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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)

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  • $\begingroup$ By third or fourth order I mean that there are many similar problems where usually the powers go up to that order $\endgroup$ – Euler_Salter Dec 29 '18 at 14:08
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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 $$f(x_1,x_2)\geq \frac{1}{16}$$

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  • $\begingroup$ yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake $\endgroup$ – Euler_Salter Dec 29 '18 at 14:13
  • $\begingroup$ Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues $\endgroup$ – Euler_Salter Dec 29 '18 at 14:14
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    $\begingroup$ Ok, nice that i could help you! $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '18 at 14:14

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