Is the Set S Defined by F(x,y)=0 a Smooth Curve?

[kingwinner]

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!

 

[Mute]

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.)

 

[kingwinner]

x=y=2 satisfies your last equation,
but it doesn't satisfy 2xy + y² - y = 0 and x² + 2xy - x = 0

How come?

 

[Mute]

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

$$y = \frac{1}{2} + \left|x - \frac{1}{2} \right|$$

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

$$2x^2 - x^2 + x = x^2 -x = 0$$ and you see you need x = 0 or 1,

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

$$2x(1-x) -x^2 + x = -3x^2 + 3x = 0$$, 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.

 

[Xevarion]

It looks like $$S$$ is a union of three lines: $$x=0, y=0, x + y = 1$$. That's definitely not smooth; you have sharp angles at the intersection points of the lines.

The solution to the other system is $$(0, 0), (1, 0), (0, 1), (1/3, 1/3)$$.

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.