2-D elements - LMU

1 1 Finite element method basis functionsFinite elements : Basis functions1-D elements coordinate transformation 1-D elements linear basis functions quadratic basis functions cubic basis functions2-D elements coordinate transformation triangular elements linear basis functions quadratic basis functions rectangular elements linear basis functions quadratic basis functionsScope: Understand the origin and shape of basis functions used in classical finite element techniques. 2 Finite element method basis functions1-D elements : coordinate transformationWe wish to approximate a function u(x) defined in an interval [a,b] by some set of basis functions ==niiicxu1)( where i is the number of grid points (the edges of our elements ) defined at locations xi.

2 As the basis functions look the same in all elements (apart from some constant) we make life easier by moving to a local coordinate system iiixxxx =+1 so that the element is defined for x=[0,1].3 Finite element method basis functions1-D elements linear basis functionsThere is not much choice for the shape of a (straight) 1-D element! Notably the length can vary across the domain. We require that our function u( ) be approximated locally by the linear function 21)(ccu+=Our node points are defined at 1,2 =0,1 and we require that 2121121211uucucccucu+ == += =Auc= =11-01A4 Finite element method basis functions1-D elements linear basis functionsAs we have expressed the coefficients ci as a function of the function values at node points 1,2 we can now express the approximate function using the node values )()()1()()(21121211 NNuuuuuuu+=+ =+ +=.

3 And N1,2 (x) are the linear basis functions for 1-D element method basis functions1-D quadratic elementsNow we require that our function u(x) be approximated locally by the quadratic function 2321)( cccu++=Our node points are defined at 1,2,3 =0,1/2,1 and we require that ++=++==Auc= =242143001A6 Finite element method basis functions1-D quadratic basis again we can now express our approximated function as a sum over our basis functions weighted by the values at three node note that now we re using three grid points per element ..Can we approximate a constant function?

4 ==+ + ++ =++=312322212321)()2()44()231()(iiiNuuuu cccu 7 Finite element method basis functions1-D cubic basis using similar arguments the cubic basis functions can be derived as324323322321342321)(23)(2)(231)()( + = =+ =+ =+++= note that here we need derivative information at the boundaries ..How can we approximate a constant function?8 Finite element method basis functions2-D elements : coordinate transformationLet us now discuss the geometry and basis functions of 2-D elements , again we want to consider the problems in a local coordinate system, first we look at trianglesP3P2P1xyP3P2P1 beforeafter9 Finite element method basis functions2-D elements : coordinate transformationAny triangle with corners Pi (xi ,yi ), i=1,2,3 can be transformed into a rectangular, equilateral triangle withP3P2P1 P1 (0,0)P3 (0,1)P2 (1,0) )()()()(1312113121yyyyyyxxxxxx + += + +=using counterclockwise numbering.

5 Note that if =0, then these equations are equivalent to the 1- D tranformations. We seek to approximate a function by the linear form 321),(cccu++=we proceed in the same way as in the 1-D case10 Finite element method basis functions2-D elements : and we obtainP3P2P1 P1 (0,0)P3 (0,1)P2 (1,0).. and we obtain the coefficients as a function of the values at the grid nodes by matrix inversion31321211)1,0()0,1()0,0(ccuuccuu cuu+==+====Auc= =101011001 Acontaining the 1-D case =11-01A11 Finite element method basis functionstriangles: linear basis functionsfrom matrix A we can calculate the linear basis functions for trianglesP3P2P1 P1 (0,0)P3 (0,1)P2 (1,0) == =),(),(1),(321 NNN12 Finite element method basis functionstriangles.

6 Quadratic elementsAny function defined on a triangle can be approximated by the quadratic function26524321),(yxyxyxyxu +++++=and in the transformed system we obtain26524321),( ccccccu+++++=as in the 1-D case we need additional points on the P1 (0,0)P3 (0,1)P2 (1,0)P5 (1/2,1/2)P4 (1/2,0)P6 (0,1/2)++++++P5P4P613 Finite element method basis functionstriangles: quadratic elementsTo determine the coefficients we calculate the function u at each grid point to obtain63166543215421463134212116/12/14/1 4/14/12/12/14/12/1cccuccccccucccucccuccc ucu++=+++++=++=++=++==P3P2P1 P1 (0,0)P3 (0,1)P2 (1,0)P5 (1/2,1/2)P4 (1/2,0)P6 (0,1/2)++++++ and by matrix inversion we can calculate the coefficients as a function of the values at PiAuc=14 Finite element method basis functionstriangles.

7 Basis functions =400202444004004022400103004013000001AP3 P2P1 P1 (0,0)P3 (0,1)P2 (1,0)P5 (1/2,1/2)P4 (1/2,0)P6 (0,1/2)++++++ to obtain the basis functionsAuc=)1(4),(4),()1(4),()12(),()1 2(),()221)(1(),(254321 == = = = = and they look like ..15 Finite element method basis functionstriangles: quadratic basis functionsP3P2P1 P1 (0,0)P3 (0,1)P2 (1,0)P5 (1/2,1/2P4 (1/2,0)P6 (0,1/2)++++++P5P4P6 The first three quadratic basis functions ..16 Finite element method basis functionstriangles: quadratic basis functionsP3P2P1 P1 (0,0)P3 (0,1)P2 (1,0)P5 (1/2,1/2P4 (1/2,0)P6 (0,1/2)++++++ and the rest.))

8 17 Finite element method basis functionsrectangles: transformationLet us consider rectangular elements , and transform them into a local coordinate systemP3P2P1xyP3P2P1 beforeafterP4P418 Finite element method basis functionsrectangles: linear elementsWith the linear Ansatz 4321),(ccccu+++=we obtain matrix A as =1111100100110001 Aand the basis functions )1(),(),()1(),()1)(1(),(4321 == = =NNNN19 Finite element method basis functionsrectangles: quadratic elementsWith the quadratic Ansatz282726524321),( ccccccccu+++++++=we obtain an 8x8 matrix A.

9 And a basis function looks likeP3P2P1 P4++++P5P6P7P8)1)(1(4),()221)(1)(1(),(51 = =NNN1N220 Finite element method basis functions1-D and 2-D elements : summaryThe basis functions for finite element problems can be obtained by: Transforming the system in to a local (to the element) system Making a linear (quadratic, cubic) Ansatzfor a function defined across the element. Using the interpolation condition (which states that the particular basis functions should be one at the corresponding grid node) to obtain the coefficients as a function of the function values at the grid nodes.

10 Using these coefficients to derive the nbasis functions for the n node points (or conditions).