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How to solve this non-linear system?

  1. Dec 2, 2007 #1
    Given that F(x,y)=xy(x+y-1). Determine whether S={(x,y) | F(x,y)=0} is a smooth curve.

    Now, F(x,y)=x2y + xy2 - xy
    Let gradient F = (2xy + y2 - y, x2 + 2xy - x) = 0
    So 2xy + y2 - y = 0 and x2 + 2xy - x = 0
    In order to determine where S is smooth, I have to solve these 2 equations for x and y, and I am stuck right here. How can I solve this non-linear system?

    Thanks if someone could help me!
  2. jcsd
  3. Dec 2, 2007 #2


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    Homework Helper

    Well, your two equations both have 2xy in them, so you might start by solving for that:

    2xy = -y^2 + y
    2xy = -x^2 + x

    So, y^2 - y = x^2 -x

    Writing y^2 - y -( x^2 - x) = 0 and completing the square, we get (y - 1/2)^2 -(x -1/2)^2 = 0.

    Hence, you get (y - 1/2)^2 = (x-1/2)^2; any pair of points (x,y) satisfying this equation will satisfy your system of equations. (Of course, double check everything to make sure I'm right, etc.)
  4. Dec 2, 2007 #3
    x=y=2 satisfies your last equation,
    but it doesn't satisfy 2xy + y2 - y = 0 and x2 + 2xy - x = 0

    How come?
  5. Dec 2, 2007 #4


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    Hm. I guess I hadn't fully thought this through. I suspect the trick is that if 2xy + y^2 - y = 0 and 2xy + x^2 - x = 0, then the final equations I wrote must be true. The converse isn't true. What the last equations tell you, I think, is that any solution of your first equation is also constrained by the last equations I wrote down. So, you should be able to write

    [tex]y = \frac{1}{2} + \left|x - \frac{1}{2} \right|[/tex]

    This gives two solutions for y in terms of x. Either y = x, in which case you have to solve

    [tex]2x^2 - x^2 + x = x^2 -x = 0[/tex] and you see you need x = 0 or 1,

    or y = 1 - x, in which case you have to solve

    [tex]2x(1-x) -x^2 + x = -3x^2 + 3x = 0[/tex], in which case again x = 0 or 1.

    As you can see, x = 0, 1 satisfy your original system.

    So, that might be the whole story, but again, check to make sure I didn't miss anything.
  6. Dec 2, 2007 #5
    It looks like [tex]S[/tex] is a union of three lines: [tex]x=0, y=0, x + y = 1[/tex]. That's definitely not smooth; you have sharp angles at the intersection points of the lines.

    The solution to the other system is [tex](0, 0), (1, 0), (0, 1), (1/3, 1/3)[/tex].

    I found all of that just by factoring the appropriate expressions and noting that at least one of the factors must be 0 in order for the product to be 0.
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