10 basis functions in the reference tetrahedro (fem)












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I'm writting a Finite Element code and I need the 10 basis functions in $mathbb{P}^2(hat K)$ (polynomial of degree less than or equal to 2 defined over the reference tetrahedro $hat K$) in the reference tetrahedro with vertices:



$(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$.



Do you know a book or link that contains this?



The local numbering of the corners (vertices) and the second-order nodes of a tetrahedron










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    0












    $begingroup$


    I'm writting a Finite Element code and I need the 10 basis functions in $mathbb{P}^2(hat K)$ (polynomial of degree less than or equal to 2 defined over the reference tetrahedro $hat K$) in the reference tetrahedro with vertices:



    $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$.



    Do you know a book or link that contains this?



    The local numbering of the corners (vertices) and the second-order nodes of a tetrahedron










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm writting a Finite Element code and I need the 10 basis functions in $mathbb{P}^2(hat K)$ (polynomial of degree less than or equal to 2 defined over the reference tetrahedro $hat K$) in the reference tetrahedro with vertices:



      $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$.



      Do you know a book or link that contains this?



      The local numbering of the corners (vertices) and the second-order nodes of a tetrahedron










      share|cite|improve this question











      $endgroup$




      I'm writting a Finite Element code and I need the 10 basis functions in $mathbb{P}^2(hat K)$ (polynomial of degree less than or equal to 2 defined over the reference tetrahedro $hat K$) in the reference tetrahedro with vertices:



      $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$.



      Do you know a book or link that contains this?



      The local numbering of the corners (vertices) and the second-order nodes of a tetrahedron







      finite-element-method






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 23 '14 at 17:10







      yemino

















      asked Apr 23 '14 at 17:03









      yeminoyemino

      2831314




      2831314






















          3 Answers
          3






          active

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          1












          $begingroup$

          In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:



          A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
          $$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
          The degrees of freedom $L_j$, $jin{1,cdots!,10}$, are the nodal values and the edge middle point values:
          $$L_1(v) = v(0,0,0);quad L_2(v)=v(1,0,0);quad cdots quad L_{10}(v)=v(0,0.5,0.5).$$
          Note that you may choose the order as you wish. Then the Vandermonde matrix is
          $$begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & cdots & L_1(x_3^2) \
          L_2(1) & L_2(x_1) & L_2(x_2) & cdots & L_2(x_3^2) \
          L_3(1) & L_3(x_1) & L_3(x_2) & cdots & L_3(x_3^2) \
          vdots & vdots & vdots & ddots & vdots \
          L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & cdots & L_{10}(x_3^2) end{pmatrix}.$$
          By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
          $$begin{aligned}
          phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\
          phi_2&=-x_1+2x_1^2,\
          phi_3&=-x_2+2x_2^2,\
          phi_4&=-x_3+2x_3^2,\
          phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\
          phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\
          phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\
          phi_8&=4x_1x_2,\
          phi_8&=4x_1x_3,\
          phi_9&=4x_2x_3.\
          end{aligned}
          $$
          I did not verify the shape functions in any way. Thus, you may want to redo the calculations.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            On page 134 of this thesis, the 10 base functions on the reference element are listed.
            10 base functions on P_2



            Also, I wrote some code in order to find the base functions for arbitrary dimension and order: http://pastebin.com/KmcgBNTw






            share|cite|improve this answer











            $endgroup$





















              0












              $begingroup$

              You can use Volume coordinates of the Linear Tetrahedron (4 Nodes), you have first to compute your volume coordinates L1, L2, L3, L4.



              Then you use them in order to obtain the basis function for higher order tetrahedrons (Quadratic and even cubic).



              The expression of The basis functions in function of Volume coordinates have many advantages, it helps you a lot when you are faced with Integration Calculus (Mass and stiffness matrix).



              In addition, if you use the formula of "Eisenberg and Malvern" it will give you exact integration.



              I will let you see the pictures and the link below



              Quadratic Tetrahedron Basis Functions



              Cubic Tetrahedron Basis Functions



              Linear Tetrahedron Element



              Higher Order Tetrahedron Element






              share|cite|improve this answer









              $endgroup$













                Your Answer





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                3 Answers
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                active

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                3 Answers
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                active

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                1












                $begingroup$

                In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:



