2 semi discrete galerkin problems 3 as usual we apply

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Unformatted text preview: kin Problems 3 As usual, we apply the divergence theorem to the second-derivative terms in L to reduce the continuity requirements on u. When L has the form of (9.2.1b), the Galerkin problem 1 consists of determining u 2 HE (t > 0) such that (v ut) + A(v u) = (v f )+ < v ; u > 8v 2 H01 t > 0: (9.2.2a) The L2 inner product, strain energy, and boundary inner product are, as with elliptic problems, (v f ) = A(v u) = ZZ ZZ vfdxdy (9.2.2b) p(vxux + vy uy ) + vqu]dxdy (9.2.2c) and < v pun >= Z @N vpunds: (9.2.2d) The natural boundary condition (9.2.1e) has been used to replace pun in the boundary inner product. Except for the presence of the (v ut) term, the formulation appears to the same as for an elliptic problem. Initial conditions for (9.2.2a) are usually determined by projection of the initial data (9.2.1c) either in L2 (v u) = (v u0) 8v 2 H01 t=0 (9.2.3a) or in strain energy A(v u) = A(v u0) 8v 2 H01 t = 0: (9.2.3b) Example 9.2.1. We analyze the one-dimensional heat conduction problem ut = (pux)x + f (x t) 0<x<1 t>0 u(x 0) = u0(x) 0x1 u(0 t) = u(1 t) = 0 t>0 thoroughly in the spirit that we did in Chapter 1 for a two-point boundary value problem. A Galerkin form of this heat-conduction problem consists of determining u 2 H01 satisfying t>0 (v ut) + A(v u) = (v f ) 8v 2 H01 4 Parabolic Problems U(x,t) cj c1 cN-1 x 0 = x0 x1 xj xN-1 xN = 1 Figure 9.2.1: Mesh for the nite element solution of Example 9.2.1. (v u) = (v u0) where A(v u) = Z 8v 2 H01 1 0 t=0 vxpuxdx: Boundary terms of the form (9.2.2d) disappear because v = 0 at x = 0 1 with Dirichlet data. We introduce a mesh on 0 x 1 as shown in Figure 9.2.1 and choose an approxiN mation U of u in a nite-dimensional subspace S0 of H01 having the form U (x t) = X N ;1 j =1 cj (t) j (x): Unlike steady problems, the coe cients cj , j = 1 2 : : : N ;1, depend on t. The Galerkin N nite element problem is to determine U 2 S0 such that ( j Ut ) + A( j U ) = ( j f ) t>0 ( j U ) = ( j u0 ) t=0 j = 1 2 : : : N ; 1: Let us chose a piecewise-linear polynomial basis 8 x;x ; > x ;x ; <; (x) = > xx ;xx k :0 k1 k k1 k+1 k+1 k if xk;1 < x xk if xk < x xk+1 : otherwise This problem is very similar to the one-dimensional elliptic problem considered in Section 1.3, so we'll skip several steps and also construct the discrete equations by vertices rather than by elements. 9.2. Semi-Discrete Galerkin Problems Since j 5 has support on the two elements containing node j we have A( j U ) = Zx j 0 pU j xj ;1 x dx + where ( )0 = d( )=dx. Substituting for j and Ux Z xj 1 cj ; cj;1 Z A( U ) = p(x)( )dx + j xj ;1 hj hj where Zx j +1 0 pU j xj xj +1 x dx ; h 1 p(x)( cj+1 ; cj )dx h j +1 xj j +1 hj = xj ; xj;1: Using the midpoint rule to evaluate the integrals, we have j ; A( j U ) pjh1=2 (cj ; cj;1) ; ph+1=2 (cj+1 ; cj ) j j +1 where pj;1=2 = p(xj;1=2 ). Similarly, ( or ( j Ut ) = Zx j xj ;1 j Ut ) = Zx j xj ;1 U dx + jt c _ j (_j ;1 j ;1 + cj j )dx + Zx j +1 U dx jt xj Zx j +1 j xj (_j j + cj+1 c _ j +1 )dx where (_) = d( )=dt. Since the integrands are quadratic functions of x they may be integrated exactly using...
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.

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