Unformatted text preview: n in the form Y (x) = N
X ci 0 (x)
i i=1 (6.4.2) and is shown in Figure 6.4.2. (In this case, the dimension of the subspace M and the
number of subintervals N are identical however, this need not be so.)
Y c1
c2
φ 0 (x)
1 1 cN x
x0 x1 x i1 x1
i xN Figure 6.4.2: Piecewise constant basis element 0(x) and the resulting piecewise constant
i
approximation Y (x).
Using (6.4.1) and (6.4.2) we see that Y (x) = ci x 2 xi;1 xi): We may interpret ci as Y (xi;1=2 ) however, this is not necessary. As shown in Figure
6.4.2, the basis 0(x) 2 C ;1 a b) thus, Y (x) 2 C ;1 a b).
i
Of course, the basis (6.4.1) doesn't satisfy any continuity requirements however, it
can be used to generate a continuous basis. More generally, a piecewisepolynomial basis
whose continuity increases with increasing degree can be constructed by integrating a
linear combination of basis elements of a piecewisepolynomial space having one degree
30 less than the desired degree. For example, let us construct a C 0 piecewiselinear basis by
integrating 0 and 0+1 as
i
i
Zx
1
0
0
i (x) =
i (s) + i+1 (s)]ds:
x0 The appropriate continuity will be automatically obtained by the integration. The result
for this example is
8
if x < xi;1
>0
>
< (x ; x )
if xi;1 x < xi
i;1
1
i (x) = > hi + (x ; xi ) if xi x < xi+1
>
: h+ h
if x
x
i i+1 i+1 where hi = xi ; xi;1 : (6.4.3) The parameters and are at our disposal. Let us pick them so that the resulting
approximation has compact support, i.e., so that
i (x) = 0 x xi+1 : 1 Let us, furthermore normalize the basis so that
i (xi ) = 1:
1 Then, we nd
i and the nal result =1
hi i =; 1
hi+1 8 x;xi;1
< x hi;x if xi;1 x < xi
1
i+1
if xi x < xi+1 :
i (x) = : hi+1
0
otherwise (6.4.4) The approximation Y (x) has the form Y (x) = N
X
i=0 31 ci 1(x):
i (6.4.5) Y
Y(x) c1
ci cN c0
φ 1 (x)
1 1 x
x0 x1 x i1 x1
i x i+1 xN Figure 6.4.3: Piecewiselinear basis element 1(x) and the resulting piecewiselinear api
proximation.
The basis element 1(x) and the piecewiselinear approximation Y (x) are shown in Figure
i
6.4.3. Using (6.4.4) we see that 1 (xj ) = ij where ij is the Kronecker delta. Using this
i
with (6.4.5) yields Y (xi ) = ci, i = 0 1 : : : N . The restriction of Y (x) to the subinterval
xi;1 xi) is the linear function Y (x) = ci;1 1;1(x) + ci 1(x)
i
i x 2 xi;1 xi): We'll continue by constructing a C 1 piecewisequadratic polynomial basis as
Zx
2
1
1
i (x) =
i;1 (s) + i (s)]ds:
x
0 Proceeding in three steps, we see that
8
if x < xi;1
>0
> (x;xi;1 )2
>
Zx
<
if xi;1 x < xi
1
(s)ds = > hi 2hihi+1 (xi+1;x)2
i
if xi x < xi+1
x0
> 2 + 2 ; 2hi+1
> hi hi+1
:+
if xi+1 x
2
2
and
80
if xi;2 < x
>
> (x;xi;2 )2
>
>
if xi;2 x < xi;1
>
1 < hi;1
(xi ;x)2
(x;xi;1 )2
2
if xi;1 x < xi :
i (x) = 2 > hi;1 + hi ; hi ] +
> (h + h ) + h + h ; h(ixi+1 ;x)2 ] if x x < x
> i;1 i
>
i
i+1
i
i+1
>
hi+1
:
(hi;1 + hi ) + (hi + hi+1)
if xi+1 x
32 Enforcing the condition that 2(x) = 0, x xi+1 implies
i
2
= ;h + h :
= h 2+ h
i;1
i
i
i+1
Thus,
8 (x;x )2
;
> hi;1 (hii;12+hi )
if xi;2 x < xi;1
>
>
>
< 1 ; (xi ;x)2 ; (x;xi;1 )2 if x
i;1 x < xi
2
hi i
:
i (x) = > (xi+1(hx;21 +hi ) hi (hi +hi+1)
;)
> hi+1 (hi +hi+1 )
if xi x < xi+1
>
>
:0
otherwise (6.4.6a) Normalizing the result so that 2(xi;1=2 ) = 1 yields
i
= 1; ;1
h
hi
; 4(h +ih ) :
4(hi;1 + hi)
i
i+1 (6.4.6b) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
−1 −0.5 0 0.5 1 1.5 2 Figure 6.4.4: Quadratic spline basis element 2(x) relative to a mesh with xi;2 = ;1,
i
xi;1 = 0, xi = 1, and xi+1 = 2.
The basis f 2(x)gN de nes a quadratic spline. The element
i
i=1
6.4.4. Evaluating (6.4.6a,b) at a node xj yields
8 hi;1
> hi;1 +hi if j = i ; 1
<
2
+1
(xj ) = > hihihi+1 if j = i
:
i
: 0+
otherwise
33 2 i is shown in Figure (6.4.6c) The basis simpli es when the mesh is uniform thus, if hi = h, i = 1 2 : : : N ,
8 x;xi;2 2
if xi;2 x < xi;1
>( h )
>
;
2 < 2 ; ( xih x )2 ; ( x;xi;1 )2 if xi;1 x < xi :
2
h
(6.4.6d)
i (x) = 3 > ( xi+1 ;x )2
if xi x < xi+1
>h
:0
otherwise
The quadratic spline approximation is written in the form
N
X2
Y (x) = ci i (x)
i=1 (6.4.7a) and its restriction to xi;1 xi ) is
82
x 2 x0 x1 )
< c1 1 (x) + c2 2 (x)
2
Y (x) = : ci;1 2;1(x) + ci 2(x) + ci+1 2+1(x) x 2 xi;1 xi ) i 6= 1 N : (6.4.7b)
i
i
i
cN ;1 2 ;1(x) + cN 2 (x)
x 2 xN ;1 xN )
N
N
Thus, three elements of the spline basis are nonzero on any interval except the rst and
the last where there are only two nonzero elements.
Using (6.4.6) with (6.4.7b) we see that
8 c1 h0
> h0 +h1
< c h +c h if i = 1
i
i+1
Y (xi ) = > i h+1 hi+1 i if i = 2 3 : : : N ; 1 :
(6.4.7c)
i+
: cN hN
if i = N
hN +hN +1 One could take h0 = h1 and hN +1 = hN .
When solving BVPs or interpolation and approximation problems, it's convenient to
introduce an extra basis element and unknown at e...
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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, Boundary value problem, di erences, di erence

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