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# 5994944.ppt

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# 5994944.ppt

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### 5994944.ppt

1. 1. Finite element method 1 Finite Elements  Basic formulation  Basis functions  Stiffness matrix  Poisson‘s equation  Regular grid  Boundary conditions  Irregular grid  Numerical Examples Scope: Understand the basic concept of the finite element method with the simple-most equation.
2. 2. Finite element method 2 Formulation Let us start with a simple linear system of equations | * y and observe that we can generally multiply both sides of this equation with y without changing its solution. Note that x,y and b are vectors and A is a matrix. b Ax  n y yb yAx     We first look at Poisson’s equation ) ( ) ( x f x u    where u is a scalar field, f is a source term and in 1-D 2 2 2 x      
3. 3. Finite element method 3 Poisson‘s equation fv uv    We now multiply this equation with an arbitrary function v(x), (dropping the explicit space dependence) ... and integrate this equation over the whole domain. For reasons of simplicity we define our physical domain D in the interval [0, 1].      D D fv uv dx fv dx uv      1 0 1 0 Das Reh springt hoch, das Reh springt weit, warum auch nicht, es hat ja Zeit. ... why are we doing this? ... be patient ...
4. 4. Finite element method 4 Discretization As we are aiming to find a numerical solution to our problem it is clear we have to discretize the problem somehow. In FE problems – similar to FD – the functional values are known at a discrete set of points. ... regular grid ... ... irregular grid ... Domain D The key idea in FE analysis is to approximate all functions in terms of basis functions j, so that i N i i c u u j     1 ~
5. 5. Finite element method 5 Basis function where N is the number nodes in our physical domain and ci are real constants. With an appropriate choice of basis functions ji, the coefficients ci are equivalent to the actual function values at node point i. This – of course – means, that ji=1 at node i and 0 at all other nodes ... Doesn’t that ring a bell? Before we look at the basis functions, let us ... i N i i c u u j     1 ~
6. 6. Finite element method 6 Partial Integration ... partially integrate the left-hand-side of our equation ... dx fv dx uv      1 0 1 0   dx u v uv dx v u            1 0 1 0 1 0 ) ( we assume for now that the derivatives of u at the boundaries vanish so that for our particular problem dx u v dx v u          1 0 1 0 ) (
7. 7. Finite element method 7 … ... so that we arrive at ... ... with u being the unknown. This is also true for our approximate numerical system dx fv dx v u      1 0 1 0 ... where ... i N i i c u j    1 ~ was our choice of approximating u using basis functions. dx fv dx v u      1 0 1 0 ~
8. 8. Finite element method 8 Partial integration ... remember that v was an arbitrary real function ... if this is true for an arbitrary function it is also true if ... so any of the basis functions previously defined ... j v j  ... now let’s put everything together ... dx fv dx v u      1 0 1 0 ~
9. 9. Finite element method 9 The discrete system The ingredients: k v j  i N i i c u j    1 ~ dx fv dx v u      1 0 1 0 ~ dx f dx c k k n i i i j j j              1 0 1 0 1 ... leading to ...
10. 10. Finite element method 10 The discrete system dx f dx c k k i n i i j j j        1 0 1 0 1 ... the coefficients ck are constants so that for one particular function jk this system looks like ... k ik i g A b  ... probably not to your surprise this can be written in matrix form k i T ik g b A 
11. 11. Finite element method 11 The solution ... with the even less surprising solution   k T ik i g A b 1   remember that while the bi’s are really the coefficients of the basis functions these are the actual function values at node points i as well because of our particular choice of basis functions. This become clear further on ...
