Knowing the greens function we can construct test

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Unformatted text preview: 14a) (9.5.14b) (9.5.14c) 34 Parabolic Problems 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Figure 9.5.2: Canonical basis element ^(s) for = 0, 10, and 100 (increasing steepness). is the cell Peclet number. The value of ! will remain unde ned for the moment. The canonical basis element ^(s) is illustrated in Figure 9.5.2. As ! 0 the basis (9.5.14b) becomes the usual piecewise-linear hat function 8 1+s < ^(s) = 1 1 ; s 2: 0 if ; 1 s < 0 if 0 s < 1 otherwise As ! 1, (9.5.14b) becomes the piecewise-constant function ^(s) = 1 if ; 1 < s 0 : 0 otherwise The limits of this function are nonuniform at s = ;1 0. N We're now in a position to apply the Petrov-Galerkin method with U 2 S0 and ^ V 2 S0N to (9.5.5). The trial space S N will consist of piecewise linear functions and, for the moment, the test space will remain arbitrary except for the assumptions j (x) 2 H 0 1] 1 j (xk ) = Z jk 1 ;1 ^(s)ds = 1 j k = 1 2 : : : N ; 1: (9.5.15) 9.5. Convection-Di usion Systems 35 The Petrov-Galerkin system for (9.5.5) is ( i0 U 0 ) + ( i !U 0) + ( i qU ) = ( i f ) i = 1 2 : : : N ; 1: (9.5.16) Let us use node-by-node evaluation of the inner products in (9.5.16). For simplicity, we'll assume that the mesh is uniform with spacing h and that ! and q are constant. Then ( 0 U 0) = i Z h 1 ;1 ^0(s)U 0 (s)ds ^ ^ where U (s) is the mapping of U (x) onto the canonical element ;1 s 1. With a ^ piecewise linear basis for U and the properties noted in (9.5.15) for j , we nd ( i0 U 0 ) = ; h 2ci: (9.5.17a) ci = ci+1=2 ; ci;1=2 (9.5.17b) ci = ( ci) = ci+1 ; 2ci + ci;1 : (9.5.17c) We've introduced the central di erence operator for convenience. Thus, 2 Considering the convective term, !( i U 0) = ! Z 1 ;1 ^(s)U 0 (s)ds = !( ; ^ 2 =2)ci (9.5.18a) where is the averaging operator ci = (ci+1=2 + ci;1=2 )=2: (9.5.18b) ci = ( ci) = (ci+1 ; ci;1)=2: (9.5.18c) Thus, Additionally, =; Similarly q( i U ) = qh Z 1 ;1 Z 1 0 ^(s) ; ^(;s)]ds ^(s)U (s)ds = qh(1 ; ^ + (9.5.18d) 2 =2)ci (9.5.19a) 36 Parabolic Problems where = Z 1 ;1 =; Z jsj ^(s)ds 1 ;1 (9.5.19b) s ^(s)ds: (9.5.19c) Finally, if f (x) is approximated by a piecewise-linear polynomial, we have ( i f ) h(1 ; + 2 =2)fi (9.5.20) where fi = f (xi ). Substituting (9.5.17a), (9.5.18a), (9.5.19a), and (9.5.20) into (9.5.16) gives a di erence equation for ck , k = 1 2 : : : N ; 1. Rather than facing the algebraic complexity, let us continue with the simpler problem of Example 9.5.1. Example 9.5.3. Consider the boundary value problem (9.5.2). Thus, q = f (x) = 0 in (9.5.17-9.5.20) and we have ( i0 U 0 ) + !( i U 0) = ; h 2ci + !( ; =2)ci i = 1 2 ::: N ;1 (9.5.21a) or, using (9.5.14c), (9.5.17c), and (9.5.18c) 1 ; 2 ( + 2 )(ci+1 ; 2ci + ci;1) + ci+1 ; ci;1 = 0 2 This is to be solved with the boundary conditions i = 1 2 : : : N ; 1: (9.5.21b) c0 = 1 2 cN = 2: (9.5.21c) The exact solution of this second-order constant-coe cient di erence equation is ;i ci = 1 + 11; N i = 0 1 : : : N: (9.5.22a) where...
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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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