\item{}
(This is all one polynomial, multiplied together.)
Now remember that we are looking for real, positive solutions of the AC equations
in the case that the masses $m_i > 0$ for all $i$.  
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\noindent
{\it Step 3 -- Breakdown into cases}
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\noindent
At this point (already!), we are ready to break our problem down into several
cases.  The Gr\"obner basis we computed also generates the AC ideal, so 
any solution we are looking for must also be a solution of all of the equations
obtained by setting the Gr\"obner basis polynomials equal to zero.
\bigskip  
\item{Q1.}  A number of the factors in the special factored polynomial
above {\it cannot} be zero at the solutions we are interested in.  Which ones
are those?  Only four of the factors here can actually be zero at the solutions
we are interested in.  Which are those?
\item{Q2.}  Using Sage, show that 
$$(r_{12} - r_{13} - r_{23})(r_{12} + r_{13} - r_{23})
(r_{12} - r_{13} + r_{23})(r_{12} + r_{13} + r_{23})$$
is (possibly up to a sign) the Cayley-Menger determinant associated to 
the three points:
$$\Delta_{CM} = \det \pmatrix{0 & 1 & 1 & 1\cr
                1 & 0 & r_{12}^2 & r_{13}^2\cr
                1 & r_{12}^2 & 0 & r_{23}^2\cr
                1 & r_{13}^2 & r_{23}^2 & 0\cr}$$
 \item{Q3.}  What is the geometric meaning, for instance, of the equation
 $r_{12} - r_{13} - r_{23} = 0$?
(Recall, these variables represent the {\it mutual distances} between pairs of
the point masses.)  
\bigskip
\noindent
For each of the factors that can be zero here, take that factor,
adjoin it to the other generators of the AC ideal, and consider those
new cases.
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\noindent
{\it Step 4 -- Analyzing the cases separately -- Case I}
\bigskip
For the case $r_{23} - 1 = 0$,  a good strategy will be to 
\bigskip
\item{$\bullet$} Compute a Gr\"obner basis obtained by adjoining 
that polynomial to the AC ideal, 
\item{$\bullet$} Look for any new Gr\"obner basis polynomials
that factor in ``interesting'' ways
\item{$\bullet$} Try adjoining those factors to the Gr\"obner
basis, and recompute a Gr\"obner basis from the resulting
sets of polynomials.
\bigskip
What are all of the central configurations that fall into 
this case?  What are they geometrically?  How do the
masses figure in?  (For instance, is there any restriction
on the $m_1,m_2,m_3$?)  Also, are there solutions of the 
AC equations in this cases that do not correspond to 
central configurations?  Explain.  (Recall the $r_{ij}$ and the 
masses $m_i$ must be real and positive.)
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\noindent
{\it Step 5 -- The other case(s)}
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\item{Q4.}  First, exactly how many other cases do we need to consider (at
least if we are allowed to label the locations of the masses
any way we want)?  Is it one case or more than one?  
\bigskip
\noindent
Let's try an alternative strategy.
Adjoin an appropriate 
factor from the factored form of the special polynomial (or
the Cayley-Menger determinant found above), then force
those polynomials into a new ring with a {\it lex} order
designed to eliminate $r_{12}$ and $r_{13}$ and give
an element of the ideal that is a univariate polynomial in $r_{23}$
whose coefficients are polynomials in the masses $m_1,m_2,m_3$.
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\item{Q5.}  Compute the {\it lex} Gr\"obner basis and 
find the univariate polynomial in $r_{23}$.  What is its
degree in $r_{23}$ (after removing any factors depending only on the masses
and any factor that is a pure power of $r_{23}$)?  Do you 
notice anything special about the remaining polynomial after 
those nonvanishing factors are removed?  There a way, for instance, to 
simplify the form (lower the degree), while keeping
the number of real roots the same.  What is that?
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\item{Q6.}  Try substituting at least 4 different
combinations of rational
values for the masses and finding the approximate roots of 
your simplified polynomial from Q4.  (Force the substituted
univariate polynomial into a single-variable ring, and
use the method {\tt .roots(ring=CC)} to find numerical approximations
to all of the complex roots (Note: {\tt CC} is Sage's floating point
complex number system).  How many real roots
do you find in each case?  Where are they located?

