4
$\begingroup$

Let ${\boldsymbol \theta}=(\theta_1,\theta_2,\ldots,\theta_n) \in{\mathbb T}^n$ and $P:{\mathbb T}^n\rightarrow {\mathbb R}$ be a function defined on $n$-torus as $$ P({\boldsymbol \theta}) = \sum_{i<j}(1+\cos(\theta_i-\theta_j))^2. $$ What are local maximum points of $P$?

One can simply show that the global maximum is ${\boldsymbol \theta} = (\theta,\theta, \ldots,\theta)$ for all $\theta \in {\mathbb S}^1$, but the question is regarding the local maximum points of it.

Improve this question
$\endgroup$
6
  • $\begingroup$ certainly, $x=(a,\dots,a)$ is a global maximum for any $a$, not only for $a=0$. $\endgroup$
    – Dima Pasechnik
    Commented Aug 18, 2013 at 6:27
  • $\begingroup$ @DimaPasechnik, Thanks! I have edited it. $\endgroup$
    – Mohammad Khosravi
    Commented Aug 18, 2013 at 6:59
  • 1
    $\begingroup$ @TheoJohnson-Freyd This is a potential function originated from a research problem in theory of differential equation for a biological system! $\endgroup$
    – Mohammad Khosravi
    Commented Aug 18, 2013 at 18:22
  • 2
    $\begingroup$ @TheoJohnson-Freyd Is it really that easy to explicitly locate all the local maxima? There are vaguely similar innocuous functions on higher tori where locating the minima could solve the notorious circulant Hadamard conjecture. I vote to re-open $\endgroup$
    – Yemon Choi
    Commented Aug 18, 2013 at 19:38
  • 2
    $\begingroup$ @YemonChoi: That's quite possible — I didn't try to answer the question myself. I keep firmly to a rule that I prefer questions that have motivation and background and so on, and I vote to "put on hold" questions that lack them. If OP includes some background and discussion into the question — what he's tried, what kind of answer he's looking for — then I'll join you in voting to reopen. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 19, 2013 at 2:07

1 Answer 1

Reset to default
1
$\begingroup$

$$ \frac{\partial P}{\partial x_k} = 2 \sum_i \sin(x_i-x_k)\left(1 + \cos(x_i-x_k)\right ) $$ We know, local maximums satisfy the equality : $\nabla P({\rm x}) = {\rm 0}$

Some obvious solutions are $x_i - x_j = k\pi, k \in \mathbb Z $

Improve this answer
$\endgroup$
9
  • $\begingroup$ How can you solve this system of nonlinear equations? besides you need $D^2f({\rm x}) \ge 0$. I have tried these all! $\endgroup$
    – Mohammad Khosravi
    Commented Aug 18, 2013 at 9:08
  • $\begingroup$ @MohammadKhosravi Are numerical solutions suitable for you? $\endgroup$
    – Mahdi Khosravi
    Commented Aug 18, 2013 at 9:21
  • $\begingroup$ @MohammadKhosravi For the case when $n=3$ and if we set $x_1 = 1$ plotting the function using wolframalpha gives this. I think it could give some ideas to guess solutions. $\endgroup$
    – Mahdi Khosravi
    Commented Aug 18, 2013 at 9:26
  • $\begingroup$ I have guessed the solutions numerically! but I can't prove it! The solution is ${\rm x}_i -{\rm x}_j = 0, \mod \pi$. $\endgroup$
    – Mohammad Khosravi
    Commented Aug 18, 2013 at 9:42
  • 1
    $\begingroup$ @MohammadKhosravi Those solutions are obvious solutions of the gradient equation (since $\sin(x_i - x_j) = 0$). If you are sure that there aren't any other local maximum, so you've done! and what are you looking for? $\endgroup$
    – Mahdi Khosravi
    Commented Aug 18, 2013 at 10:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .