Introduction to PDEs: Assignment 9: Finite elements in 1D

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1 Institute of Scientific Computing Technical University Braunschweig Dr. Elmar Zander Dr. Noemi Friedman Winter Semester 213/14 Assignment 9 Due date: Introduction to PDEs: Assignment 9: Finite elements in 1D In the current assignment the Poisson equation 1 shall be solved on the unit interval using the finite element method. The problem is given by u (x) = sin(πx), u() = u(1) =. The interval shall be divided into 8 equally spaced elements, i.e. we have 9 nodes at which the function value u i has to be determined with linear interpolation in between. To make things more realistic, i.e. like in a real FEM program, we consider an arbitrary ordering of the elements and the nodes instead of the usual linear ordering in 1D; This will also translate better to higher dimensions later. The numbers of the elements and the nodes should be assigned as follows: x: 1/ /8 1 elem: node: Your task will be to create appropriate data structures to store the element and node information, assemble the system matrix and right hand side, solve the system, and display the solution. Exercise 1: Definition and test of data structures (12 points) The first thing to do is to set up two data structure to store the information as to the nodes belong to each element and to the positions of each node and to initialize them according to the scheme given above. The two data structures: 2/8 coord of node nodes of elelement nodes of elelement 2 coord of node 2 4/8 coord of node nodes of elelement nodes = coord = 9 4 3/ /8 6/ /8 6 1 nodes of elelement 8 7/8 coord of node 9 And we also define a vector with the restricted nodes (where boundary conditions has to be enforced): restr_nodes = [ ] Only the special case where the right hand side is zero is called Laplace equation. Still, some authors use errorneously the name Laplace equation also in the inhomogenous case considered here.

2 The corresponding matlab code: %Set i n i t i a l parameters numel =8; numnode=numel +1; h=1/numel ; %Set data s t r u c t u r e nodes =[8,7; 5, 3 ; 1, 5 ; 3, 8 ; 9, 4 ; 7, 9 ; 2, 6 ; 6, 1 ] ; node_order = [ 2 ; 6 ; 1 ; 5 ; 3 ; 8 ; 7 ; 9 ; 4 ] ; [ nn, IX]= s o r t ( node_order ) ; x=l i n s p a c e (, 1, numnode ) ; coord=x ( IX ) ; restr_nodes = [ 2, 4 ] ; Test your data structures with the following tests: Check for element 7 that the left node is at x = and for element 5 that the right node is at x = 1. Check one node of three more elements of your choice. For all elements check that it has the correct size h, i.e. coord(nodes(i,2))-coord(nodes(i,1)) should be equal to h within some reasonable accuracy (supposing you have the coordinates belonging to each node in coords and the nodes belonging to each element in nodes). i f testing_ mode %Check coord o f l e f t node o f element 7 and o f r i g h t node o f element 5 i f ~ ( coord ( nodes ( 7, 1 ) ) == ) e r r o r ( TESTING FAILED :... ) e l s e d i s p l a y ( TESTING... ok ) i f ~ ( coord ( nodes ( 5, 2 ) ) == 1) e r r o r ( TESTING FAILED :... ) e l s e d i s p l a y ( TESTING... ok ) %Check wether each element has length 1/8=,125 within a t h r e s h o l d eps eps =1^ 5; element_lengths=coord ( nodes (:,2)) coord ( nodes ( :, 1 ) ) ; err_length=abs ( element_lengths h ones ( s i z e ( element_lengths ) ) ) ; i f any ( err_length > eps ones ( s i z e ( err_length ) ) ) e r r o r ( TESTING FAILED :... ) e l s e d i s p l a y ( TESTING... ok )

