• Keine Ergebnisse gefunden

Unsymmetric Weak Recovery

Im Dokument Sampling Inequalities and Applications (Seite 105-110)

7.4 Unsymmetric Weak Recovery

In this section we show how suitable sampling inequalities can be used to derive upper bounds for a weak unsymmetric recovery process from [47]. We briefly recall this process.

Since we are going to apply a non-symmetric method, we have to distinguish carefully between the tRial and the teSt side. As outlined in Section 2.3.1 we shall denote the test kernel withS:Rd×Rd→Rand the discretization byXs={x1, . . . , xNs} ⊂Ωwith the associated fill distancehs. On the tRial side we denote the kernel withR:Rd×Rd→R and the discretization withYr = {y1, . . . , yNr} ⊂ Ωwith fill distancehr. According to [45, Chapters 3.4 and 3.5] we use weak data of the form given in (2.1.7), where we replace KbyS, namely

λj(u) = Z

S(x, xj)u(x)dx , forxj ∈Xs, (7.4.1) where the kernelS:Rd×Rd→Rneeds to fulfill the following properties,

1. R

S(x, xj)dx= 1 for all xj ∈Xs,

2. supp(S(·, xj)) =Vj , diam(Vj) =δVj ≈hs, 3. kS(·, xj)kLp(Ω) ≤CδVjd/q≈hsd/q, 1p+ 1q = 1.

Under these conditions we may apply Theorem 7.2.1, i.e., there is a sampling inequality of the form

|u|Wp|α|(Ω) ≤Chτ+d/p−|α|s |u|Wpk+s(Ω)+ ˜Ch−|α|s max

1jNj(u)|

for all u ∈ W2τ(Ω)and all discrete test sets Xs. Now we consider the following prob-lem. An unknown function f ∈ W21(Ω) has to be recovered approximately from its data (λ1(f), . . . , λN(f))T. From the previous chapters we know that there is a good but unknown approximation. It is given by the best approximation from the trial space sf ∈ VR,Yr = span{R(· −yj) |yj ∈Yr} to the functionf ∈ W21(Ω). In Chapter 8 we obtain under certain conditions an error estimate of the form

kf −sfkL2(Ω)≤hrkfkW21(Ω) , (7.4.2) showing linear approximation order, which is optimal [48]. Unfortunately, this best appro-ximation is in general numerically unavailable. The idea from [47] is to solve the system

findsr∈VR,Yrj(f−sr) = 0for all1≤j≤Ns

to produce an approximation. Since this, however, is an unsymmetric kernel method, we first have to prove that this system is solvable. To do so, we proceed along the lines of [47], which we briefly outlined in Section 2.3.1. We assume an inverse inequality of the form

kskW21(Ω) ≤Cγ(Yr)kskL2(Ω) for alls∈VR,Yr .

Unfortunately, the value ofγ(Yr)is in general not known. There is a result in this direction in Chapter 6, namely ifRis a radial basis function with algebraic smoothnessτ > d/2and

ifYr is separated from the boundary thenγ(Yr) ≈ qr−τ. However, we can always make sure that the test-meshnorm is small enough to stabilize the reconstruction, i.e., we shall assume

hsγ(Yr)≤ 1

2C , C >1. (7.4.3)

Now we can prove the full-rank of the unsymmetric reconstruction of a function f ∈ W21(Ω)by an approximationsr ∈VR,Yr.

Lemma 7.4.1 Under the condition (7.4.3) the system

λj(f−sr) = 0, for all1≤j≤Ns. (7.4.4) is of full rank.

Proof:We have forsr=PNr

ℓ=1aΦ (· −y)the equivalence λj(f−sr) = 0for all1≤j≤Ns

λj(f) =

Nr

X

ℓ=1

aλj(Φ (· −y)) for all1≤j≤Ns. (7.4.5) If we introduce the notation

A := (Aℓ,j)1≤ℓ≤Nr 1j≤Ns :=

Z

R(x−y)S(x−xj)dx

1≤Nr 1j≤Ns

∈RNr×Ns ,

F :=



λ1(f) ... λNs(f)

∈RNs ,

equation (7.4.5) leads to the unsymmetric linear system

Aa=F . (7.4.6)

Now we can compute

kurkL2(Ω) ≤ C

hskurkW21(Ω)+ max

1jNsj(ur)|

≤ C

hsγ(Xr)kurkL2(Ω)+ max

1jNsj(ur)|

≤ 1

2kurkL2(Ω)+C max

1jNsj(ur)| . This yields

kurkL2(Ω)≤2C max

1jNsj(ur)| .

7.4. UNSYMMETRIC WEAK RECOVERY 103 Therefore,max1jNsj(ur)| = 0implieskurkL2(Ω) = 0. Sinceur ∈ VR,Yr we have ur ≡0. Thus rank ofA= #Yr ≤#Xs, which means that the system is of full rank. 2 Equation (7.4.2) implies that we can theoretically solve the system (7.4.4) to some pre-scribed accuracyδ(r, s). This is due to the fact sf is a good candidate for a solution, and by choosinghssmall enough we can achieve every prescribed valueδ(r, s). Practically we can apply some residual minimization techniques to find an approximate solution, such that the residual is smaller thanδ(r, s). We denote this approximate solution byur,s ∈ VR,Xr, i.e., we have

j(f−ur,s)| ≤δ(r, s)for all1≤j≤Ns. The exact error bound is given in the following theorem.

