Classifying the stationary points of $f(x, y) = 4xy-x^4-y^4 $

Question

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

Answer 1

Note that:

• $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $x\in\left(-\sqrt 2,\sqrt2\right)\setminus\{0\}$;
• $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

Answer 2

Note that the gradient of $f$ is given by $\nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=\begin{pmatrix}-12x^2 &4\\4 &-12y^2\end{pmatrix}$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=\begin{pmatrix}0 &4\\4 &0\end{pmatrix}, H_f(1,1)=\begin{pmatrix}-12 &4\\4 &-12\end{pmatrix},H_f(-1,-1)=\begin{pmatrix}-12 &4\\4 &-12\end{pmatrix}$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.