3 Note: The testing section should be inside an if block looking something like: if testing_mode, then some lines of code doing the test, and then an statement. That way you can decide whether to run the tests by setting testing_mode to either or 1. Exercise 2: Write a plot function and test it (12 points) Before we solve the equation, we should make sure that if we have solution we can also plot it otherwise we wouldn t know whether the error is in the solution algorithm or in the plotting function. The function to be plotted is given by f(x) = sin(2πx). Store the values into a vector v such that v(i) is the value of node i i.e. v i = f(x i ). Now plot the solution using the line drawing mode x- in the plot command. Note that you first have to sort the coordinates and then the function values according to the previous sorting of the coordinates (read the help for the sort command, especially with two output arguments). Make sure that it looks like a sin function of one wavelength without any zigzagging lines. The matlab code for plotting the testfunction (or any other function): f u n c t i o n p l o t f u n c t i o n ( f, coord ) i f i s a ( f, function_handle ) v=f e v a l ( f, coord ) ; e l s e v=f ; [X, IX]= s o r t ( coord ) ; V_i=v ( IX ) ; p l o t (X, V_i, r x ) f u n c t i o n v=t e s t f u n c t i o n ( x ) v=s i n (2 pi x ) ; The plotfunction can be tested (when in testing_mode) with the command: i f testing_ mode p l o t f u n c t i o n coord ) ; The plot should give the following result when tested:

4 1.8 testing f=sin(2*pi*x) Exercise 3: Assemble the matrix (14 points) Now assemble the stiffness matrix K and the right hand side vector F. You can calculate the right by assembling the whole mass matrix M and calculating the right hand side like F = M f, or which is more efficient assemble it directly in the assembly loop by summing up local contributions using only the local mass matrix. The element stiffness matrix and mass matrix are given by ( ) ( ) K e = 1 h 1 1 M e = h respectively. Matlab code to assemble the global stiffness matrix and the right hand side F: %Define element s t i f f n e s s matrix : K_e=1/h [1, 1; 1,1]; M_e=h / 6 [ 2, 1 ; 1, 2 ] ; %Assemble g l o b a l s t i f f n e s s matrix (K) K=z e r o s ( numnode, numnode ) ; F=z e r o s ( numnode, 1 ) ; f o r i =1:numel K( nodes ( i, : ), nodes ( i, : ) ) =K( nodes ( i, : ), nodes ( i, : ) ) +K_e ( :, : ) ;

5 f_e=source_function ( coord ( nodes ( i, : ) ) ) ; F_e=M_e f_e ; F( nodes ( i, : ) ) =F( nodes ( i, : ) ) + F_e ; f u n c t i o n v=source_function ( x ) v=s i n ( pi x ) ; Test the correctness of the assembled matrix: The diagonal entries of K have to be 2/h for all inner nodes. In each row corresponding to an inner node there should be exactly three non-zero elements where the off-diagonal entries have the value 1/h. The stiffness matrix has to be singular before application of the boundary conditions and non-singular afterwards. The nonreduced stiffnes matrix: K = F = The singularity of the non-reduced stiffness matrix can be tested: i f testing_ mode i f ~ det (K)== e r r o r ( TESTING FAILED :... ) e l s e d i s p l a y ( TESTING... ok ) The given boundary condition can be enforced by changing the rows, corresponding to the restricted nodes, of F and of K (in our case row 2 and row 4): %Change K and F in accordance with the boundary c o n d i t i o n s f o r j =1: s i z e ( restr_nodes, 2 ) rn=restr_nodes ( j ) ; K( rn, : ) = z e r o s ( 1, numnode ) ; K( rn, rn )=1; F( rn )=;

6 Which gives K = F = From the resulted matrices the solution can be calculated. However it is recommed to check whether the siffness matrix is not singular: i f testing_ mode i f det (K)== e r r o r ( TESTING FAILED :... ) e l s e d i s p l a y ( TESTING... ok ) Plot the right hand side F. It should look like a smoothed version of the original function f(x) = sin(πx).

