# Finding all stationary points quickly

I need to find all the stationary points of the following function defined on $$\mathbb{R}^2$$: $$f(x_1, x_2) = (x_2 - x_1^2)(x_1 - x_2^2)$$ I would like to know if in general there is a quick way of solving the resulting non linear systems in $$\mathbb{R}^2$$ where powers go up to $$3$$rd or $$4$$th order.

How I Solved it

Obviously the first step is to find the gradient and set it to zero to find the system of equations. $$\nabla f(x_1, x_2) = \left(x_2-3x_1^2+2x_1x_2^2\,\, , \,\, x_1-3x_2^2+2x_1^2x_2\right)^T =0$$ Then since it looks symmetric in $$x_1$$ and $$x_2$$ I guessed a solution for $$x_1=x_2$$, thus: $$\begin{cases} x_1-3x_1^2+2x_1x_1^2 = 0 \\ x_1-3x_1^2+2x_1^2x_1 = 0 \end{cases} \Longrightarrow x_1(2x_1-3x_1 +1)=0 \Longrightarrow x_1(2x_1-1)(x_1-1)=0$$ From which we obtain $$A=(0,0)^T$$, $$B=\left(\frac{1}{2},\frac{1}{2}\right)^T$$, $$C=(1, 1)^T$$. Calculating the Hessian we see that two of the are saddle points and one is a maximum. However now I don't know if there are any more saddle points, and in general when using symmetry arguments I don't know how to prove that these are all the possible solutions. I know there is a different method to solve this almost as quickly as this but that guarantees to find all solutions (there has to cause the exercise uses a non-symmetric method)

• By third or fourth order I mean that there are many similar problems where usually the powers go up to that order – Euler_Salter Dec 29 '18 at 14:08

Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(\frac{1}{2};\frac{1}{2})$$ is a maximum with $$f(x_1,x_2)\geq \frac{1}{16}$$
• yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake – Euler_Salter Dec 29 '18 at 14:13