12. 12. Finite element method 12 Basis functions ... otherwise we are free to choose any function ... The simplest choice are of course linear functions: + grid nodes blue lines – basis functions ji 1 2 3 4 5 6 7 8 9 10 we are looking for functions ji with the following property        i j x x for x x for x j i i , 0 1 ) ( j
13. 13. Finite element method 13 Basis functions - gradient To assemble the stiffness matrix we need the gradient (red) of the basis functions (blue) 1 2 3 4 5 6 7 8 9 10
14. 14. Finite element method 14 Stiffness matrix Knowing the particular form of the basis functions we can now calculate the elements of matrix Aij and vector gi dx f dx c k k i n i i j j j        1 0 1 0 1 dx A k i ik     1 0 j j k ik i g A b  dx f g k k j   1 0 Note that ji are continuous functions defined in the interval [0,1], e.g.                         elsewhere x x x for x x x x x x x for x x x x x i i i i i i i i i i i 0 ) ( 1 1 1 1 1 1 j Let us – for now – assume a regular grid ... then
15. 15. Finite element method 15 Stiffness matrix –regular grid ... where we have used ...                          elsewhere x x x for x x x x x x x for x x x x x i i i i i i i i i i i 0 ) ( 1 1 1 1 1 1 j                  elsewhere dx x for dx x x dx for dx x x i 0 ~ 0 ~ 1 0 ~ 1 ~ ) ~ ( j 1 ~      i i i x x dx x x x dx xi ji
16. 16. Finite element method 16 Regular grid - gradient              elsewhere dx x for dx x dx for dx x i 0 ~ 0 / 1 0 ~ / 1 ) ~ ( j 1 ~      i i i x x dx x x x dx xi ji 1/dx -1/dx
17. 17. Finite element method 17 Stifness matrix - elements dx A k i ik     1 0 j j 1 2 3 4 5 6 7 8 9 10 ... we have to distinguish various cases ... e.g. ... dx dx dx dx dx dx dx dx A dx dx x x dx x x 1 1 1 1 0 2 1 1 1 0 1 1 11 1 1 1 1                  j j j j dx dx dx dx dx dx dx dx A dx dx dx x x x dx x 2 1 1 0 2 0 2 2 2 2 2 1 0 2 2 22 2 2 2 2                     j j j j j j
18. 18. Finite element method 18 Stiffness matrix dx A k i ik     1 0 j j 1 2 3 4 5 6 7 8 9 10 ... and ... dx dx dx dx dx dx dx dx A dx dx x x dx x x 1 1 1 1 0 2 2 1 1 0 2 1 12 1 1 1 1                   j j j j 12 21 A A  ... and ... ... so that finally the stiffness matrix looks like ...
19. 19. Finite element method 19 Stiffness matrix dx A k i ik     1 0 j j 1 2 3 4 5 6 7 8 9 10                        1 1 1 2 1 1 2 1 1 1 1  dx Aij ... so far we have ignored sources and boundary conditions ...
20. 20. Finite element method 20 Boundary conditions - sources ... let us start restating the problem ... ) ( ) ( x f x u    ... which we turned into the following formulation ... dx f dx c k k i n i i j j j        1 0 1 0 1 ... assuming ... i N i i c u j    1 ~ with b.c. N i N i i u u c u j j j ) 1 ( ) 0 ( ~ 1 1 2       where u(0) and u(1) are the values at the boundaries of the domain [0,1]. How is this incorporated into the algorithm?
