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Subject: Re: Non linear inequalities with 2 variables
Date: Sat, 24 Nov 2012 22:32:07 +0000 (UTC)
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"Narasimha Aditya " <ngollapu@asu.edu> wrote in message <k8rbq8$1nd$1@newscl01ah.mathworks.com>...
> Is there a way to get the possible range for variables with help of a set of non linear inequalities? As in
> 
> a1*x^2+b1*y^2+c1*x+d1*y+e1<0
> a2*x^2+b2*y^2+c2*x+d2*y+e2<0
> a3*x^2+b3*y^2+c3*x+d3*y+e3<0
> 0<=x<=1
> 0<=y<=1
- - - - - - - - - -
  The answer I give here will probably not please you.  If your first three inequalities are changed to equalities, they would each represent a conic section, either an ellipse, a parabola, a hyperbola, or degenerate versions of these depending on the values of their coefficients and all aligned with respect to the x-y axes.  As inequalities they represent open areas bounded by such conics.  To have all five inequalities hold true you are dealing with the intersection of a unit square and areas bounded by three different conics.  It can be a very complicated point set and it is not easy to obtain a precise limit for x and y coordinates within it.  To do so you would have to look for all possible intersections among pairs from among the five corresponding equalities as well as certain extreme values for x and y such as the ends of ellipses, vertices of parabolas, etc.  There seems no easy 
way to obtain an answer to all of these by way of straightforward symbolic manipulation.  Each different combination of conics would require a somewhat different approach. 

  A possible crude approach is to fill the square with a very tightly-packed grid of points and to determine the range intervals approximately defined by the pairs (x,y) which satisfy all five inequalities.  Are you interested in the code for such a operation?

Roger Stafford