we know that an interpolant of the form

N

y

su, = ± j » j (·, y)

j=1

(16.12)

n N

= ± j L 2 (·, x j ) + ± j B2 (·, x j )

j=1 j=n+1

uniquely exists that satis¬es

Lsu, (x j ) = f (x j ), 1 ¤ j ¤ n,

Bsu, (x j ) = g(x j ), n + 1 ¤ j ¤ N.

16.3 Solving PDEs by collocation 297

The subscripts 2 on B and L denote again the fact that these operators act with respect to

the second argument.

In the rest of this section we want to analyze the error for our approximation by giving

bounds on both L(u ’ su, )(x), x ∈ , and B(u ’ su, )(x), x ∈ ‚ . In the case of an

elliptic problem with Dirichlet boundary conditions these two bounds lead to a bound for

u ’ su, on . We will end the section with such an example.

To derive our estimates we have to prove a couple of results that are interesting in

themselves. Even if we have a differential operator L in mind, all results hold in the more

general case of a linear operator L with δx —¦ L ∈ N ( )— , x ∈ . Thus L could be for

example an integral operator of Volterra type. We will concentrate mainly on the analysis

for L. Similar steps have to be undertaken for B.

Theorem 16.7 Suppose that H is a real Hilbert space with reproducing kernel . Let

», μ ∈ H — . Then » y (·, y) ∈ H and

»( f ) = ( f, » y (·, y))

for all f ∈ H . Moreover

(», μ) H — = »x μ y (x, y).

Proof Riesz™ representation theorem guarantees the existence of an h » such that ( f, h » ) =

»( f ) for all f ∈ H . Since f x := (·, x) is an element of H we ¬nd that

»( f x ) = ( f x , h » ) = (h » , f x ) = (h » , (·, x)) = h » (x)

by the reproducing property of the kernel. This establishes » y (·, y) = h » ∈ H and the

¬rst property. For the second, note that

(», μ) H — = (h » , h μ ) = (h μ , h » ) = »(h μ ) = »x μ y (x, y).

the functionals δx —¦ L are in N ( )— . Thus

Under our assumptions on L (or B) and

we can de¬ne

y) := (δx —¦ L)u (δ y —¦ L)v (u, v), x, y ∈ .

L (x, (16.13)

By Theorem 16.7 we know that L is symmetric. It is also positive de¬nite under the

following mild condition: from now on we will assume that

{δx —¦ L : x ∈ } is linearly independent over N ( ). (16.14)

Theorem 16.8 Suppose that ∈ C( — ) is positive de¬nite. Suppose further that the

linear operator L satis¬es δx —¦ L ∈ N ( )— for x ∈ . Finally, assume that L satis¬es

(16.14). Then L is positive de¬nite on .

298 Generalized interpolation

Proof Let » = ± j δ y j with pairwise distinct y j ∈ be given. Then

j

± j ±k (δ y j —¦ L)u (δ yk —¦ L)v (u, v)

± j ±k L (y j , yk ) =

j,k j,k

= (» —¦ L)u (» —¦ L)v (u, v)

= »—¦L N ( )— ,

which is clearly nonnegative and vanishes only if ± = 0, by the conditions imposed

on L.

Our next result enlightens the connection between the kernels and L . Now that we

know that L is also a positive de¬nite function we have the whole machinery of native

spaces, power functions, etc. at hand.

∈ C( — ) is a positive de¬nite kernel and that

Theorem 16.9 Suppose that

L : N ( ) ’ C( ) satis¬es (16.14). Then L(N ( )) = N L ( ), and the following

mappings,

N ( )’N f ’ L f,

L: ( ),

L

— —

N ( ) ’N ( ) , » ’ » —¦ L,

L

( )— then L f ∈

are isometric isomorphisms. In particular, if f ∈ N ( ) and » ∈ N L

N L ( ) and » —¦ L ∈ N ( )— with

=f » = »—¦L N ( )— .

Lf and ( )—

N N() N

()

L L

Proof The proof will be given in several steps. First of all let us introduce the following

spaces:

L0 := span{δx : x ∈ },

L0 := {» —¦ L : » ∈ L0 } ⊆ N ( )— ,

T

F ( ) = {» y (·, y) : » ∈ L0 } ⊆ N ( ),

FT := {(» —¦ L) y (·, y) : » ∈ L0 } ⊆ N ( ).

0

When equipped with the inner product of the space N ( )— , the linear space L0 becomes

a subspace of N ( )— ; we will use the notation L0, to make this clear. Each of the other

spaces is assumed to carry the inner product of the space of which it is a subspace. The

space F ( ) is not new to us at all. We have already used this space to construct the native

space N ( ). We now proceed as follows.

(1) Since F ( ) is dense in N ( ) by construction, Riesz™ representation theorem en-

sures that L0, is dense in N ( )— . One of our goals here is to show that FT is also dense

0

in N ( ).

(2) Let us introduce the isometric isomorphism

T : F L ( ) ’ FT , »y y) ’ (» —¦ L) y (·, y),

0

L (·,

T — : L0, ’ L0 , » ’ » —¦ L,

T

L

16.3 Solving PDEs by collocation 299

induced by the operator L. Both mappings are indeed isometric, because

(» —¦ L)x (» —¦ L) y (x, y) = »u »v (δu —¦ L)x (δv —¦ L) y (x, y) = »u »v L (u, v)

for all » ∈ L0 . Theorem 16.7 and the norm-preserving property of the Riesz representer are

also helpful. Moreover, both mappings are surjective by construction.

(3) The restriction of the Riesz mapping

R |L0 : L0 ’ FT , » —¦ L ’ (» —¦ L) y (·, y)

0

T T

is also an isomorphic isomorphism. Thus we can describe the situation so far by the following

commutative diagram:

T*

L0 , ¦ L L0

T