HAL Id: hal-01266859
https://hal.archives-ouvertes.fr/hal-01266859
Submitted on 3 Feb 2016
HAL
is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from
L’archive ouverte pluridisciplinaire
HAL, estdestinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de
A smooth fictitious domain/multiresolution method for elliptic equations on general domains
Ping Yin, Jacques Liandrat
To cite this version:
Ping Yin, Jacques Liandrat. A smooth fictitious domain/multiresolution method for elliptic equations
on general domains. Numerical Algorithms, Springer Verlag, 2016, �10.1007/s11075-015-0063-x�. �hal-
01266859�
A smooth fictitious domain/multiresolution method for elliptic equations on general domains
Ping Yin
∗Science college, Jiangnan University, Wuxi 214122, China
Jacques Liandrat
Centrale Marseille, I2M, UMR 7373, CNRS, Aix-Marseille Univ., Marseille, 13453, France
Abstract
We propose a smooth fictitious domain/multiresolution method for enhanc- ing the accuracy order in solving second order elliptic partial differential equations on general bivariate domains. We prove the existence and unique- ness of the solution of corresponding discrete problem and the interior error estimate which justifies the improved accuracy order. Numerical experiments are conducted on a cassini oval.
Keywords: Elliptic partial differential equations, Smooth fictitious domain methods, Multiresolution discretization, Interior error estimate
1. Introduction
Recently, multiresolution methods have became a new powerful tool in the numerical approximation of partial differential equations due to their high accuracy and related fast algorithms [1, 2]. However, it is difficult to construct scaling and wavelet functions adapted to the geometry of general domains. This is an important issue, especially in the case of moving bound- ary problems.
∗Corresponding author
Email addresses: wanderapple@gmail.com(Ping Yin), jliandrat@centrale-marseille.fr(Jacques Liandrat)
Fictitious domain methods were introduced to encounter this problem.
They consist in solving a problem raised on a general domain ω by merging ω into a larger and simpler domain Ω. Fictitious domain methods differ fun- damentally in the way how the boundary conditions are enforced on γ, the boundary of ω. In penalization methods, additional terms are added in Ω\ ω.
These terms are usually weighted with a coefficient
1². To the limit when ² goes to zero, the boundary conditions are enforced [3, 4]. The Lagrange mul- tipliers methods emerge from the optimization techniques under constraints.
A minimization problem on a subset is transformed into a saddle point prob- lem on a larger set. Lagrange multipliers as supplementary unknowns that may live on ω, on its boundary γ or on Ω \ ω [5, 6].
In this paper, the starting point is a fictitious domain method coupled with Lagrange multipliers on the boundary. By shifting the Lagrange mul- tipliers on a control boundary Γ located away from γ in the outer normal direction(as proposed in [7]), this new approach is shown to preserve the good approximation properties of multiresolution methods and to improve the accuracy order of the numerical solution on ω.
This work is a generalization to bivariate framework of the results pre- sented in [8, 9]. The paper is organized as follows. Section 2 presents the for- mulation of our smooth fictitious domain method and gives the approximate controllability result. Section 3 contains the multiresolution discretization and proves the existence of an unique solution of the discrete saddle point problem. Section 4 gives the main theorem of interior error estimate which theoretically justifies the improved accuracy order. We list the important aspects of numerical implementation in Section 5. Some numerical examples are presented in Section 6 to verify the accuracy of the proposed method.
Section 7 concludes this paper.
