From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.symbolic
Subject: Re: Help!
Date: 10 Mar 1998 20:03:23 GMT
In article <3504D5D9.54B6BFB1@adam.montreal.ca>,
root wrote:
>I am trying to find solutions to the following system of equations.
>
>Let g[u1, u2, q]= u1^8 (u1 - q^2 u2)(u1 u2 - q^2 ) - (u2 - q^2 u2)(1 -
>q^2 u1 u2)
>
>The system is then
>{g[u1, u2, q] = 0, g[u2, u1, q] =0}
Let me analyze the solution in some detail, in case I've misunderstood
what your real goal is.
Your two equations describe a curve in u1-u2-q space. I'm not sure
what exactly you mean by "find the solutions", but I'm guessing
you want to know u1 and u2 as functions of q. Your other comments
suggest you want to be able to compute them numerically for
given numerical q.
(If that's true, I can't see why you refer to q since only Q:= q^2 is
involved. I will refer only to Q in the sequel.)
Maybe it's easier if you first look at the projection of your curve to
u1-u2 space: I used maple.
>readlib(eliminate):
>eliminate({g,h},Q);
This gives a formula expression Q as a rational function of u1, u2, and
an equation describing the curve in the u1-u2 plane. That curve includes
a few easy components (u1= +- u2 and u1 = +- 1/u2) as well as one
complicated one:
14 14 7 15 15 7 11 11 9 13 13 9 9 15
u1 + u2 + u2 u1 + u2 u1 + u2 u1 + u2 u1 + u2 u1 - u2 u1
11 13 9 15 11 13 13 2 14 6 10
- u2 u1 - u1 u2 - u1 u2 - u2 u1 - 2 u1 u2 - 4 u2 u1
12 4 8 8 10 6 12 4 11 3
- 3 u1 u2 - 5 u2 u1 - 4 u2 u1 - 3 u2 u1 - 2 u1 u2
6 8 8 6 10 4 15 9 5 2 12
+ 4 u2 u1 + 4 u2 u1 + 3 u2 u1 + u1 u2 - 3 u2 u1 + 2 u1 u2
4 10 12 2 11 5 3 13 9 7
+ 3 u2 u1 + 2 u1 u2 + 3 u1 u2 + 2 u1 u2 + 4 u2 u1
7 9 3 11 5 11 16 16 13 15
+ 4 u2 u1 - 2 u1 u2 + 3 u1 u2 - u2 - u1 - u1 u2 + u2 u1
13 3 14 2 7 7 5 9
+ 2 u1 u2 - 2 u1 u2 - 4 u2 u1 - 3 u2 u1 = 0
(The equation Q=... then allows one to reconstruct the curve in 3-dimensional
space as a covering of this curve in the plane.)
I'll discuss this complicated curve carefully; however, we will eventually
discover it contains _none of the points of interest in the original problem_.
I observe that maple's implicitplot feature has some difficulty
discerning the behaviour of this curve in a few regions; let's analyze
the curve carefully. In order to visualize this planar curve, we can
convert to polar coordinates. I will avoid trigonometry for a moment:
> subs({u1=t*v1,u2=t*v2}, crv):
> simplify("/t^14):
> simplify(", {t^2=T}):
> c2:=convert(factor(taylor(",T,6)),polynom);
which gives a quintic in T:
2 2 2 2 6 3 3 6 4 4
c2 := (v2 - v2 v1 + v1 ) (v2 + v1 ) (v2 - v1 v2 + v1 ) (v2 + v1 ) -
...
9 9 2 2 4 4 5
- v2 v1 (v2 + v1 ) (v2 + v1 ) T
Thus it shouldn't be hard to draw this curve in polar coordinates:
for any point (v1,v2) on the unit circle, there should be at most 5 points
on the curve which lie on the ray through that point. Actually the number
of points per ray is locally constant since there is no bifurcation of the
roots:
> c3:=subs({v1=cos(H),v2=sin(H)},c2):
> eliminate({c3,diff(c3,T)},T);
yields an immense polynomial in cos(H), sin(H) which never vanishes.
