P3 Cubic Element in 2D

We explain degree of freedoms for the cubic element on triangles There are three types of dofs: vertex type, edge type and element type Given a mesh, the required data structure can be constructured by

[elem2dof,elem2edge,edge,bdDof,freeDof] = dofP3(elem)

Contents

help dofP3

  DOFP3 dof structure for P3 element.
 
   [elem2dof,elem2edge,edge,bdDof] = DOFP3(elem) constructs the dof
   structure for the quadratic element based on a triangle. elem2dof(t,i)
   is the global index of the i-th dof of the t-th element.
 
   The global indices of the dof is organized  according to the order of
   nodes, edges and elements. To be consistent, the dof on an edge depends
   on the orientation of edge only. 
 
   See also dofP2, dof3P3.
   
   Documentation: <a href="matlab:ifem dofP3doc">P3 Cubic Element in 2D</a>
 
   Created by Jie Zhou. M-lint by Long Chen. 
 
  Copyright (C) Long Chen. See COPYRIGHT.txt for details.

Local indexing of DOFs

node = [0 0; 1 0; 0.5 0.5*sqrt(3)];
elem = [1 2 3];
lambda = [1 0 0; 0 1 0; 0 0 1; ...  % 1,2,3 three vertices
          0 1/3 2/3; 0 2/3 1/3; ... % 4, 5 first edge
          2/3 0 1/3; 1/3 0 2/3; ... % 6, 7 second edge
          1/3 2/3 0; 2/3 1/3 0; ... % 8, 9 third edge
          1/3 1/3 1/3];             % 10   center of element
dofNode = lambda*node;
figure;
set(gcf,'Units','normal');
set(gcf,'Position',[0,0,0.3,0.3]);
showmesh(node,elem);
hold on;
findnode(dofNode);

A local basis of P3

The 10 Lagrange-type basis functions are denoted by $\phi_i, i=1:10$, i.e $\phi_i(x_j)=\delta _{ij},i,j=1:10$. In barycentric coordinates, they are

$$ \phi_1 = 1/2(3\lambda_1-1) (3\lambda_1-2)\lambda_1,\quad \nabla \phi_1 = (27/2 \lambda_1 \lambda_1-9 \lambda_1+1) \nabla \lambda_1,$$

$$ \phi_2 = 1/2(3\lambda_2-1) (3\lambda_2-2)\lambda_2,\quad \nabla \phi_2 = (27/2 \lambda_2 \lambda_2-9 \lambda_2+1) \nabla \lambda_2,$$

$$ \phi_3 = 1/2(3\lambda_3-1) (3\lambda_3-2)\lambda_3,\quad \nabla \phi_3 = (27/2 \lambda_3 \lambda_3-9 \lambda_3+1) \nabla \lambda_3,$$

$$ \phi_4 = 9/2\lambda_3\lambda_2 (3\lambda_2-1),\quad \nabla\phi_4 = 9/2 ((3 \lambda_2 \lambda_2-\lambda_2) \nabla \lambda_3+ \lambda_3 (6 \lambda_2-1) \nabla \lambda_2)$$

$$ \phi_5 = 9/2\lambda_3\lambda_2 (3\lambda_3-1),\quad \nabla\phi_5 = 9/2 ((3 \lambda_3 \lambda_3-\lambda_3) \nabla \lambda_2+ \lambda_2 (6 \lambda_3-1) \nabla \lambda_3),$$

$$ \phi_6 = 9/2\lambda_1\lambda_3 (3\lambda_3-1),\quad \nabla\phi_6 = 9/2 ((3 \lambda_3 \lambda_3-\lambda_3) \nabla \lambda_1+ \lambda_1 (6 \lambda_3-1) \nabla \lambda_3),$$

$$ \phi_7 = 9/2\lambda_1\lambda_3 (3\lambda_1-1),\quad \nabla\phi_7 = 9/2 ((3 \lambda_1 \lambda_1-\lambda_1) \nabla \lambda_3+ \lambda_3 (6 \lambda_1-1) \nabla \lambda_1),$$

$$ \phi_8 = 9/2\lambda_1\lambda_2 (3\lambda_1-1),\quad \nabla\phi_8 = 9/2 ((3 \lambda_1 \lambda_1-\lambda_1) \nabla \lambda_2+ \lambda_2 (6 \lambda_1-1) \nabla \lambda_1),$$

$$ \phi_9 = 9/2\lambda_1\lambda_2 (3\lambda_2-1),\quad \nabla\phi_9 = 9/2 ((3 \lambda_2 \lambda_2-\lambda_2) \nabla \lambda_1+ \lambda_1 (6 \lambda_2-1) \nabla \lambda_2),$$

$$ \phi_{10} = 27\lambda_1\lambda_2\lambda_3, \quad \nabla \phi_{10} = 27 (\lambda_1 \lambda_2 \nabla \lambda_3+\lambda_1 \lambda_3 \nabla \lambda_2+ \lambda_3 \lambda_2 \nabla \lambda_1).$$

When transfer to the reference triangle formed by $(0,0),(1,0),(0,1)$, the local bases in x-y coordinate can be obtained by substituting

$$\lambda _1 = x, \quad \lambda _2 = y, \quad \lambda _3 = 1-x-y.$$

Global indexing of DOFs

The global indices of the dof is organized  according to the order of
nodes, edges and elements. To be consistent, the dof on an edge depends
on the orientation of edge only.


Published with MATLAB® R2015b