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I have a function

$$f(x,y) = x^4 + x^2 - 6xy + 3y^2$$

for all $x$ and $y$ in $R^2$

so

$f_x = 4x^3 + 2x - 6y$

$f_y = -6x + 6y$

$f_{xx} = 12x^2 + 2$

$f_{yy} = 6$

I'm looking for the extrema here and attempting to classify them. Solving the first derivatives set equal to 0, I get the points $(0,0,0); (1,1,-1); (-1,-1,-1)$

Plugging them into the second derivatives, I get positive values for all the points.

Wolfram Alpha tells me that only (1,1) and (-1,-1) are global mininma. Why is (0,0) not considered a local min?

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  • 1
    $\begingroup$ Probably because $f_x(0) \neq 0$? $\endgroup$ – baudolino Sep 5 '13 at 3:46
  • $\begingroup$ Plugging $x=0$ and $y=0$ into expression for $f_x$ shows that it is equal to $-6$, not $0$. So, that's the reason why $(0, 0)$ isn't an extremum. $\endgroup$ – Evgeny Sep 5 '13 at 3:47
  • $\begingroup$ It's just a typo. $f_x = 4x^3+2x-6y$, so $f_x(0,0) = 0$. $\endgroup$ – Kaster Sep 5 '13 at 3:48
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Let $f$ be a function of two variables that has continuous second partial derivatives on a rectangular region $Q$ and let:

$$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$$

for every $(x,y)$ in $Q$.

Let $f$ be a function of two variables that has continuous second partial derivatives on a rectangular region $Q$ and let:

$$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$$

for every $(x,y)$ in $Q$.

If $(a,b)$ is in $Q$ and $f_x(a,b) = 0, f_y(a,b) = 0$, then:

  • (i) $f(a, b)$ is a local maximum of $f$ if $g(a,b) \gt 0$ and $f_{xx}(a, b) \lt 0$.
  • (ii) $f(a, b)$ is a local minimum of $f$ if $g(a,b) \gt 0$ and $f_{xx}(a, b) \gt 0$.
  • (iii) $f(a,b)$ is not an extremum of $f$ if $g(a,b) \lt 0$

Given:

$$f(x,y) = x^4 + x^2 - 6xy + 3y^2$$

  • $f_x = 4x^3 + 2x - 6y$
  • $f_y = -6x + 6y$
  • $f_{xx} = 12x^2 + 2$
  • $f_{yy} = 6$
  • $f_{xy} = -6$
  • $f_{yx} = -6$
  • Critical Points $= (0,0), (1,1), (-1,-1)$

Now, lets classify them.

  • $(0,0) \rightarrow f_{xx}f_{yy}-(f_{xy})^2 = (2)(6) - (-6)^2 = -20 < 0$, thus, this is not an extremum, but is a saddle point.
  • For $(\pm 1, \pm 1) \rightarrow g(\pm 1, \pm 1) = 48 > 0 \rightarrow~~$ minimum.

We can also look at a plot of this:

enter image description here

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  • $\begingroup$ from which software have you graphed this. $\endgroup$ – Shobhit Sep 5 '13 at 4:38
  • $\begingroup$ @Shobhit: It is a feature that Google added. If you type "plot xy^2 - 3*xy$ in the Google search bar, it should get you to the tool. Regards $\endgroup$ – Amzoti Sep 5 '13 at 4:40
  • $\begingroup$ @Shobhit: You are very welcome. Regards $\endgroup$ – Amzoti Sep 5 '13 at 4:43
  • $\begingroup$ should the $f_{xx} < 0$ and $f_{xx} > 0$ bits be $f_{xy}$ or $f_{yx}$?? $\endgroup$ – Nick C Sep 5 '13 at 5:44
  • $\begingroup$ No, they are correct as shown. This is known as the second derivative test. For example, see: math.ucsd.edu/~wgarner/reference/math10c_su10/lectures/… Regards $\endgroup$ – Amzoti Sep 5 '13 at 5:46
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Analysis at $(0,0)$ only: $$H(x,y) = \left ( \begin{array}{cc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right ) = \left ( \begin{array}{cc} 2 & -6 \\ -6 & 6 \end{array} \right )$$ so, determinant of Hessian $$ \det(H) = 12 - 36 = -24 $$ is negative, and therefore $(0,0)$ is a saddle point.

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First of all, the question contains a slight error; we have

$f_x = 4x^3 + 2x - 6y, \tag{1}$

not

$f_x = 4x^3 + 2x - 6 \tag{2}$

as stated in the text of the question. All the other derivatives in the question look fine. So, setting $f_x, f_y = 0$, we see that

$4x^3 + 2x - 6y = 0, \tag{3}$

$-6x + 6y = 0. \tag{4}$

Thus, from (4), $x = y$ and so (1) becomes

$4x^3 + 2x - 6x = 4x^3 -4x = 0, \tag{5}$

or

$x(x^2 - 1) = x(x + 1)(x - 1) = 0, \tag{6}$

whence $x = -1, 0, 1$ at the critical points. Using $x = y$ we obtain the full $x$-$y$ coordinates of each such point: $(-1, -1), (0, 0), (1, 1)$. In addition,

$f_{xy} = f_{yx} = -6, \tag{7}$

which allows us to see immediately that the matrix of second derivatives, also sometimes called the Hessian, of $f$ at $(-1, -1)$ and $(1, 1)$ is given by

$H = \begin{bmatrix} 14 & -6 \\ -6 & 6 \end{bmatrix}. \tag{8}$

The eigenvalues of this $H$ are both positive, $2$ and $24$ to be precise, so these points are both local minima. At $(0, 0)$, we have

$H = \begin{bmatrix} 2 & -6 \\ -6 & 6; \end{bmatrix}. \tag{9}$

the characteristic equation is

$\lambda^2 -8 \lambda -24 = 0; \tag{10}$

the roots are $\lambda = 4 \pm 2\sqrt{10}$, one positive, one negative: $(0, 0)$ is a saddle.

It's worth noting that the saddle is necessary for topological reasons; since the flow of $-\nabla f$ has two attractors, it must have a saddle lest $\Bbb R^2$ (a connected set) become "pulled apart" by the flow of $-\nabla f$. The saddle allows this since it "absorbs" (that is, is a limit point of) the trajectories tangent to it's negative eigenvector; points flowing into the saddle never reach it or either of the other critical points; the topological integrity of the plane remains intact. NOTA BENE: speaking intuitively, of course!

ADDED IN EDIT: the surface $z = f(x, y)$ looks like a bowl with two "bottoms", the minima. If you imagine a thin sheet of oil or water flowing down this surface under the influence of gravity, you will see that almost all the fluid ends up going down one of the "drains", the attractors of $-\nabla f$. But there are two trajectories which approach $(0, 0)$ as $t \to \infty$; the flow "splits" along them into three "pieces", two heading into the minima, and the third (which consists of the two disjoint curves) heading into the saddle. If you think of what happens for very large $t$, this can be easily visualized. Sorry, no graphics SW at present!

I thank you, and Dr. Morse thanks you!

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