Scientic Computing 2013 Computer Classes: Worksheet 11: 1D FEM and boundary conditions

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1 Scientic Computing 213 Computer Classes: Worksheet 11: 1D FEM and boundary conditions Oleg Batrashev November 14, 213 This material partially reiterates the material given on the lecture (see the slides) and then presents dierent types of boundary conditions and how to apply them in FEM. 1 Finite Element Method The main idea of FEM is to replace the space of all functions V with much smaller subspace a space of piecewise linear functions v V h (see Figure 1). Provided the points x i, where linearity breaks, it is possible to express all such functions using basis functions ϕ i ('hat' functions) v(x) = η i ϕ i (x) Variational formulation Another idea is to rewrite a dierential equation using weak formulation, which has almost the same solution as the original problem. Particularly, Poisson's equation in 1D u (x) = f(x), x [, 1] may be rewritten using variational formulation u vdx = fvdx, v V v(x) 1 ϕ 2 1 Figure 1: Test function ϕ 2 and approximation v(x) from the set V h 1

2 where test functions v are usually taken to be the same as basis functions ϕ i above. It is not convenient to have second-order derivative u, so it is possible to replace it with the rst-order derivatives using integration by parts 1.1 Dirichlet boundary condition u vdx = u v 1 + u v dx (1) If we a given Dirichlet boundary condition u() = u(1) =, then Equation 1 looses u v (v is from the same space as u, thus v() = v(1) = ). The M M stiness matrix A = (a ij ) and right-hand side b are calculated with a ij = ϕ iϕ jdx, b i = fϕ i dx, i, j = 1,..., M Task 1 Given x i and f(x) write a script that solves the Poisson equation with Dirichlet boundary condition using FEM. Hints: ˆ the initialization may look like N = 1 xs = linspace (,1, N +1) M = len ( xs ) -2 A = np. asmatrix ( np. zeros ((M,M ))) def f(x ): return -.4+ x 1 ˆ ϕiϕjdx may be calculated directly using h values (see lecture slides) 1 ˆ fϕi could be calculated using scipy.integrate.quad() phi = np. vectorize ( get_phi ( i )) xa = xs [i -1] xb = xs [i +1] r = integrate. quad ( lambda x:f(x )* phi (x),xa, xb )[] ˆ where get_phi(j) returns ϕ j function def get_phi (j ): def _phi (x ): if x <= xs [j -1] or x >= xs [j +1]: return. elif x <= xs [j ]: return (x - xs [j -1])/( xs [j]- xs [j -1]) else : return ( xs [j +1] - x )/( xs [j +1] - xs [j ]) return _phi 2

3 (a) with uniform h (b) non-uniform h Figure 2: Solution for Poisson's equation with f(x) =.4 x Example. The following output shows the matrix A, the right-hand side b and the solution u (ys) for the uniform h =.1 and f(x) =.4 x: Listing 1: k = 5 and u () =.2 xs = [ ] A [[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]] b = [ ] ys = [. e e e e e e e e e e -2. e +] Figures 2a and 2b show solutions correspondingly for the uniform h =.1 and the following non-uniform case xs = np. array ([.,.33,.66,.1,.125,.15,.166,.2,.3,.4,.5,.6,.7,.8,.9, 1.]) 2 Other boundary conditions FEM is well applicapable with dierent boundary conditions and complex geometries. Here we try the former and the example of complex geometry can be presented with 2D FEM. 3

4 2.1 Mixed boundary condition Mixed boundary condition means that part of the boundary Γ (e.g left in our case) has Neumann boundary condition (u (x) = c) while other part (right) has Dirichlet boundary condition (u(x) = d). Particularly, in our case we try u () = c, u(1) = First observation is that the test function ϕ is now non-zero because left edge is not xed, so we have to introduce new unknown u. The test function is however only half of a typical test function. Another observation comes from Equation 1 where u v with x =. This adds u ()v() = c 1 to the right-hand side b. Task 2 Write a script that given x i and f(x) calculates u i using mixed boundary condition (left Neumann and right Dirichlet). Hints: ˆ matrix A and RHS vector b now have one more row (rst one), look carefully through all the indices and expressions in the code; ˆ A, is however only half of other diagonal values (because ϕ function has twice smaller non-zero domain); ˆ b must be updated with c. Example. With f(x) = 1, uniform h =.1 and u () = c =.5 A [[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]] b = [ ] ys = [ e e -2-8.e e e e e e e e -2. e +] Notice, that if you try Neumann BC on both sides you will probably get singular matrix A. This is because there are heat ows given at both sides and heat generation/consumption on domain [, 1] but no anchor temperature, so the heat equation either gives indenite temperature growth or has no meaningful solution. 2.2 Robin boundary condition Robin boundary condition can be given with u (x) = k (u(x) u D (x)) x Γ 4

5 Figure 3: Solution for Poisson's equation with f(x) = 1 and mixed BC This condition may be used when there is external temperature u D and diusion on the boundary at rate k. The temperature at the boundary tries to achieve u D unless the diusion is bad. Here we try Robin boundary condition on the left side x =. The matrix is almost the same as with Mixed boundary condition. From Equation 1 we get kϕ ()ϕ () = k must be added to A, and ku D ϕ () = ku D to b. Task 3 Write a script that given x i and f(x) calculates u i using Robin boundary condition (left Robin and right Dirichlet). Hints: ˆ generation of matrix A is the same is with Mixed boundary condition, except diagonal value in the rst row must be updated Example. With f(x) = 1, Robin condition on the left boundary u () = k (u() u D ), k = 5, u D =.2 and Dirichlet on the right boundary u(1) = we get the following A [[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]] b = [ ] ys = [ ] Figure 4 shows corresponding examples with dierent k values. Notice, that value on the boundary converges to the external value u D as the coecient k increases. 5

6 (a) k = 5 and u D () = (b) k = 5 and u D () =.2 Figure 4: Solution for Poisson's equation with f(x) = 1 and Robin BC 6

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