Critical points of $f(x,y) = \sin x \sin y, -\pi<x<\pi, -\pi<y<\pi$

Question

$\frac{df}{dx} = \cos x\sin y = 0$

$\frac{df}{dy} = \cos y\sin x = 0$

$\cos x\sin y = \cos y\sin x$

$\frac{\cos x}{\sin x} = \frac{\cos y}{\sin y}$

$\cot x = \cot y$

$P_1 = (\pi/2,\pi/2)$

$P_2 = (-\pi/2,\pi/2)$

$P_3 = (\pi/2,-\pi/2)$

$P_4 = (-\pi/2,-\pi/2)$

$P5 = (0,0)$

This gives me that $P_1$ and $P_4$ are maximum, $P_2$ and $P_3$ are minimum and $P_5$ is saddle. But when testing with Wolfram Alpha I get something different:

http://www.wolframalpha.com/input/?i=local+maxima+senxseny+%2C+-pi%3Cx%3Cpi+%2C+-pi%3Cy%3Cpi

http://www.wolframalpha.com/input/?i=local+minima+senxseny+%2C+-pi%3Cx%3Cpi+%2C+-pi%3Cy%3Cpi

Did I do anything wrong?

EDIT: Can people please stop removing the second wolfram link? They aren't the same, one is for maximum point and the other is for minimum.

Answer 1

The only dubious thing I can see is the introduction of the equation $\cot x = \cot y$; this appears problematic since (i.) the cotangent functions have singularities in $(-\pi, \pi)$; and (ii.) I can see no way to infer the values of $x$ and $y$ given by $P_1$-$P_5$ from this equation; indeed, all it directly tells us is that $x = y$, $x = y + \pi$ or $x = y -\pi$ on $(-\pi, \pi) \setminus \{ 0 \}$. As far as issues with Wolfram Alpha are concerned, I can’t say much as I don’t know that much about it; but I did notice xsr’s input syntax appears different than Mary Star’s, so perhaps there is a clue to any mysteries.

These things being said, here’s my analysis of this problem:

As our O.P. xsr correctly showed,

$f_x =\dfrac{\partial f}{\partial x} = (\cos x)(\sin y), \tag{1}$

and similarly,

$f_y = (\sin x)(\cos y). \tag{2}$

Thus we seek those points

$(x, y) \in (-\pi, \pi) \times (-\pi, \pi) \tag{3}$

such that

$(\cos x)(\sin y) = 0, \tag{4}$

and

$(\sin x)(\cos y) = 0. \tag{5}$

Then from (4),

$\cos x = 0 \;\; \text{or} \;\; \sin y = 0, \tag{6}$

whilst from (5)

$\sin x = 0 \;\; \text{or} \;\; \cos y = 0. \tag{7}$

We recall that $\sin x$, $\cos x$ cannot simultaneously vanish, and likewise for $\sin y$, $\cos y$; thus if

$\cos x = 0, \tag{8}$

then $\sin x \ne 0$, so we must have

$\cos y = 0; \tag{9}$

it follows from (8) and (9) that

$x, y = \pm \dfrac{\pi}{2}. \tag{10}$

Next, if

$\cos x \ne 0, \tag{11}$

(4) yields

$\sin y = 0, \tag{12}$

hence

$\cos y \ne 0, \tag{13}$

hence, from (5),

$\sin x = 0. \tag{14}$

The only solutions to (12), (14) with $x, y \in (-\pi, \pi)$ are

$x = y = 0. \tag{15}$

We have thus covered all possible cases according to whether $\cos x = 0$ or not; we see the critical points are

$(x, y) = \left(\pm \dfrac{\pi}{2}, \pm \dfrac{\pi}{2}\right), (0, 0); \tag{16}$

we next proceed to classify them according to type. We do this by evaluating the Hessian of $f$, $H_f$:

$H_f(x, y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -(\sin x)(\sin y) & (\cos x)(\cos y) \\ (\cos x)(\cos y) & -(\sin x)(\sin y) \end{bmatrix}. \tag{17}$

From (17), we calculate that

$H_f\left(\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) = H_f\left(-\dfrac{\pi}{2}, -\dfrac{\pi}{2}\right) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}; \tag{18}$

$H_f\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) = H_f\left(\dfrac{\pi}{2}, -\dfrac{\pi}{2}\right) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \tag{19}$

and

$H_f(0, 0) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}. \tag{20}$

We immediately see from (18), (19) that the points $P_1$, $P_4$ are local maxima, whilst $P_2$, $P_3$ are local minima; these assertions follow from the manifest fact that the eigenvalues of $H_f(P_1)$ and $H_f(P_4)$ are $-1, -1$, and those of $H_f(P_2)$, $H_f(P_3)$ are $1, 1$; as for $P_5 = (0, 0)$͵ the characteristic polynomial of $H_f(P_0)$ is

$\det(H_f(0, 0) - \lambda I) = \det\left(\begin{bmatrix} -\lambda & 1 \\ 1 & -\lambda \end{bmatrix}\right)$ $= \lambda^2 - 1; \tag{21}$

the eigenvalues are thus $\lambda = \pm 1$, being the roots of $\lambda^2 - 1 = 0$; $P_5$ is indeed a saddle point.

Answer 2

It is correct!!

For the minimum

For the maximum

For the saddle points

Answer 3

Locating maximum and minimum values are straight forward as also replied by others.

About saddle points:

One half of your surface has positive Gauss curvature K, the other half has negative K.

The second half area has saddle points everywhere. Prominent central lines associated with these regions where normal curvatures vanish ... are asymptotic lines.

I made the above remark because max/min points are single points locally whereas saddle points are distributed in the negative K region.. for any surface.

There are no single unique stand alone saddle points for any surface.