4
$\begingroup$

Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $(0,0)$ is a critical point. Observe, \begin{align*} f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\ f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\ f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\ f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\ f_{yy} & = 4, f_{yy}=4 \end{align*}

So, in order to determine the nature of the above critical point, we need to check the Hessian at $(0,0)$ which is $0$ and hence the test is inconclusive. $$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$So, I tried to see the function on slices like $y=0$ and $y=x$ but nothing worked. So please suggest me how do I find the nature of the critical point in this case?

$\endgroup$
  • $\begingroup$ @GeorgeDewhirst Still the Hessia at $(0,0)$ is $0$ and which is again not useful. Do you mean $f(x,x^2)$, because $f(x^2,y)$ for sure is not working? Even the first one is also not working. $\endgroup$ – Sachchidanand Prasad Jan 4 at 17:50
  • $\begingroup$ you have a typo: last equality should be $f_{yy} = 4$. $\endgroup$ – peek-a-boo Jan 4 at 17:53
  • $\begingroup$ @peek-a-boo Yeah thanks for that. $\endgroup$ – Sachchidanand Prasad Jan 4 at 17:56
  • $\begingroup$ @GeorgeDewhirst The Hessian is coming $0$ and therefore it can not be positive definite. Please correct me if I am wrong. $\endgroup$ – Sachchidanand Prasad Jan 4 at 17:57
  • $\begingroup$ @GeorgeDewhirst that is definitely not positve definite, since it isn't invertible. (more directly, if you call the matrix $H$ then $(e_1)^t H e_1 = 0$ which is not positive) (I think OP meant to say it has determinant zero... i'm not sure though) $\endgroup$ – peek-a-boo Jan 4 at 18:00
3
$\begingroup$

You have that $$ g_a(x)=f(x,ax^2 ) = 2\left( {a^2 - a} \right)x^4 + \left( {1 - a} \right)x^6 $$ With $0<a<1$ the function $g_a(x)$ has a local maximum. With $a>1$ the function has a local minimum. This means that $(0,0)$ is a saddle point.

$\endgroup$
1
$\begingroup$

You might note that your function factors as

$$(x^2-y)(x^4-2y).$$

So there are easy to find regions in the $xy$-plane where the function is positive an negative. Close to the origin and between the curves $y=x^2$ and $y=x^4/2$, the function is negative. This suggests trying the limit along the curve $y=x^3$. It's not too horrible to analyse

$$f(x,x^3) = 3x^6-2x^5 -x^7$$

around $x=0$ to see that it's negative there.

Comparing with the curve given by $y=0$, we get a saddle point at $(0,0).$

$\endgroup$
0
$\begingroup$

With $g(x) = x^2$ we have

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

or

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

and the hessian of $f(g,y)$ is

$$ H(g,y) = \left( \begin{array}{cc} 6 g-2 y & -2 g-2 \\ -2 g-2 & 4 \\ \end{array} \right) $$

and also

$$ H(0,0) = \left( \begin{array}{cc} 0 & -2 \\ -2 & 4 \\ \end{array} \right) $$

with eigenvalues

$$ \left\{2 \left(1+\sqrt{2}\right),2 \left(1-\sqrt{2}\right)\right\} $$

characterizing a saddle point.

$\endgroup$
0
$\begingroup$

You have $f(0,0)=0$. You can find both positive and negative values of $f(x,y)$ in any region around $(0,0)$, which means that you don't have a local extremum at this point: $$f(x,x^3)= -x^5(1-x)(2-x)<0\quad\text{for}\,\, |x|<1, x\ne 0$$ and $$f(x,0)=x^6>0\quad\text{for}\,\, x\ne 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.