
 FlexPDE User's Forum: DTANGENTIAL BC in cylindrical coordinates 
  However, for cylindrical coordinates (r,z), and vector potential A (having azimuthal component only), the axial value of the curl is given by (1/r)*Dr(r*A), so that if A is specified as constant (= c), there is a nonzero axial component of (CURL(A))z (= c/r). 
  In (r,z) coordinates, the azimuthal component of the potential, A = c/r, on the boundary has zero tangential derivative, whereas in Cartesian geometry, A = c on the boundary is the correct value that gives DTANGENTIAL(A) = 0. 
  To reiterate, I suspect that the implementation of DTANGENTIAL is incorrect for cylindrical 2D coordinates. 
 www.pdesolutions.com /cgibin/discus/show.cgi?tpc=4&post=821 (837 words) 
