Local Maxima/ Minima of $F(x,y)=(x^2+y^4-1)^2$

Question

Searching for Local Minima/ Maxima of the function $F(x,y)=(x^2+y^4-1)^2$ I found out that $(0,0)$ and $(\pm \sqrt{1-y^4},y)$ are possible candidates. Looking up in Wolfram-Alpha that seems to be right.

My problem is, that the Hessian-Matrix is just semi-definite in both cases (I assume this is due to the fact that these points aren't strict local Minimums).

Is there any way to show that this points are local Minima without using the Hessian-Matrix? In some way it seems to be obvious that this points have to be local Minima since the function is a parabola. How do we handle such cases?

PS: The Wolfram-Alpha plot: https://www.wolframalpha.com/input/?i=%28x%5E2+%2By%5E4+-1%29%5E2+

Answer 1

The partial derivatives are $$F_x(x,y)=4x(x^2+y^4-1)\\ F_y(x,y)=8y^3(x^2+y^4-1) $$ Setting them equal to $0$, we obtain that the critical points are the set $x^2+y^4-1=0$ and $(0,0)$. It immediately follows that all points in the set $x^2+y^4-1=0$ are local (and absolute) minima because $F(x,y)\geq 0$ for all $x,y\in\mathbb{R}$, and you have $F(x,y)=0$ for those points. So that gives you $(\pm\sqrt{1-y^4},y)$ for $-1\leq y\leq 1$.

As for $(0,0)$, you have that $F(0,0)=1$. The inequality $F(x,y)\leq 1$ is equivalent to $$(x^2+y^4-2)(x^2+y^4)\leq 0 $$ Notice that it's true for $|x|\leq 1, |y|\leq 1$, i.e. a region around $(0,0)$. This means that you have a local maximum at $(0,0)$.