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

The function is given by:

\begin{equation} f(a, b, c) = a^2b + b^2c + c^2a, \end{equation}

hence the first order conditions are:

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

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

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

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

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:

\begin{equation} 2b2c2a - |H| \end{equation}

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

At a stationary point I would have

\begin{equation} 0 - 0\geq 0 \end{equation}

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...

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

enter image description here

  • $\begingroup$ so it’s not possible to classify it? Also what would be the general rules for concave and convex function in case of more than 2 variables. It’s not part of my textbook and on internet I found only answers for special cases $\endgroup$ – 1muflon1 Sep 30 '18 at 16:54

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 $f(\boldsymbol{x}_2)<f(\boldsymbol{x}_0)$.

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.


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