                A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
                $$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
                The degrees of freedom $L_j$, $jin{1,cdots!,10}$, are the nodal values and the edge middle point values:
                $$L_1(v) = v(0,0,0);quad L_2(v)=v(1,0,0);quad cdots quad L_{10}(v)=v(0,0.5,0.5).$$
                Note that you may choose the order as you wish. Then the Vandermonde matrix is
                $$begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & cdots & L_1(x_3^2) \
                L_2(1) & L_2(x_1) & L_2(x_2) & cdots & L_2(x_3^2) \
                L_3(1) & L_3(x_1) & L_3(x_2) & cdots & L_3(x_3^2) \
                vdots & vdots & vdots & ddots & vdots \
                L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & cdots & L_{10}(x_3^2) end{pmatrix}.$$
                By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
                $$begin{aligned}
                phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\
                phi_2&=-x_1+2x_1^2,\
                phi_3&=-x_2+2x_2^2,\
                phi_4&=-x_3+2x_3^2,\
                phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\
                phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\
                phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\
                phi_8&=4x_1x_2,\
                phi_8&=4x_1x_3,\
                phi_9&=4x_2x_3.\
                end{aligned}
                $$
                I did not verify the shape functions in any way. Thus, you may want to redo the calculations.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:



                  A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
                  $$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
                  The degrees of freedom $L_j$, $jin{1,cdots!,10}$, are the nodal values and the edge middle point values:
                  $$L_1(v) = v(0,0,0);quad L_2(v)=v(1,0,0);quad cdots quad L_{10}(v)=v(0,0.5,0.5).$$
                  Note that you may choose the order as you wish. Then the Vandermonde matrix is
                  $$begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & cdots & L_1(x_3^2) \
                  L_2(1) & L_2(x_1) & L_2(x_2) & cdots & L_2(x_3^2) \
                  L_3(1) & L_3(x_1) & L_3(x_2) & cdots & L_3(x_3^2) \
                  vdots & vdots & vdots & ddots & vdots \
                  L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & cdots & L_{10}(x_3^2) end{pmatrix}.$$
                  By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
                  $$begin{aligned}
                  phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\
                  phi_2&=-x_1+2x_1^2,\
                  phi_3&=-x_2+2x_2^2,\
                  phi_4&=-x_3+2x_3^2,\
                  phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\
                  phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\
                  phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\
                  phi_8&=4x_1x_2,\
                  phi_8&=4x_1x_3,\
                  phi_9&=4x_2x_3.\
                  end{aligned}
                  $$
                  I did not verify the shape functions in any way. Thus, you may want to redo the calculations.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:



                    A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
                    $$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
                    The degrees of freedom $L_j$, $jin{1,cdots!,10}$, are the nodal values and the edge middle point values:
                    $$L_1(v) = v(0,0,0);quad L_2(v)=v(1,0,0);quad cdots quad L_{10}(v)=v(0,0.5,0.5).$$
                    Note that you may choose the order as you wish. Then the Vandermonde matrix is
                    $$begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & cdots & L_1(x_3^2) \
                    L_2(1) & L_2(x_1) & L_2(x_2) & cdots & L_2(x_3^2) \
                    L_3(1) & L_3(x_1) & L_3(x_2) & cdots & L_3(x_3^2) \
                    vdots & vdots & vdots & ddots & vdots \
                    L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & cdots & L_{10}(x_3^2) end{pmatrix}.$$
                    By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
                    $$begin{aligned}
                    phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\
                    phi_2&=-x_1+2x_1^2,\
                    phi_3&=-x_2+2x_2^2,\
                    phi_4&=-x_3+2x_3^2,\
                    phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\
                    phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\
                    phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\
                    phi_8&=4x_1x_2,\
                    phi_8&=4x_1x_3,\
                    phi_9&=4x_2x_3.\
                    end{aligned}
                    $$
                    I did not verify the shape functions in any way. Thus, you may want to redo the calculations.






                    share|cite|improve this answer









                    $endgroup$



                    In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:



                    A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
                    $$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
                    The degrees of freedom $L_j$, $jin{1,cdots!,10}$, are the nodal values and the edge middle point values:
                    $$L_1(v) = v(0,0,0);quad L_2(v)=v(1,0,0);quad cdots quad L_{10}(v)=v(0,0.5,0.5).$$
                    Note that you may choose the order as you wish. Then the Vandermonde matrix is
                    $$begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & cdots & L_1(x_3^2) \
                    L_2(1) & L_2(x_1) & L_2(x_2) & cdots & L_2(x_3^2) \
                    L_3(1) & L_3(x_1) & L_3(x_2) & cdots & L_3(x_3^2) \
                    vdots & vdots & vdots & ddots & vdots \
                    L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & cdots & L_{10}(x_3^2) end{pmatrix}.$$
                    By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
                    $$begin{aligned}
                    phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\
                    phi_2&=-x_1+2x_1^2,\
                    phi_3&=-x_2+2x_2^2,\
                    phi_4&=-x_3+2x_3^2,\
                    phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\
                    phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\
                    phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\
                    phi_8&=4x_1x_2,\
                    phi_8&=4x_1x_3,\
                    phi_9&=4x_2x_3.\
                    end{aligned}
                    $$
                    I did not verify the shape functions in any way. Thus, you may want to redo the calculations.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 26 '14 at 17:31









                    knlknl

                    850611




                    850611























                        0












                        $begingroup$

                        On page 134 of this thesis, the 10 base functions on the reference element are listed.
                        10 base functions on P_2



