Evaluating an extremal value if the hessian matrix has at least one eigenvalue which is zero

Question

$$f(x,y) = 2x^4-3x^2y + y^2$$

We want to find all extremal values:

$$df(x,y)=(8x^3-xy,-3x^2+2y)\overset{!}{=}0 \quad \Rightarrow \quad p=(0,0)$$

$$H_f(x,y)=\begin{pmatrix}24x^2-6y& -6x\\-6x & 2 \end{pmatrix}$$

So we get:

$$H_f(0,0)=\begin{pmatrix}0&0\\0&2\end{pmatrix}$$

This is a positive-semi-definit matrix with determinant 0. So apparently I can't make any statement using e.g. sylvester. Why exactly can't I just say: Since the matrix isn't positive definit nor negative definit, it has to be indefinit, so $p$ is a saddle point?

Now, let's just accept the fact that we can't make an statement about it. The only thing I can think of is checking the neighborhood, but I don't know how to do that with a multi-variabled function.

Answer 1

Note that $f(x,y)=2\left(x^2-\frac{3y}4\right)^2-\frac{y^2}8$. Therefore:

• If $y=0$ and $x$ is close to but different from $0$, $f(x,y)>0$.
• If $x$ is close to but different from $0$ and $y=\frac{4x^2}3$, $f(x,y)<0$.

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