2
$\begingroup$

I have

$$f(x_1,x_2) = 2x^4_1 + 2x_1x_2 + 2x_1 + (1+x_2)^2$$

How can I determine the nature of the stationary points?

I know;

$$f_{x_1,x_1}(x) = 24x_1^2$$ $$f_{x_2,x_2}(x) = 2$$ $$f_{x_1,x_2}(x) = 2$$

This gives me a hessian:

$$ \nabla^2 f(x) = \left[ \begin{array}{ c c } 24x_1^2 & 2 \\ 2 & 2 \end{array} \right] $$

By examining the leading principal minors; $$D_1 = \| (24x_1^2)\| \geq 0$$ and $$D_2 = \det\left|\array{24x_1^2&2\\2&2}\right| = 48x_1^2 -4$$

Since I have no condition on $x_1^2$ then $D_2$ can be positive, negative or even 0.

So then does this mean this matrix is indefinite, and hence all the stationary points here are saddle points?

Have I made a mistake in the Hessian? I feel the $24x_1^2$ term is out of place.

any help is VERY MUCH appreciated.

$\endgroup$
  • $\begingroup$ The stationary points satisfy $\nabla f(x_1,x_2) = 0$; that should tell you what $x_1$ is at each. $\endgroup$ – Matthew Leingang Jan 22 '15 at 2:34
  • $\begingroup$ I know what each of the stationary points are, but to determine the type of stationary point, I have to examine the hessian - is that not correct? The stationary points I worked out as $(0,-1), (\frac{1}{2}, \frac{-3}{2}), (\frac{-1}{2}, \frac{1}{2})$ It was my understanding to figure out the nature of the stationary points, one must examine the 2nd derivatives. $\endgroup$ – diabloescobar Jan 22 '15 at 2:50
  • $\begingroup$ Yes, the $x_1$ in $D_2$ is the first coordinate of the stationary point. So you will get a number at each point. $\endgroup$ – Matthew Leingang Jan 22 '15 at 2:54
1
$\begingroup$

We have \begin{align*} f_{x_1} (x_1,x_2) &= 2x_1^4 + 2x_2 +2 \\ f_{x_2} (x_1,x_2) &= 2x_1 + 2(x_2 + 1) \\ \end{align*} So if $(x_1,x_2)$ is a critical point we have \begin{align} 8x_1^3 + 2x_2 +2 &= 0 \\ 2x_1 + 2x_2 + 2 &= 0 \end{align} Subtracting the two gives $$ 8x_1^3 - 2x_1 = 0 = 4x_1(4x_1^2-1) $$ therefore $x_1 = 0, \pm \frac{1}{2}$. The critical points are $(0,-1)$, $\left(\frac{1}{2},-\frac{3}{2}\right)$, and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.

You have already computed that that $D_1 > 0$ if $x_1 \neq 0$, and in this case $D_2 = 48 \cdot \frac{1}{4} - 4 = 8$. So these two points are local minima. $D_2 < 0$ at $(0,0)$ so this critical point is a saddle point.

Here is a contour plot to see what's going on:

contour plot of $f$

$\endgroup$
  • $\begingroup$ I did all the work, and completely forgot to substitute in the points - I am not wise. Thanks matthew, you legend! $\endgroup$ – diabloescobar Jan 22 '15 at 3:06
  • $\begingroup$ @diabloescobar nope, just a professional calculus teacher. $\endgroup$ – Matthew Leingang Jan 22 '15 at 3:28

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.