2. The smooth fictitious domain method
We consider a Dirichlet boundary-value problem on an open, bounded domain ω ⊂ R
2with Lipschitz boundary γ,
( u ˆ − ν4ˆ u = f in ω, ˆ
u = g on γ, (1)
where ν > 0, f ∈ L
2(ω) and g ∈ H
12(γ). In order to separate the difficulty of
approximating the differential operator and the integration of the boundary
condition as well as to overcome the difficulty of constructing basis functions adapted to the complicated domain ω, a fictitious domain/Lagrange mul- tiplier method (FDLM) [10] can be defined. It introduces a rectangular Ω containing ω, ¯ ω ⊂ Ω, and uses the Lagrange multipliers λ on the bound- ary γ to enforce the boundary conditions. Thus the extended problem has a solution u on Ω such that u|
ω= ˆ u [11]. Since the auxiliary variables λ represent the normal derivative jump of u at γ by green formula [12], the regularity of u is limited in the vicinity of γ such that u ∈ H
3/2−²(Ω), for any ² > 0. It is well known that the smoothness degree of the solution of a boundary-value problem does affect the order of convergence of a numerical solution [13]. J.Haslinger et al. [7] investigated a new formulation of ficti- tious domain methods connecting in putting the Lagrange multiplier λ on a control boundary Γ located outside of ¯ ω to enforce the boundary condition on γ(see in Figure 1). We will call it the smooth fictitious domain/Lagrange multiplier method (SFDLM). The Dirichlet boundary value problem (1) is
Figure 1: Geometry: ωis the original domain,γis its boundary. Γ is the control boundary and Ξ is the control domain with Ξ⊃ω. Ω represents the fictitious domain with Ω¯ ⊃Ξ.¯
then rewritten as:
Find (u, λ) ∈ H
1(Ω) × H
−1/2(Γ) such that, a(u, v) + b
2(v, λ) = ( ˜ f , v), ∀v ∈ H
1(Ω), b
1(u, µ) =< g, µ >
γ, ∀µ ∈ H
−1/2(γ),
(2)
where, if the trace operators B and B are defined, Bu := u|
γ∈ H
1/2(γ), Bu := u|
Γ∈ H
1/2(Γ), ∀u ∈ H
1(Ω) then
b
1(u, q) :=< Bu, q >
H1/2(γ)×H−1/2(γ):=< Bu, q >
γ, ∀q ∈ H
−1/2(γ),
b
2(u, p) :=< Bu, p >
H1/2(Γ)×H−1/2(Γ):=< Bu, p >
Γ, ∀p ∈ H
−1/2(Γ).
The bilinear form a(·, ·) reads a(u, v) =
Z
Ω
(uv + νOuOv)dx, ∀(u, v) ∈ H
1(Ω) × H
1(Ω),
and ˜ f is an extension of f from ω to Ω. Note that the bilinear forms b
1and b
2are distinct. This kind of problem can be considered as a generalization of saddle point problems which have been studied in [14, 15]. Since the derivative jump λ lays on Γ at a positive distance from γ, this new method is expected to produce a smoother solution in the control domain Ξ which motivates the proof of interior error estimate depending on the smoothness of u in Ξ. For any λ ∈ H
−1/2(Γ)(if H
p1(Ω) stands for the sobolev space of order 1 of periodic functions on Ω), there exists a unique solution ¯ u of the
problem (2’): (
Find ¯ u ∈ H
p1(Ω) such that a(¯ u, v) = ( ˜ f , v) − b
2(v, λ).
However, the extended problem (2) does not have a solution in general [16].
We will use the following approximate controllability result [7].
Lemma 1. Defining the linear mapping Φ : H
−1/2(Γ) 7→ H
1/2(γ) by Φ(λ) =
¯
u|
γ, where u ¯ is the solution of (2’). Then for any ² > 0, there exists a value of λ ¯ and the corresponding solution u ¯ associated to ¯ g = Φ(¯ λ) in (2’) such that
kˆ u − uk ¯
H1(ω)≤ ². (3) 3. The multiresolution discretization
The finite dimensional subspaces used in the numerical approximation of problem (2) include the multiresolution analysis on Ω, Γ, γ denoted by U
hΩ⊂ H
1(Ω), Q
Γh0⊂ H
−1/2(Γ) ,Q
γh00⊂ H
−1/2(γ), and the subspace V
hΩwhich is the image of U
hΩby the operator (I −ν4)
−1, i.e., V
hΩ= (I −ν4)
−1U
hΩ. This space is spanned by biorthogonal vaguelettes [17]. We call it the Petrov-Galerkin and wavelet-vaguelette method [18]. Then the discretization formulation of (2) reads
Find (u
h, λ
h0) ∈ U
hΩ× Q
Γh0, such that a(u
h, v
h) + b
2(v
h, λ
h0) = ( ˜ f , v
h), ∀v
h∈ V
hΩ, b
1(u
h, µ
h00) =< g, µ
h00>
γ, ∀µ
h00∈ Q
γh00.
(4)
We assume that Γ and γ are piecewise regular in the sequel. The definition of the various spaces reads as follows:
U
hΩ= span{Φ
Ωα, α ∈ K
j:= {(j, k
1, k
2), 0 ≤ k
1, k
2≤ 2
j− 1, (j, k
1, k
2) ∈ N
3}}, where ∀f ∈ L
2(R
2), f
j,k1,k2(x, y) = 2
jf(2
jx − k
1, 2
jy − k
2). It is an orthogonal m-order(m ∈ N) multiresolution
1on Ω with h = 2
−j(see [17]).
