Second derivative test $\sin x + \sin y + \sin(x-y)$

Question

Consider the function $g(x,y) = \sin(x) + \sin(y) + \sin(x-y) $

Using the second derivative test, determine all(whenever applicable) of the local maximum, minimum and saddle point of $g$ in the region $0\leq x \leq2\pi$ and $0\leq y \leq2\pi$.

Really appreciate it if someone can help me with this question.

Answer 1

First of all let's write

$$\sin(x-y) = \sin(x)\cos(y) - \cos(x)\sin(y)$$

In order to have a better writing for the function:

$$g(x, y) = \sin(x)(1 + \cos(y)) + \sin(y)(1 - \cos(x))$$

Now this is a function in two variable so we have to start with the gradient

$$ \begin{cases} \nabla_x g(x, y) = 0\\\\ \nabla_y g(x,y) = 0 \end{cases} $$

Hence (derivative are easy)

$$ \begin{cases} \cos(x) + \cos(x-y) = 0 \\\\ \cos(y) - \cos(x-y) = 0 \end{cases} $$

That is we find those solutions:

$$x = 2k\pi ~~~~~~~~~~~ y = -2\pi(k+1)$$

$$x = -\frac{2\pi}{3} + 2k \pi ~~~~~~~~~~~ y = -\frac{\pi}{3} - 2k\pi$$

$$x = \frac{2\pi}{3} + 2k\pi ~~~~~~~~~~~ y = \frac{\pi}{3} - 2k\pi$$

Where $k\in\mathbb{Z}$

Since we have to restrict to $[0, 2\pi]$ we only take into account solutions for $k = 0, 1$ hence we only remain with one point

$$\boxed{z_1 = (x_1, y_1) = \left(\frac{2\pi}{3}, \frac{\pi}{3}\right)}$$

Now it's about making second derivatives and mixed one:

$$\nabla^2_x g(x, y) = -\sin(x) - \sin(x-y)$$

$$\nabla^2_y g(x. y) = \sin(y) - \sin(x-y)$$

$$\nabla_x\nabla_y g(x, y) = \nabla_y \nabla_y g(x, y) = \sin(x-y)$$

And now the Hessian Matrix

$$\mathcal{H} = \begin{pmatrix} \nabla^2_x g(x, y) & \nabla_x\nabla_y g(x, y)\\\\ \nabla_y \nabla_y g(x, y)& \nabla^2_y g(x, y) \end{pmatrix} ~~~~~ = \begin{pmatrix} -\sin(x) - \sin(x-y) & \sin(x-y) \\\\ \sin(x-y) & \sin(y) - \sin(x-y) \end{pmatrix} $$

Now we evaluate the matrix in the point we found, simply substituting it

$$ \mathcal{H}(x_1, y_1) = \begin{pmatrix} \left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) & - \frac{\sqrt{3}}{2}\\\\ - \frac{\sqrt{3}}{2} & 0 \end{pmatrix} $$

The determinant is

$$\text{Det}\ \mathcal{H} = -\frac{3}{4}$$

Since it's negative, the point $z_1$ we found is a Saddle Point

Answer 2

COMMENT.-For saddle points you have to calculate the Hessian matrix $$\begin{bmatrix} \frac{\partial^2f}{\partial x^2}& \frac{\partial^2f}{\partial x\partial y}\\\frac{\partial^2f}{\partial y\partial x}&\frac{\partial^2f}{\partial y^2}\end{bmatrix}=\begin{bmatrix} -\sin x-\sin(x-y)& \sin(x-y)\\\sin(x-y)&-\sin y-\sin(x-y)\end{bmatrix}$$ of $f(x,y)=\sin x+\sin y+\sin(x-y)$

I add the graph of the function for $(x,y)\in [0,2\pi]^2$ so you can guide in what you do.