Theorem 7.4.2 We denote byur,s ∈ VR,Yr the approximate solution of the system (7.4.4) to the accuracyδ(r, s). Then we have an error bound of the form

kf −ur,skL2(Ω)≤C

2hs+ 1 C1hr

kfkW21(Ω)+δ(r, s). If we chooseδ(r, s) =

2hs+ C1

1hr

kfkW21(Ω), we get a final bound of the form kf−ur,skL2(Ω)≤C

2hs+ 1 C1hr

kfkW21(Ω) .

Proof:Let us denote byurthe unknown best trial approximation. Then we can compute 1

C kf−ur,skL2(Ω) ≤hskf−ur,skW21(Ω)+ max

1jNsj(f −ur,s)|

≤ hskfkW21(Ω)+hskur,skW21(Ω)+ max

1jNsj(f −ur,s)|

≤ hskfkW21(Ω)+hskurkW21(Ω) + +hskur,s−urkW21(Ω)+ max

1jNsj(f−ur,s)|

≤ 2hskfkW21(Ω)+hsγ(Xr)kur,s−urkL2(Ω)+ max

1jNsj(f−ur,s)|

≤ 2hskfkW21(Ω)+ 1

2C1kur,s−urkL2(Ω)+ max

1≤j≤Nsj(f−ur,s)|

≤ 2hskfkW21(Ω)+ 1 2C1

kur,s−fkL2(Ω)+kur−fkL2(Ω)

+

+ max

1jNsj(f−ur,s)|

≤ 2hskfkW21(Ω)+ 1

2C1 kur,s−fkL2(Ω)+ 1

2ChrkfkW21(Ω)

+ max

1≤j≤Nsj(f−ur,s)| . This shows

kf−ur,skL2(Ω)≤C

2hs+ 1 2C1

hr

kfkW21(Ω)+ max

1≤j≤Nsj(f −ur,s)| ,

giving the expected error bound kf −ur,skL2(Ω)≤C

2hs+ 1 2C1

hr

kfkW21(Ω)+δ(r, s).

The second inequality follows from the special choice forδ(r, s). 2 This generalizes the results of [47] in two ways. Our error analysis does not assumef to be known on a slightly larger domainΩ. Furthermore, we obtain results for the˜ L2-norm, not for negative order Sobolev norms.

Chapter 8

Galerkin Methods

In this chapter we shall combine the well-known theory of variational problems with the idea of sampling inequalities. Galerkin methods are a powerful tool for solving elliptic partial differential equations. We shall extend the theory given in [62] and show a new application in the theory of weak recovery [47]. Unfortunately, the error estimates in [62]

are given in theW21(Ω)-norm, and they require the solutionuto be in a native space of a radial basis function, which automatically forcesuto be continuous. In contrast, we shall give error estimates in theL2-norm, hence we do not assumeuto be continuous.

8.1 Model Problem: Elliptic Partial Differential Equations

We shall concentrate on problems of the following form. LetΩbe a bounded domain with boundary∂Ω∈C1. We shall consider partial differential equations of the form

− Xd i,j=1

∂xi

ai,j(x) ∂u

∂xj(x)

+c(x)u(x) = f(x) , x∈Ω Xd

i,j=1

ai,j|∂Ω(x)∂u(x)

∂xj νi(x) +h(x)u(x) = g(x) , x∈∂Ω, (8.1.1) where we assumeai,j, c ∈L(Ω),f ∈ L2(Ω),ai,j|∂Ω, h ∈L(∂Ω)andg ∈ L2(∂Ω) fori, j = 1, . . . , d. Further, we denote byν the outward pointing unit normal vector to the boundary∂Ω.

The weak formulation of (8.1.1) is given in terms of the bilinear form apde(u, v) :=

Z

 Xd i,j=1

ai,j∂u(x)

∂xj

∂v(x)

∂xi

+cuv

dx+ Z

∂Ω

huvdS (8.1.2)

and the linear form

Fpde(v) :=

Z

f vdx+ Z

∂Ω

gvdS (8.1.3)

105

for a functionv ∈W21(Ω). The matrixApde(x) := (ai,j(x))is assumed to be uniformly elliptic onΩ, i.e., there is a constantγpdesuch that

γpdekαk22 ≤αTA(x)α

for allα∈Rdand allx∈Ω. This gives rise to the variational problem

find u∈W21(Ω) : apde(u, v) =Fpde(v) for allv∈W21(Ω) . (8.1.4) If furtherc, h ≥0, and if at least one of them is uniformly bounded from zero on a subset of nonzero measure of Ω or ∂Ω, we know that apde(·,·) : W21(Ω)×W21(Ω) → R is strictly coercive, and thatF :W21(Ω)→Ris continuous. We point out that the boundary conditions are part of the weak formulation. This unfortunately excludes Dirichlet boundary conditions at the moment.

Im Dokument Sampling Inequalities and Applications (Seite 105-110)