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.

  • $\begingroup$ certainly, $x=(a,\dots,a)$ is a global maximum for any $a$, not only for $a=0$. $\endgroup$ Aug 18, 2013 at 6:27
  • $\begingroup$ @DimaPasechnik, Thanks! I have edited it. $\endgroup$ 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! $\endgroup$ 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
    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. $\endgroup$ Aug 19, 2013 at 2:07

1 Answer 1


$$ \frac{\partial P}{\partial x_k} = 2 \sum_i \sin(x_i-x_k)\left(1 + \cos(x_i-x_k)\right ) $$ 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 $

  • $\begingroup$ How can you solve this system of nonlinear equations? besides you need $D^2f({\rm x}) \ge 0$. I have tried these all! $\endgroup$ Aug 18, 2013 at 9:08
  • $\begingroup$ @MohammadKhosravi Are numerical solutions suitable for you? $\endgroup$ 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. $\endgroup$ Aug 18, 2013 at 9:26
  • $\begingroup$ I have guessed the solutions numerically! but I can't prove it! The solution is ${\rm x}_i -{\rm x}_j = 0, \mod \pi$. $\endgroup$ 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? $\endgroup$ Aug 18, 2013 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.