On the other hand, the polynomial drops degree precisely at the axes,
so the number of roots can change there. Taking a few test values of H
shows there is one point per ray in the first and third quadrants and
two per ray in the other quadrants. (There are three real roots T in
those quadrants, but only two are positive, so t is one of two pairs of
positive/negative values.)
Thus we can ask maple to plot some points, e.g.:
> for ii from 1 to 100 do fsolve(evalf(subs(H=ii*Pi/400,c3))); od;
(By symmetry of the curve we need only the T for H in [-Pi/4, Pi/4]).
The values of T rise monotonically from 1 to 2 in this interval.
As H varies from 0 to -Pi/4, one values of T rises from 1 to
about 1.568670203; the other positive one decreases (from +infinity) to
about 4.096085035. A plot of these points shows a closed loop which
roughly resembles the outline of a rounded square with one pair of
opposite corners pulled away from the square. The unbounded portion of
the curve resembles a hyperbola xy=-2.
The bounded and unbounded portions of this curve describe the "nontrivial"
combinations of u1 and u2 for which a Q exists satisfying the
original equations g=h=0. As I remarked at the beginning, there
are also the curves u1=u2, u1=-u2, u1=1/u2, and u1=-1/u2.
Now, this is just the projection of the original curve to the u1-u2 plane.
The original curve in R^3 is easily found from this since we know how to
lift the projection back up. Indeed, elimination of Q at the beginning
provided a rational function which enables us to write Q =
4 4 4 3 2 2 3 4
(v2 + v1 ) - v2 v1 (v2 - v2 v1 + v2 v1 - v2 v1 + v1 )
4 3 2 2 3 4 2 8 8 5
(v2 + v2 v1 + v2 v1 + v2 v1 + v1 ) T + v2 v1 T
----------------------------------------------------------------
4 4 2 2 6 3 3 6
(v2 + v1 ) - (v2 - v2 v1 + v1 ) (v2 - v1 v2 + v1 ) T
2 2 4 4 2 7 7 4
- v2 v1 (v2 + v1 ) T + v2 v1 T
where the dependence of T on the point (v1,v2) on the unit circle has
been discussed above. Here the denominator is only zero when v1=v2 or
v1 v2 = 0 (as can be seen by eliminating T between the denominator of
Q and the equation c2 which defines T). In the former case, the
numerator is also zero, and in fact the expression has a limit of 1; the
curve in 3 dimensions is locally isomorphic to its projection in R^2 here.
On the other hand, as either v_i -> 0 we get Q -> +- oo (e.g. the
near H=0, Q is (1/H)(1 - H^2/3 + ...)
We can similarly show Q is never zero. In particular, Q never
changes sign along the curve except at those four asymptotes. A quick
test of a few values shows that Q is negative on the unbounded portion
of the curve, and on the bounded portion when u1 u2 < 0.
In the original application, Q=q^2 is positive, we see (u1, u2) will have
to lie in the first or third quadrants and on the bounded portion of the
curve. However, in those regions, we find the minimum value of Q to
be 1, so since -1 < q was originally assumed, we conclude
* No points of interest in the original problem lie on the curve
* discussed so far; all must instead project to the "trivial"
* curves u1 = +- u2 or u1 = +- 1/u2.
So now we return to the original problem and look for solutions with
these extra conditions.
If u1=1/u2, the only solutions are those with Q=1 (in which
case u2 is arbitrary).
If u1=-1/u2, the only solutions are those with Q=-1, in which case again
u2 is arbitrary.
If u1=-u2, then Q is now variable, but u2 must be determined from
2 8 10
(- 1 + Q) + Q (- 1 + Q) u2 + Q (1 + Q) u2 + (1 + Q) u2 = 0
If u1=u2, then either Q=1 (u2 arbitrary), u2=0, 1, or -1 (Q arbitrary), or
8 6 6 4 4 2 2
u2 + u2 - u2 Q + u2 - u2 Q + u2 - u2 Q + 1 = 0
Perhaps Mathematica was misbehaving because the algebraic variety in
question has several distinct components, and it was unable to figure out
how to pick the right one on which to perform its business.
dave