On Wed, Jul 15, 2009 at 7:17 PM, <alejandro.martinez@...> wrote:
> The jacobian for a 2d triangular element is a constant. I heard that for a
> quadratic 2d triangular element the jacobian should be linear but I'm
> getting a quadratic one (I'm using classical coordinate transformations: x =
> ax' + by' + cx'y' + dx'^2 + ey'^2). It should be the same in natural
> coordinates. Should the final result be linear?
The 2D Jacobian is a 2x2 matrix given by
[dx/dxi, dx/deta]
[dy/dxi, dy/deta]
Expressing x (resp y.) in terms of the quadratic basis functions phi:
x = \sum_i x_i \phi_i
and differentiating, you should be able to convince yourself that the
entries of the Jacobian for the quadratic triangular element are
linear in xi and eta, the coordinates of the reference element.
Give it a try, the basis functions for the TRI6 can be found in
src/fe/fe_lagrange_shape_2D.C starting at line 158.

John