The basis function Φ
Ωαis the tensor product of φ
[0,Lj,k11](x) and φ
[0,Lj,k22](y) with Ω = [0, L
1] × [0, L
2], where φ
[0,Lj,k11](x)(φ
[0,Lj,k22](y)) is the scaling function on the interval [0, L
1](resp [0, L
2]) in the x(resp y) direction.
V
hΩ= span{θ
αΩ, α ∈ K
j}. θ
Ωα= (I − ν4)
−1Φ
Ωα. For any ², there exists an integer r > 0 and g ∈ U
˜hΩ, ˜ h = 2
−j−rsuch that kθ
αΩ− gk ≤ ² [19]. Thus we can approximate θ
αΩby an element in U
h˜Ωup to arbitrary precision.
Q
Γh0= span{φ
Γα0, α
0∈ K
j0:= {(j
0, k
0), (j
0, k
0) ∈ N
2}}. It is an orthogo- nal m’-order(m
0∈ N) multiresolution on Γ constructed by a map from the multiresolution on [0, 1] with h
0= 2
−j0[20].
Q
γh00= span{φ
γα00, α
00∈ K
j00:= {(j
00, k
00), (j
00, k
00) ∈ N
2}}}. It is an orthog- onal m”-order(m
00∈ N) multiresolution on γ with h
00= 2
−j00.
The matrix form of (4) is
Find U
h∈ R
4j, Λ
h0∈ R
2j0, such that µ I C
D 0
¶ µ U
hΛ
h0¶
= µ F
G
¶
, (5)
where the off-diagonal matrices C
α,α0=
Z
Γ
θ
αΩφ
Γα0ds, D
α00,α= Z
γ
Φ
Ωαφ
γα00ds, and the vectors at the right side
(F )
α= Z
Ω
f θ ˜
Ωαdx, (G)
α00= Z
γ
gφ
γα00ds.
Assuming the discretization level j
0and j
00are equal, the matrix involved in (5)
A :=
µ I C D 0
¶ .
1The multiresolution is of order mif it produces polynomials of degree up tom−1.
is square. To prove that there exists a unique solution of (4), we only need to show that A is invertible. We will use the following lemma [21].
Lemma 2. A is invertible if and only if C has a full-column rank and Ker(D) ∩ R(C) = {0},
where Ker(D) is the null space of matrix D and R(C) is the range space of matrix C.
Therefore, if the conditions in Lemma 2 are satisfied, (4) has a unique solu- tion.
Theorem 3. Let U
hΩ, V
hΩ, Q
Γh0and Q
γh00be constructed as previously. If Q
Γh0∈ H
s0−12(Γ) with s
0> 2, Q
γh00∈ H
s00−12(γ) with s
00> 1 such that j(s
0− 2) − j
0(s
0− 1) > B where B is a positive constant, then there exists a unique solution of (4).
Proof. We mimic the proof of Theorem 3.1 in [22]. The condition j(s
0− 2) − j
0(s
0− 1) > B ensures that the following LBB condition holds [23], i.e., there exist constants β
1, β
2such that
sup
vh∈VΩ vh6=0h
b
2(v
h, p
h0)
kv
hk
H2(Ω)≥ β
1kp
h0k
H−12(Γ)
, ∀p
h0∈ Q
Γh0, (6)
sup
uh∈UΩ uh6=0h
b
1(u
h, q ˜
h00)
ku
hk
H1(Ω)≥ β
2k˜ q
h00k
H−12(γ)
, ∀˜ q
h00∈ Q
γh00. (7) We then estimate the condition number of C
TC, D
TD and (DC)
TDC as follows.
-estimation of C
TC
According to the definition of the matrix C, it follows
∀w ∈ R
2j0, (C
TCw, w) = X
α∈Kj
( X
α0∈K0j
C
α,α0w
α0)
2= X
α∈Kj
| Z
Γ
θ
αΩX
α0∈Kj0
w
α0φ
Γα0|
2.
Substituting v
h= P
α∈Kj
z
αθ
αΩ, p
h0= P
α0∈Kj0
w
α0φ
Γα0in (6), we have sup
z∈R4j
R
Γ
P
α∈Kj
z
αθ
αΩP
α0∈Kj0
w
α0φ
Γα0k P
α∈Kj
z
αθ
Ωαk
H2(Ω)≥ β
1k X
α0∈Kj0
w
α0φ
Γα0k
H−1/2(Γ), ∀w
α0∈ R
2j0.