7 1.9 F sin(pi*x) Exercise 4: Solve the system (12 points) Now solve the system Ku = F for u to get the desired solution; plot it. Calculate the exact solution and compare it to your numerical result graphically (put both graphs into one diagram). The analytical solution of the PDE is: u a = 1 π 2 sin(πx) And the numerical solution can be calculated by solving the system of equations: Ku = F Using the matlab command: v=k\f Where the elements of u give the value of the solution in the distinct nodes. The function of the solution can be calculated from the ansatz function: u(x) = i v i N i (x) where N i (x) are the linear nodal basis functions. Plotting both, the analytical and the numerical solutions on one figure should give similar results:

8 .12.1 FEM solution analytical solution Now compute the error in different norms. Use the L norm (or absolute value norm) and the L 2 norm (or Euclidian norm). For simplicity use the midpoint rule for integration in each element. Computing error in the L norm means we have to calculate: u a (x) u(x) = max u a (x) u(x) We will approximate this error by linearizing the analytical solution in between the nodal values, and correspondingly we take only the maximal value of the difference in between the nodal values: u_an=( a n a l y t i c a l _ s o l u t i o n ( coord ) ) ; e r r o r _ i n f=norm ( ( u_an v ), 1 ) r e l _ e r r o r _ i n f=e r r o r _ i n f /norm (u_an, 1 ) f u n c t i o n v=a n a l y t i c a l _ s o l u t i o n ( x ) v=s i n ( pi x ) / ( pi ) ^ 2 ; Which will give the maximal nodal error:.65 And the relative error (the given error devided by the L norm of the analytical solution) is.128, thus around one percent. For the L 2 norm the following expression has to be calculated:

9 1 1 u a (x) u(x) 2 = u a (x) u(x) 2 dx We will approximate the integral by the mid-point rule: u a (x) u(x) 2 dx h [ u a ( x 1 ) u( x 1 ) 2 + u a ( x 2 ) u( x 2 ) u a ( x N ) u( x N ) 2] where x i is the coordinate of the midpoint of the elements, and N is the number of the elements (numnode). It is somewhat easy to calculate the value of numerical solution at the midpoints, as the approximated ansatzfunctions were linear, so the values can be calculated from the avarage of the values at the nodes. However, the midpoint values of the analytical solution have to be calculated from the midpointcoordinates x i.: L2_norm=; ref_norm =; f o r i =1:numel v_mid=mean( v ( nodes ( i, : ) ) ) ; coord_mid= mean( coord ( nodes ( i, : ) ) ) ; u_an_mid=a n a l y t i c a l _ s o l u t i o n ( coord_mid ) ; L2_norm=L2_norm+ h ( v_mid u_an_mid ) ^ 2 ; ref_norm=ref_norm+h u_an_mid ^ 2; error_2= s q r t ( L2_norm ) rel_error_2=error_2 / s q r t ( ref_norm ) This will give the absolute error.23, and the relative error.317. It would be somewhat faster to calculate the L2 norm using the trapezoidal rule, that is by avaraging the nodal values, that would give smaller errors. (8.962e-4, rel error:.128) Exercise 5: Optional: 2 δ-function on right hand side (2 points) (a) Modify your program in such a way that you can put a δ function on the right hand side of the original PDE. Note that you have to calculate the right hand side vector F now analytically, and not like in exercise 3, by interpolation of f. Place the position of the δ peak once on x = 1/2, i.e. exactly on a node, and once on x = 7/16, i.e. in between nodes. Give plots of both solutions. The integral to get the right hand side of the week form: F i = δ(x x 3 )N i (x)dx Ω When the dirac delta pic is at the midpoint of the bar(over node 3), the integral gives zero everywere, except when integrating with the basis function N 3 (x), corresponding to the hat function, taking unit value at the 3rd node. In the second case, only those integrals will not vanish, which involves the basis functions that are not zero on element 2, that is, the basis functions N 3 (x) and N 5 (x). Both basis function take value 1/2 at x = 7/16 and accordingly the two nonzero element of the right 2 You can gain extra points by doing this exercise, if not, this will not be counted for eligibility to the exam.

10 hand side are: F 5 = Ω 5 δ(x x 3 )N 5 (x)dx = 1/2 F 3 = Ω 3 δ(x x 3 )N 3 (x)dx = 1/2 Correspondingly the right hand side in the two cases needs to be modified to: F 1 =. 1 F 2 =..5.5 (1 points) (b) Compare the results from (a) with the exact solution. Explain why the numerical solution coincides with the exact solutions in one case, but not in the other.

11 .25 Solution with Dirac delta with midpic FEM solution analytical solution

12 .25 FEM solution analytical solution (1 points)