Extreme Points of $f(x,y) = \sin(x)+\sin(y)+\sin(x+y)$.

Question

$f: (0,\frac{\pi}{2}) \times (0,\frac{\pi}{2}) \mapsto \mathbb{R}$

$\qquad (x,y) \mapsto \sin(x)+\sin(y)+\sin(x+y)$

Looking for the extreme points of the mentioned function. First compute the critical points:

$\nabla f(x) = \begin{pmatrix} \cos(x)+\cos(x+y)\\ \cos(y) + \cos(x+y)\\ \end{pmatrix} \overset{!}{=} \begin{pmatrix} 0\\ 0\\ \end{pmatrix} $

Here is the point where my my problem arises:

The soloution of the equation is $y = -2y$ or $x = \frac{-y}{2}$

I can conclude that this is not a critical point but a critical "line". Plotting the function and my line yields nonsense. How should I proceed?

Answer 1

Hint: you should start from $\cos(x) = -\cos (x+y), \; \cos(y) = -\cos(x+y) \Rightarrow \cos(x) = \cos(y)$. This gives you that $x = y + 2\pi k$ or $x = 2\pi k-y$. After that you have to solve two equations: $\cos(x) + \cos(2x) = 0$ and $\cos(x) + 1 = 0$. Combining solutions of these equations with corresponding first step gives you critical points of your function.

Answer 2

This is quite an interesting function, since it factors to a smooth functions on the torus ($\mathbb{R}^2/2\pi \mathbb{Z}^2$) with 3 critical points, which proves that the lower bound of the critical points of a Morse function, given by the Lusternik-Schnirelmann category is actually good (since it also evaluates to 3 on the 2-torus).

Pointing this out, you see that there should be in fact precisely 3 solutions for a vanishing derivative!

Short answer is, that you computation of the solution is unfortunately wrong and does not seem to use the properties of sin,cos.

In the following I will try to give you some advice to solve your problem.

You have to ask yourself three things:

• Which real solutions does cosx=-cos(x+y), cos(y)=-cos(x+y) have? (hint: look at the cos and figure out where this is the case. Namely for a certain translation by some multiples of $\pi$), also don't forget to add or subtract them and see what happens.
• Which of those solutions lie actually in $(0,\pi/2)^2$, since only those are interesting (why?)
• how does the function behave near the "boundary", are some local extrema from the above maybe not global? How does the function behave near the critical points?

Again - I think it is worth thinking about it, so try to solve it yourself!

Hope it helps, Good luck!

If you want to verify your solutions look at:

Wolfram Alpha real solutions

(also look at the plot in wolfram alpha by entering "sinx+siny+sin(x+y)" to see what happens in this situation. Note that this plot will be defined in $\mathbb{R}^2$. I have not enough reputation to post a 3rd link)

Answer 3

You may want to have a look at what your friend Wolfram Alpha tells you about this set of equations. Agreed, using it for computing solutions to homework is counterproductive at least, but I believe it is useful for intermediate steps.

On the plotted graph, the two colors represent the two solution sets. Indeed they are lines, and countably infinitely many lines at that, but they are not parallel and they meet at points. As in extreme points. When restricted to the domain of your function, you find the extremum poolpt already found.

I suspect you can arrive at the same solution by carefully solving the equations and not assuming $0 \leq x \leq \frac{\pi}{2}$ or $0 \leq y \leq \frac{\pi}{2}$.