{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Knowing the greens function we can construct test

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}