Solve an ODE on a torus
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I have an ODE that I need to solve on a torus. Namely, I have many "point charges" that I need to put on the square  , and I need to study a motion under the action of said points. The problem is, I'd really like to set it up as if the square
, and I need to study a motion under the action of said points. The problem is, I'd really like to set it up as if the square  were a torus, so that when I exit from one side I pop up from the opposite one. However, I do not know how to impose this. Notice, also, that I am using the brand new tool "Solve" from the latest release since it seems to speed things up quite a bit (basically because MATLAB knows better than me what method to use...).
 were a torus, so that when I exit from one side I pop up from the opposite one. However, I do not know how to impose this. Notice, also, that I am using the brand new tool "Solve" from the latest release since it seems to speed things up quite a bit (basically because MATLAB knows better than me what method to use...).  
 , and I need to study a motion under the action of said points. The problem is, I'd really like to set it up as if the square
, and I need to study a motion under the action of said points. The problem is, I'd really like to set it up as if the square  were a torus, so that when I exit from one side I pop up from the opposite one. However, I do not know how to impose this. Notice, also, that I am using the brand new tool "Solve" from the latest release since it seems to speed things up quite a bit (basically because MATLAB knows better than me what method to use...).
 were a torus, so that when I exit from one side I pop up from the opposite one. However, I do not know how to impose this. Notice, also, that I am using the brand new tool "Solve" from the latest release since it seems to speed things up quite a bit (basically because MATLAB knows better than me what method to use...).  0 件のコメント
回答 (1 件)
  John D'Errico
      
      
 2024 年 6 月 5 日
        
      編集済み: John D'Errico
      
      
 2024 年 6 月 5 日
  
      Simple enough. Just transform the problem. So if [u,v] live on [0,1]x[0,1], then
    s = (cos(2*pi*u)+1)/2
    t = (cos(2*pi*v)+1)/2
also lives on [0,1]x[0,1], but (s,t) now behave as you wish.
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