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Unformatted text preview: ntity h=2 is called the cell Peclet or cell Reynolds number. If h=2
= 1 + h + O(( h )2) = eh= + O(( h )2) (9.5.4f)
1, then 30 Parabolic Problems which is the correct solution. However, if h=2 1, then
;1 and
1 if i is even
ci
2 if i is odd
when N is odd and
(N + i)=N if i is even
ci
O(1= )
if i is odd
when N is even. These two obviously incorrect solutions are shown with the correct
results in Figure 9.5.1.
Let us try to remedy the situation. For simplicity, we'll stick with an ordinary di erential equation and consider a twopoint boundary value problem of the form L u] = ; u00 + !u0 + qu = f 0<x<1 (9.5.5a) u(0) = u(1) = 0:
(9.5.5b)
Let us assume that u v 2 H01 with u0 and v0 being continuous except, possibly, at
x = 2 (0 1). Multiplying (9.5.5a) by v and integrating the second derivative terms by
parts yields (v L u]) = A(v u) + u0v]x= (9.5.6a) A(v u) = (v0 u0) + (v !u0) + (v qu) (9.5.6b) Q]x= = lim Q( + ) ; Q( ; )]:
!0 (9.5.6c) where We must be careful because the \strain energy" A(v u) is not an inner product since
A(u u) need not be positive de nite. We'll use the inner product notation here for
convenience.
Integrating the rst two terms of (9.5.6b) by parts
(v L u]) = (L v] u) ; (v0 u ; u0v) + !vu]x=
or, since u and v are continuous
(v L u]) = (L v] u) ; (v0u ; u0v)]x= (9.5.7a) The di erential equation L v] = ; v00 ; (!v)0 + qv: (9.5.7b) with the boundary conditions v(0) = v(1) = 0 is called the adjoint problem and the
operator L ] is called the adjoint operator. 9.5. ConvectionDi usion Systems 31 De nition 9.5.1. A Green's function G( x) for the operator L ] is the continuous function that satis es L G( x)] = ; Gxx ; (!G)x + qG = 0 x 2 (0 ) ( 1) (9.5.8a) G( 0) = G( 1) = 0 (9.5.8b) Gx( x)]x= = ; 1 : (9.5.8c) Evaluating (9.5.7a) with v(x) = G( x) while using (9.5.5a, 9.5.8) and assuming that
1) gives the familiar relationship u0(x) 2 H 1(0 u( ) = (L u] G( )) = Z 1
0 G( x)f (x)dx: (9.5.9a) A more useful expression for our present purposes is obtained by combining (9.5.7a) and
(9.5.6a) with v(x) = G( x) to obtain u( ) = A(u G( )): (9.5.9b) As usual, Galerkin and nite element Galerkin problems for (9.5.5a) would consist of
N
determining u 2 H01 or U 2 S0 H01 such that A(v u) = (v f ) 8v 2 H01 (9.5.10a) A(V U ) = (V f ) 8v 2 S0N : (9.5.10b) and
Selecting v = V in (9.5.10a) and subtracting (9.5.10b) yields A(V e) = 0 8v 2 S0N (9.5.10c) where e(x) = u(x) ; U (x): (9.5.10d) Equation (9.5.9b) did not rely on the continuity of u0(x) hence, it also holds when u
is replaced by either U or e. Replacing u by e in (9.5.9b) yields e( ) = A(e G( )): (9.5.11a) 32 Parabolic Problems Subtacting (9.5.10c) e( ) = A(e G( ) ; V ): (9.5.11b) Assuming that A(v u) is continuous in H 1, we have je( )j C kek1kG( ) ; V k1: (9.5.11c) Expressions (9.5.11b,c) relate the local error at a point to the global error. Equation
(9.5.11c) also explains superconvergence. From Theorem 7.2.3 we know that kek1 =
O(hp) when S N consists of piecewise polynomials of degree p and u 2 H p+1. The test
fun...
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 Spring '14
 JosephE.Flaherty
 The Land

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