# Classifying stationary points in 3 variable case

I have a following problem and I am not sure if I understand correctly how to classify stationary points.

The function is given by:

$$$$f(a, b, c) = a^2b + b^2c + c^2a,$$$$

hence the first order conditions are:

$$$$\label{first} \frac{\partial f}{\partial x} = 2ab + c^2 = 0$$$$

$$$$\label{second} \frac{\partial f}{\partial y} = a^2 +2bc = 0$$$$

$$$$\label{third} \frac{\partial f}{\partial z} = b^2 +2ac = 0$$$$

With one unique solution $$a=b=c=0$$.

Now so far I think I understand things, but I have now problem with classifying the stationary point. In a 2 variable case I would simply calculate second order derivatives and then the determinant of hessian at a stationary point.

\begin{aligned} H(a,b,c) & = \begin{bmatrix} 2b & 2a & 2c\\ 2a &2c& 2b\\ 2c&2b&2a \end{bmatrix} \end{aligned}

now I dont know if this is correct but just kinda trying to extend the two variable case I would calculate the following at the stationary point:

$$$$2b2c2a - |H|$$$$

where $$|H|$$ is the determinant of hessian.

At a stationary point I would have

$$$$0 - 0\geq 0$$$$

So this should not be a saddle point since the above equation is not negative, but also since the second order derivatives are exactly zero at the point it could be both convex or concave - I am completely lost at this point...

• You need to study whether the Hessian is positive definite, negative definite, etc. – Xiangxiang Xu Sep 30 '18 at 16:17
• @XiangxiangXu yes well I know that the hessian is zero at the point but I don’t know what that means exactly – 1muflon1 Sep 30 '18 at 16:18
• How about considering a similar case where $f(a, b) = a^2b + b^2a$? – Xiangxiang Xu Sep 30 '18 at 16:24
• @XiangxiangXu but I need it for this case. I know already how to do it for case of only two variables. This is completely unhelpful. – 1muflon1 Sep 30 '18 at 16:25
• They are similar. Put it aside, you may need to consider the behavior of $f(x, x, x)$ as $x \to 0$. – Xiangxiang Xu Sep 30 '18 at 16:28

The function $$f(a,b,c)$$ does not have relative minimum nor relative maximum nor saddle point at $$(0,0,0)$$ The eigenvalues for $$H_f(0,0,0) = 0_{3\times 3}$$ are all null.
Attached a plot showing the maniford $$f(a,b,c)=0$$
By definition, $$\boldsymbol{x}_0=(0,0,0)$$ is a saddle point, since it's neither a maximum nor a minimum. Indeed, in any neighborhood $$B(\boldsymbol{x}_0)$$ of $$\boldsymbol{x}_0$$, you can find $$\delta>0$$ such that $$\boldsymbol{x}_1=(\delta,\delta,\delta)\in B(\boldsymbol{x}_0)$$, $$\boldsymbol{x}_2=(-\delta,-\delta,-\delta)\in B(\boldsymbol{x}_0)$$ with $$f(\boldsymbol{x}_1)>f(\boldsymbol{x}_0)$$ and $$f(\boldsymbol{x}_2).
The same analysis goes for the stationary point $$(0, 0)$$ of function $$f(a, b) = a^2b + b^2a$$, and the stationary point $$x = 0$$ of function $$f(x) = x^3$$. In all these three cases you have Hessian matrix being equal to zero.
In short, the Hessian-based decision rule is only a sufficient condition. When the Hessian equals zero, you need to decide the type of stationary points case by case, where higher-order terms should be taken into account. A simple example is the stationary points $$x = 0$$ of functions $$f_1(x) = x^3$$, $$f_2(x) = x^4$$, and $$f_3(x) = -x^4$$, which are a saddle point, a local minimum (also global minimum), and a local maximum (also global maximum), respectively.