[Phys-L] Velocity-Verlet with dissipation

[Bernard Cleyet <bernard@cleyet.org>]

As written much on phys-l, the symplectic integrators assume no dissipation.  I’ve found this quite true for a SHO w/ linear dissipation (the term 0.01*thetadot).  However, I’ve  discovered an article that describes a modified V-V resulting is orders of magnitude better fit with the analytic solution.  It’s, as bc translated from JD’s original, here:    

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// acceleration at beginning of interval:
      
      double A = - ((4*pi*pi)/(period*period))*(1+deltag_now) * theta  - r*thetadot +da;
      
      // note the absence of the dissipation term for the n-1 force
      
      double Athetaprevious =  - ((4*pi*pi)/(period*period))*(1+deltag_now)  * theta +da;
      
      theta = theta + deltat * thetadot + A * deltat * deltat / 2 ;
      
      //acceleration (force) recalculated without dissipation
      
      
      double Atheta =  - ((4*pi*pi)/(period*period))*(1+deltag_now)  * theta +da;
      
      //note dissipation constant in new speed calculation
      
      
      thetadot = (1 / (1+ deltat*r/2))* (thetadot*(1-deltat*r/2))  +  (Atheta + Athetaprevious) * deltat/2;

      
    t = t + deltat;            // keep track of elapsed time



The “gory” detail is here:  

‎physics.bu.edu/py502/lectures3/cmotion.pdf  

http://physics.bu.edu/py502/lectures3/cmotion.pdf  

And a “slide” projection version (It’s slightly different.):  

‎physics.bu.edu/py502/slides/l06.pdf  

http://physics.bu.edu/py502/slides/l06.pdf  

About the author:  

Anders Sandvik  

http://physics.bu.edu/~sandvik/  

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The majority of my pendulum simulations use < r*thetadot*abs(thetadot) > as the dissipation term. So when I rewrote with the Sandvik solution I forgot to change it.  Curiously, the result using 1e6 steps for a one second period is very little different from the linear version.  (WRT difference between numerical resut and linear analytic)

bc several times visited a Sandvik engineer in Sandviken.  (Sweden not Finland)