By the discrete Cauchy-Schwartz inequality, it yields Z
Γ
X
α∈Kj
z
αθ
αΩX
α0∈Kj0
w
α0φ
Γα0≤ (z, z)
1/2(C
TCw, w)
1/2.
Using this inequality, it leads to β
1k X
α0∈Kj0
w
α0φ
Γα0k
H−1/2(Γ)≤ sup
z∈R4j
R
Γ
P
α∈Kj
z
αθ
ΩαP
α0∈Kj0
w
α0φ
Γα0k P
α∈Kj
z
αθ
αΩk
H2(Ω)≤ (z, z)
1/2k P
α∈Kj
z
αθ
Ωαk
H2(Ω)(C
TCw, w)
1/2.
(8)
Due to θ
αΩ= (I − ν4)
−1Φ
Ωαand the orthonormality of the bases {Φ
Ωα}
αin the norm k · k
L2(R), it follows
k X
α∈Kj
z
αθ
αΩk
H2(Ω)∼ k X
α∈Kj
z
αΦ
Ωαk
L2(Ω)= (z, z)
1/2. (9)
Applying the Bernstein inequality[24] on the left term, we have k X
α0∈Kj0
w
α0φ
Γα0k
H−1/2(Γ)≥ c2
−j0/2(w, w)
1/2, (10)
where the constant C > 0 depends on the property of the multiresolution on the boundary Γ.
Combing the (8),(9) and (9) together, it follows
(C
TCw, w) ≥ K
1,c2
−j0(w, w), ∀w ∈ R
2j0.
On the other hand, we use the continuous Cauchy-Schwartz inequality to
get
Z
Γ
X
α∈Kj
z
αθ
αΩX
α0∈Kj0
w
α0φ
Γα0≤ k X
α∈Kj
z
αθ
Ωαk
L2(Γ)k X
α0∈Kj0
w
α0φ
Γα0k
L2(Γ)≤ Ck X
α∈Kj
z
αθ
Ωαk
H1(Ω)(w, w)
1/2≤ C(z, z)
1/2(w, w)
1/2,
where C > 0 only denotes a constant, does not mean the same value.
Choosing z
α= R
Γ
θ
αΩP
α0∈Kj0
w
α0φ
Γα0, then we have (C
TCw, w) ≤ K
2,c(w, w).
Therefore,
K
1,c2
−j0(w, w) ≤ (C
TCw, w) ≤ K
2,c(w, w). (11) -estimation of D
TD
We use the definition of the matrix D to get
∀z ∈ R
4j, (D
TDz, z) = X
α00∈Kj00
( X
α∈Kj
D
α00,αz
α)
2= X
α00∈Kj00
| Z
γ
φ
γα00X
α∈Kj
z
αΦ
Ωα|
2. Taking u
h= P
α∈Kj
z
αΦ
Ωα, ˜ q
h00= P
α00∈Kj00
w
α00φ
γα00in (7), it follows sup
z∈R4j
R
γ
P
α∈Kj
z
αΦ
ΩαP
α00∈Kj00
w
α00φ
γα00k P
α∈Kj
z
αΦ
Ωαk
H1(Ω)≥ β
2k X
α00∈Kj00
w
α00φ
γα00k
H−1/2(γ). (12) By the discrete Cauchy-Schartz inequality, it follows
Z
γ
X
α∈Kj
z
αΦ
ΩαX
α00∈K00j
w
α00φ
γα00≤ (w, w)
1/2(D
TDz, z)
1/2.
Using (12) and the Bernstein inequality of the multiresolution Q
γh00, we have (D
TDz, z)
1/2≥ β
22
−j00/2k X
α∈Kj
z
αΦ
Ωαk
H1(Ω)≥ β
22
−j00/2k X
α∈Kj
z
αΦ
Ωαk
L2(Ω)= β
22
−j00/2(z, z)
1/2.
The inequality above equals to
(D
TDz, z) ≥ K
1,D2
−j00(z, z).
By the continuous Cauchy-Schwartz inequality, it follows Z
γ
X
α∈Kj
z
αΦ
ΩαX
α00∈Kj00
w
α00φ
γα00≤ k X
α∈Kj
z
αΦ
Ωαk
L2(γ)k X
α00∈Kj00
w
α00φ
γα00k
L2(γ)≤ Ck X
α∈Kj
z
αΦ
Ωαk
H1(Ω)(w, w)
1/2≤ C2
j(z, z)
1/2(w, w)
1/2,
where we use the Bernstein inequality of the multiresolution U
hΩand the trace theorem.