21. 21. Finite element method 21 Boundary conditions ) ( ) ( x f x u    ... which we turned into the following formulation ... dx f dx c k k i n i i j j j        1 0 1 0 1 ... in pictorial form ... dx u dx u dx f dx c k n k k k i n i i                 1 0 1 0 1 1 0 1 0 1 2 ) 1 ( ) 0 ( j j j j j j j = boundary condition boundary condition source heterogeneity (f) ... the system feels the boundary conditions through the (modified) source term AT b = g
22. 22. Finite element method 22 Numerical Example ) ( ) ( x f x u    Domain: [0,1]; nx=100; dx=1/(nx-1);f(x)=d(1/2) Boundary conditions: u(0)=u(1)=0 f(nx/2)=1/dx; for it = 1:nit, uold=u; du=(csh(u,1)+csh(u,-1)); u=.5*( f*dx^2 + du ); u(1)=0; u(nx)=0; end Matlab FD code % source term s=(1:nx)*0;s(nx/2)=1.; % boundary left u_1 int{ nabla phi_1 nabla phij } u1=0; s(1) =0; % boundary right u_nx int{ nabla phi_nx nabla phij } unx=0; s(nx)=0; % assemble matrix Aij A=zeros(nx); for i=2:nx-1, for j=2:nx-1, if i==j, A(i,j)=2/dx; elseif j==i+1 A(i,j)=-1/dx; elseif j==i-1 A(i,j)=-1/dx; else A(i,j)=0; end end end fem(2:nx-1)=inv(A(2:nx-1,2:nx-1))*s(2:nx-1)'; fem(1)=u1; fem(nx)=unx; Matlab FEM code
23. 23. Finite element method 23 Regular grid ) ( ) ( x f x u    Domain: [0,1]; nx=100; dx=1/(nx-1);f(x)=d(1/2) Boundary conditions: u(0)=u(1)=0 Matlab FD code (red) Matlab FEM code (blue) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 x u(x) FD (red) - FEM (blue)
24. 24. Finite element method 24 Regular grid - non zero b.c. ) ( ) ( x f x u    Domain: [0,1]; nx=100; dx=1/(nx-1);f(x)=d(1/2) Boundary conditions: u(0)=0.15 u(1)=0.05 Matlab FD code (red) Matlab FEM code (blue) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 x u(x) FD (red) - FEM (blue) -> Regular grid % Quelle s=(1:nx)*0;s(nx/2)=1.; % Randwert links u_1 int{ nabla phi_1 nabla phij } u1=0.15; s(2) =u1/dx; % Randwert links u_nx int{ nabla phi_nx nabla phij } unx=0.05; s(nx-1)=unx/dx;
25. 25. Finite element method 25 Stiffness – irregular grid 1 2 3 4 5 6 7 8 9 10 dx A k i ik     1 0 j j 21 1 0 2 1 1 2 1 1 0 2 1 12 1 1 1 1 1 1 1 1 1 1 1 1 A h dx h dx h h dx dx A h h x x h x x                    j j j j i i ii h h A 1 1 1    i=1 2 3 4 5 6 7 + + + + + + + h1 h2 h3 h4 h5 h6
26. 26. Finite element method 26 Example ) ( ) ( x f x u    Domain: [0,1]; nx=100; dx=1/(nx-1);f(x)=d(1/2) Boundary conditions: u(0)=u0; u(1)=u1 for i=2:nx-1, for j=2:nx-1, if i==j, A(i,j)=1/h(i-1)+1/h(i); elseif i==j+1 A(i,j)=-1/h(i-1); elseif i+1==j A(i,j)=-1/h(i); else A(i,j)=0; end end end i=1 2 3 4 5 6 7 + + + + + + + h1 h2 h3 h4 h5 h6 Stiffness matrix A
27. 27. Finite element method 27 Irregular grid – non zero b.c. ) ( ) ( x f x u    Domain: [0,1]; nx=100; dx=1/(nx-1);f(x)=d(1/2) Boundary conditions: u(0)=0.15 u(1)=0.05 Matlab FD code (red) Matlab FEM code (blue) + FEM grid points 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 x u(x) FD (red) - FEM (blue) FEM on Chebyshev grid
28. 28. Finite element method 28 Summary In finite element analysis we approximate a function defined in a Domain D with a set of orthogonal basis functions with coefficients corresponding to the functional values at some node points. The solution for the values at the nodes for some partial differential equations can be obtained by solving a linear system of equations involving the inversion of (sometimes sparse) matrices. Boundary conditions are inherently satisfied with this formulation which is one of the advantages compared to finite differences.