Unformatted text preview: alled quasi uniform if there
N i i i; exists a constant C > 0 independent of ;
h= bNa (8.3.3a) such that hC
i = 1 2 ::: N
N ! 1:
(8.3.3b)
h
Example 8.3.1. Consider a quadratically graded mesh on 0 x 1 (Figure 8.3.1)
i
x = ( N )2
i = 0 1 : : : N:
The mesh spacing for this example is
2 (i
2
;
h = i ; N 2; 1) = 2iN 2 1 :
Then,
h = 1=N = N
i = 1 2 : : : N:
h (2i ; 1)=N 2 2i ; 1
i i i i The smallest subinterval is the rst (i = 1) thus,
hN
i = 1 2 : : : N:
h
i 9 There is no bound as N ! 1 therefore, this sequence of quadratically graded meshes
is not quasi uniform.
The presence of a quasiuniform mesh enables us to show that the global error has
an expansion in even powers of the mesh spacing. Theorem 8.3.1. Suppose that there is a unique solution y(x) 2 C 2 K +2 (a b) of the linear BVP y = A(x)y + b(x) a<x<b 0 Ly(a) = l (8.3.4a) Ry(b) = r: (8.3.4b) Let nite di erence solutions of (8.3.4) be generated by the trapezoidal rule on a sequence
of quasiuniform meshes fa = x0 < x1 < : : : < x = bg, N ! 1. Then the global error
satis es
N y(x ) ; y =
i X
K i k =1 d (x )h2 + O(h2
k k i K i +2 ) i = 0 1 ::: N (8.3.5a) where d (x), k = 1 2 : : : K , are smooth functions satisfying
k d ; A(x)d = T (y(x)) ;
0 k k 1
X
k; k l =1 T (d ; l(x))
l (8.3.5b) k Ld (a) = Rd (b) = 0:
k (8.3.5c) k Proof. ( 1], Section 5.5). Given a trapezoidal rule solution of (8.3.4) on a mesh fx0 <
x1 < : : : < x g, let
N v = y(x ) ; y ;
i i i X
K d (x )h2 i = 0 1 : : : N: k k =1
O(h2 +2) i i k Our task is to show that jvj =
where v = v1 v2 : : : v ] . Utilizing
(8.1.2a), (8.1.3), and the linearity of the problem
K 1 i (v ) = (y(x ) ; y ) ;
i i i X
K i k =1 i (d k T N (x ))h2
i i k = i ; X
K k 10 =1 (d (x ))h2 :
k i k i i Using the expansion (8.3.2a) for the local discretization error
X
(v ) = h2 T (y(x 1 2 )) ; (d (x ))] + O(h2 +2):
K k i i k =1 (8.3.6) K k i i; = k k i Since y(x) is smooth, the data A(x) and b(x) must also be smooth. With (8.3.2b),
this further implies that T , k = 1 2 : : : K , are smooth. These results can be used with
(8.3.5b,c) to inductively show that d (x) 2 C 2( )+3. With this level of smoothness, we
can replace y(x) and K in (8.3.1) and (8.1.3) by d (x) and K ; k to obtain
X
(d (x )) = d (x ) ; A(x )d (x ) + h2 T d (x 1 2 )] + O(h2( +1)):
k K ;k k k K ;k l 0 k k i i k i k i l =1 K ;k l i k i; = Using (8.3.5b)
k (d (x )) = T (y(x 1 2 )) ;
k i k 1
X
k; i; = l T (d (x
l =1 k ;l 1 2 ))+ X2
h T (d (x 1 2 ))+ O(h2( K ;k i; = l l =1 K ;k l i k i; = Substituting this result into (8.3.6)
1
X2X
X
(v ) = h
T (d (x 1 2 )) ; h2 T (d (x 1 2 )) + O(h2(
K k; K ;k k i i k =1 l i l =1 l k ;l i; = l i =1 l k K ;k i; = +1)): +1) )]: All but the highest powers of h vanish leaving
i (v ) = O(h2 +2):
K i i Ascher et al. 1] show that the operator is stable. We'll omit their proof, but use
the result with the homogeneous boundary data (8.3.5c) to infer that v = O(h2 +2),
i = 0 1 : : : N.
i i K Remark 1. The results appear to hold for nonlinear problems with smooth solutions
and for graded meshes that are not quasi uniform.
Knowing that the global error of the trapezoidal rule has an expansion in even powers of...
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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, Boundary value problem, Piecewise linear function, Infantry regiments of Sweden, BVP

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