                        Also, I wrote some code in order to find the base functions for arbitrary dimension and order: http://pastebin.com/KmcgBNTw






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          On page 134 of this thesis, the 10 base functions on the reference element are listed.
                          10 base functions on P_2



                          Also, I wrote some code in order to find the base functions for arbitrary dimension and order: http://pastebin.com/KmcgBNTw






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            On page 134 of this thesis, the 10 base functions on the reference element are listed.
                            10 base functions on P_2



                            Also, I wrote some code in order to find the base functions for arbitrary dimension and order: http://pastebin.com/KmcgBNTw






                            share|cite|improve this answer











                            $endgroup$



                            On page 134 of this thesis, the 10 base functions on the reference element are listed.
                            10 base functions on P_2



                            Also, I wrote some code in order to find the base functions for arbitrary dimension and order: http://pastebin.com/KmcgBNTw







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jul 9 '15 at 22:38

























                            answered Jul 9 '15 at 18:22









                            AndyAndy

                            12317




                            12317























                                0












                                $begingroup$

                                You can use Volume coordinates of the Linear Tetrahedron (4 Nodes), you have first to compute your volume coordinates L1, L2, L3, L4.



                                Then you use them in order to obtain the basis function for higher order tetrahedrons (Quadratic and even cubic).



                                The expression of The basis functions in function of Volume coordinates have many advantages, it helps you a lot when you are faced with Integration Calculus (Mass and stiffness matrix).



                                In addition, if you use the formula of "Eisenberg and Malvern" it will give you exact integration.



                                I will let you see the pictures and the link below



                                Quadratic Tetrahedron Basis Functions



                                Cubic Tetrahedron Basis Functions



                                Linear Tetrahedron Element



                                Higher Order Tetrahedron Element






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  You can use Volume coordinates of the Linear Tetrahedron (4 Nodes), you have first to compute your volume coordinates L1, L2, L3, L4.



                                  Then you use them in order to obtain the basis function for higher order tetrahedrons (Quadratic and even cubic).



                                  The expression of The basis functions in function of Volume coordinates have many advantages, it helps you a lot when you are faced with Integration Calculus (Mass and stiffness matrix).



                                  In addition, if you use the formula of "Eisenberg and Malvern" it will give you exact integration.



                                  I will let you see the pictures and the link below



                                  Quadratic Tetrahedron Basis Functions



                                  Cubic Tetrahedron Basis Functions



                                  Linear Tetrahedron Element



                                  Higher Order Tetrahedron Element






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    You can use Volume coordinates of the Linear Tetrahedron (4 Nodes), you have first to compute your volume coordinates L1, L2, L3, L4.



                                    Then you use them in order to obtain the basis function for higher order tetrahedrons (Quadratic and even cubic).



                                    The expression of The basis functions in function of Volume coordinates have many advantages, it helps you a lot when you are faced with Integration Calculus (Mass and stiffness matrix).



                                    In addition, if you use the formula of "Eisenberg and Malvern" it will give you exact integration.



                                    I will let you see the pictures and the link below



                                    Quadratic Tetrahedron Basis Functions



                                    Cubic Tetrahedron Basis Functions



                                    Linear Tetrahedron Element



                                    Higher Order Tetrahedron Element






                                    share|cite|improve this answer









                                    $endgroup$



                                    You can use Volume coordinates of the Linear Tetrahedron (4 Nodes), you have first to compute your volume coordinates L1, L2, L3, L4.



                                    Then you use them in order to obtain the basis function for higher order tetrahedrons (Quadratic and even cubic).



                                    The expression of The basis functions in function of Volume coordinates have many advantages, it helps you a lot when you are faced with Integration Calculus (Mass and stiffness matrix).



                                    In addition, if you use the formula of "Eisenberg and Malvern" it will give you exact integration.



                                    I will let you see the pictures and the link below



                                    Quadratic Tetrahedron Basis Functions



                                    Cubic Tetrahedron Basis Functions



                                    Linear Tetrahedron Element



                                    Higher Order Tetrahedron Element







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 14 '18 at 9:30









                                    Walid AmehriWalid Amehri

                                    61




                                    61






























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