Find stationary points of the function $f(x,y) = (y^2-x^4)(x^2+y^2-20)$

Question

I have problem in finding some of the stationary points of the function above. I proceeded in this way: the gradient of the function is:

$$ \nabla f = \left( xy^2-3x^5-2x^3y^2+40x^3 ; x^2y+2y^3-x^4y-20y \right) $$

So in order to find the stationary points, I must resolve the system:

$$ \begin{cases} xy^2-3x^5-2x^3y^2+40x^3 = 0 \\ x^2y+2y^3-x^4y-20y = 0 \end{cases} $$

So far I've found the points:

$$ (0,0) \qquad \left(\pm 2 \sqrt{10 \over 3} , 0 \right) \qquad (0, \pm \sqrt{10}) $$

But, I'm still blocked when I've to found the points deriving by the system:

$$ \begin{cases} 2x^6 + 3x^4 +x^2 -20 = 0 \\ y^2 = \frac{1}{2} \left( x^4 - x^2 + 20 \right) \end{cases} $$

Which I don't know how to solve. Can someone help me ? Thanks.

Answer 1

As you implied, you can factor the original system as $x(3x^4-40x^2+2x^2y^2-y^2)$ and $y(x^4-x^2-2y^2+20)=0$.

Taking $y=0$ gives you $x=0,x=\pm2\sqrt{\frac{10}{3}}$. Taking $x=0$ gives you $y=\pm\sqrt{10}$.

So you are left to solve $3x^4-40x^2+2x^2y^2-y^2=0,x^4-x^2-2y^2+20=0$ and $x,y\ne0$. Substituting from the second into the first gives $2x^6+3x^4-39x^2-20=0$ (you dropped the $-40x^2$ here), which factorises as: $$(x-2)(x+2)(x^2+5)(2x^2+1)$$

Can I leave you to finish?

Answer 2

WA gets $$ \DeclareMathOperator{grad}{grad} \grad((x^2+y^2-20) (y^2-x^4)) = (-6 x^5-4 x^3 (y^2-20)+2 x y^2, 2 y (-x^4+x^2+2 y^2-20)) $$ (link) and nine real critical points (link).