For which values of $a$ there is no relative extremum?

Question

In a previous question, I explored the conditions under which the function $ f(x, y) = (y - x^2)(y - ax^3) $ does not have a relative extremum at the origin. I've found a similar but distinct problem formulation to be equally interesting. Here's the modified version:

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a function defined by $f(x,y) = (y - x^2)(y - ax^2)$ at each $ (x,y) \in \mathbb{R}^2$, where $a \in \mathbb{R}$. What are the values of $ a $ for which $ f$ does not have a relative extremum at $(0,0)$? Here is my solution attempt: To find the values of $ a$ for which $ f $ does not have a relative extremum at $ (0, 0) $, it is necessary to analyze the critical points of $ f $ at $ (0, 0) $.

$f(x, y) = \frac{y}{y^2} - axy^2 + 2xy + ax^4$

$\frac{\partial f}{\partial x} = -2(a+1)xy + 4ax^3$

$\frac{\partial f}{\partial y} = 2y - 2(a+1)x^2$

$\frac{\partial^2 f}{\partial x^2} = -2a + 12ax^2$

$\frac{\partial^2 f}{\partial y^2} = 2$

$\frac{\partial^2 f}{\partial x \partial y} = -2a + 2x$

$H = \begin{bmatrix} -2a + 12ax^2 & -2a + 2x \\ -2a + 2x & 2 \end{bmatrix}$

$H_1(0,0) = 0$, first principle minor.

$H_2(0,0) = 0$, second principle minor.

I have outlined my analysis and calculations above to determine the values of $a$ for which the function $f$ does not have a relative extremum at $(0, 0)$. But I could not reach any conclusions. I would greatly appreciate it if members of the community could review my approach and calculations. Are there any errors or aspects I might have overlooked? Thank you in advance for your insights and assistance.

Answer 1

If $a = 1$, then $f = (y - x^2)^2$ attains a global minimum at $(0, 0)$.

If $a = 0$, then $f = (y - x^2)y$, and $f(0, y) = y^2 > 0$ for any $y \ne 0$, and $f(x, x^2/4) = -\frac{3}{16}x^4 < 0$ for any $x \ne 0$. Thus, $f$ does not have a relative extremum at $(0, 0)$.

If $a = -1$, then $f = y^2 - x^4$, and $f(0, y) = y^2 > 0$ for any $y \ne 0$, and $f(x, 0) = -x^4 < 0$ for any $x \ne 0$. Thus, $f$ does not have a relative extremum at $(0, 0)$.

If $a \ne 0, 1, -1$, we have $$f\left(\frac{r(a+1)}{2a}, \frac{r^2(a+1)}{2a}\right) = -\frac{r^4(a^2 - 1)^2}{16a^3} < 0$$ for any $r > 0$, and $f(0, y) = y^2 > 0$ for any $y \ne 0$. Thus, $f$ does not have a relative extremum at $(0, 0)$.

Remark. $f$ has a local minimum at $(0, 0)$ if there exists $\delta > 0$ such that $f(0, 0) \le f(x,y)$ for all $x, y$ satisfying $x^2 + y^2 < \delta$.

Answer 2

Let $f(x, y) = (y - x^2)(y - ax^2)$

Then $f_x=4ax^3-2(a+1)xy$ and $f_y=2y-(a+1)x^2$.

So, \begin{align} f_{xx}&=12ax^2-2(a+1)y\\ f_{xy}&=-2(a+1)x\\ f_{yx}&=-2(a+1)x\\ f_{yy}&=\phantom{-}2 \end{align}

Then $D=-4[(a^2-4a+1)x^2+(a+1)y]$.

So $D(0,0)=0$ meaning that the second derivative test is inconclusive at $(0,0)$. So we have to approach the problem some other way.

Notice that $z=f(x,0)=y^2$ is concave up at $(0,0)$ with $f_yz(0,0)=0$.

Notice that $z=f(0,x)=ax^4$ is concave up at $(0,0)$ when $a>0$ and concave down at $(0,0)$ when $a<0)$ with $f_xz(0,0)=0$.

This indicates that $z=f(x,y)$ has a relative minimum at $(0,0)$ when $a>0$ and a saddle point at $(0,0)$ when $a<0$. There is no relative extremum at $(0,0)$ when $a=0$ since the surface $z=y^2$ has no relative extremum since $z=0$ along the entire $x$-axis and $z>0$ otherwise.

ADDENDUM This is the answer I should have given:

So far, this only shows that there is no relative extremum when $a\le0$. However what about when $a>0$. The second derivative test is inapplicable there as well.

Consider the parabola $y=\frac{a+1}{2}x^2$ in the $xy$ plane. If we evaluate $f(x,y)$ along that parabola we get

$f(x,y)=-(a-1)^2x^4$ which is negative for all $x$ except $x=0$ unless $a=1$. So if $a\ne1$, any $\epsilon$ neighborhood of $(0,0)$ contains a point with negative $f$ value.

Furthermore, along the line $x=0$ through the origin, $f(0,y)=y^2$. Thus every $\epsilon$ neighborhood of the origin also contains a point with a positive $f$ value.

Thus the function $f(x,y)=(y-x^2)(y-ax^2)$ contains no extremum at the origin except for $a=1$, in which case it has a relative minimum.