\bigskip
\noindent
From the evidence generated in Q6, you will probably have 
come up with a conjecture about what is going on.  But
can we {\it prove something} about the number of real 
solutions of the univariate polynomial?  There is a general
method for answering this sort of question called {\it Sturm sequences}.
Here is an outline of the way it works.  Given a polynomial $f$,
the {\it Sturm sequence} of a square-free polynomial $f$ is computed by a process similar
to the Euclidean algorithm for computation of $\gcd(f,f')$, {\it except that the negative
of the remainder} is used rather than the remainder itself:
$$\eqalign{f_0 &= f\cr
           f_1 &= f'\cr
           f_i &= -{\rm rem}(f_{i-2},f_{i-1}), i = 2,\ldots, m\cr}$$
When $f_m \ne 0$ but $-{\rm rem}(f_{m-1},f_m) = 0$ for the first time, we stop.
The polynomial sequence $[f_0,f_1,\ldots,f_m]$ is called the Sturm sequence.

Then to determine the number of roots of $f$ in any interval $(a,b)$, you 
compute $s(a) - s(b)$, where $s(a)$ is the number of sign changes in the sequence
of numbers 
$$[f_0(a),f_1(a),\ldots,f_m(a)]$$ 
and similarly for $s(b)$.  
In counting sign changes, any zero terms in the sequence are ignored.  Here is a 
simple example.   Say $f_0 = x^4-3x^2+x-2$, then computing as above, check using Sage
that the Sturm sequence is
$$[f_0,f_1,f_2,f_3,f_4] = \left[x^4 - 3x^2 + x - 2, 4x^3 - 6x + 1, {3\over 2}x^2 - {3\over 4}x + 2,
{31\over 3}x +{5\over 3}, -{8303\over 3844}\right].$$
On the interval $[a,b] = [0,2]$, note that $f(x) = 0$ has at least one real solution
by the intermediate value theorem: $f(0) = -2 < 0$, while $f(2) = 4 > 0$.  Is there more
than one?   Well the sequence of values at $x = 0$ is
$$\left[-2,1,2,{5\over 3}, {-8303\over 3844}\right]$$
which has $s(0) = 2$ sign changes.
The sequence of values at $x = 2$ is 
$$\left[4,21,{13\over 2},{67\over 3},{-8303\over 3844}\right]$$
which has $s(2) = 1$ sign change.  This gives $s(0) - s(2) = 1$, so Sturm's
theorem says that there is exactly one real root of $f(x) = 0$ on the interval
$[0,2]$.  Note that all of the above calculations are {\it exact} using 
rational arithmetic.  This is the virtue of Sturm sequences -- they are a purely
symbolic computational method yielding exact results (no use of numerical approximations).  

\bigskip 
\item{Q7.}  (This question is purposely very open-ended.  Think of 
it as a first, smaller-scale research project.)  Develop as much of a proof 
as you can to show 
that the univariate polynomial of degree 5 in $x = r_{23}^3$ obtained
above in Q5 has exactly one real root in the interval $[0,1]$ for all 
combinations of masses $m_1,m_2,m_3 > 0$ and no other real roots.  
Suggestions:  It is good to think of using a unit of mass 
defined so that the smallest mass
is 1, and the others have the form $1 + a$, $1 + a + b$ for 
some real $a,b \ge 0$.  But then of course you will need to consider
the locations of the different masses.  Which is located where? 
How many different cases are there that you need to consider?  Compute 
the Sturm sequence for the general degree 5 univariate polynomial in $x$, leaving
$a,b$ in the coefficients as symbolic constants.  You can then get
a feeling for signs in the Sturm sequence, for instance, by doing
some 3D plotting of the Sturm sequence terms evaluated at $r_{23} = 0$
and $r_{23} = 1$ for $a,b \ge 0$.  Can you prove your assertions completely?  
\bigskip
\item{Q8.}  Put your results together: What are all of the central
configurations in the Newtonian $3$-body problem?  
\bye