Choosing w
α00= R
γ
φ
γα00P
α∈Kj
z
αΦ
Ωα, we get (D
TDz, z) ≤ K
2,D2
2j(z, z).
Therefore,
K
1,D2
−j00(z, z) ≤ (D
TDz, z) ≤ K
2,D2
2j(z, z). (13) -estimation of (DC)
TDC
Taking z = cw in (13), we have
2
−j00K
1,D(Cw, Cw) ≤ (D
TDCw, Cw) ≤ 2
2jK
2,D(Cw, Cw).
By (11), it yields
2
−(j0+j00)K
1,DK
1,C(w, w) ≤ (D
TDCw, Cw) ≤ 2
2jK
2,DK
2,C(w, w).
Denote K
1= K
1,DK
1,C, k
2= K
2,DK
2,C, and here j
0= j
00, we have K
12
−2j0(w, w) ≤ (DCw, DCw) ≤ K
22
2j(w, w), ∀w ∈ R
2j0,
Thus it yields that C has a full-column rank and Ker(D) ∩ R(C) = {0}. By
Lemma 2, the proof concludes.
4. Interior error estimate
We will use the notation A . B which means that there exits a constant c > 0 such that A ≤ cB, similarly for &. Local interior error estimate deals with the error between the computed solution u
hand ¯ u, a good approxima- tion of u on ω, on ¤ a subdomain of Ξ including ¯ ω. This estimate depends on the regularity of ¯ u. We first recall the results for the bilinear form a(·, ·) [22].
Lemma 4. For U
hΩ, V
hΩconstructed as previously, there exist sup
vh∈VΩ vh6=0h
a(u
h, v
h)
kv
hk
H1(Ω)& ku
hk
H1(Ω), ∀u
h∈ U
hΩ, (14)
sup
uh∈UΩ uh6=0h
a(u
h, v
h)
ku
hk
H1(Ω)& kv
hk
H1(Ω), ∀v
h∈ V
hΩ. (15) Then, we derive the interior error estimate as follows.
Theorem 5. Let ω ⊂⊂ ¤ ⊂⊂ Ξ ⊂⊂ Ω, u ˆ (resp u
h) be the solution of problem of (1) (resp problem (4)). Moreover, (¯ u, ¯ λ) is the solution of problem (2’) associated to g ¯ = Φ(¯ λ) such that u ¯ satisfies (3). Then if u ¯ ∈ H
s(¤), 1 < s ≤ m, λ ¯ ∈ H
s0(Γ), 0 < s
0≤ m
0, and j ≥ j
0(j
0depending on ω and Ξ), we have
kˆ u − u
hk
H1(ω). 2
−j(s−1)k¯ uk
Hs(¤)+ 2
−j0s0k λk ¯
Hs0(Γ)+ ², ∀² > 0.
Proof. Let ω ⊂⊂ ¤
0⊂⊂ ¤ ⊂⊂ Ξ ⊂⊂ Ω and δ = 1 on ω, δ ∈ C
0∞(¤
0) and set ˜ u = δ u. Let ¯ T u ˜ ∈ U
hΩbe the unique solution of
a(˜ u − T u, v ˜
h) = −b
2(v
h, λ ¯ − λ), ˜ ∀v
h∈ V
hΩ, (16) where ˜ λ ∈ Q
Γh0solves k ˜ λ − λk ¯
H−1/2(Γ)= inf
λh0∈QΓh0
kλ
h0− ¯ λk
H−1/2(Γ). The existence of T u ˜ is guaranteed by Lemma 4 and Lax-Milgram lemma.
Now
k˜ u − T uk ˜
H1(Ω). sup
v∈H1(Ω)
a(˜ u − T u, v) ˜ kv k
H1(Ω). sup
v∈H1(Ω)
( a(˜ u − T u, v ˜ − P
∗v)
kvk
H1(Ω)− b
2(P
∗v, λ ¯ − ˜ λ) kvk
H1(Ω)),
(17)
where P
∗v ∈ V
hΩsatisfies
a(u
h, v − P
∗v ) = 0, ∀u
h∈ U
hΩ. (18) Furthermore, it follows from (15) that
kP
∗vk
H1(Ω). sup
uh∈UhΩ
a(u
h, P
∗v)
ku
hk
H1(Ω)= sup
uh∈UhΩ
a(u
h, v)
ku
hk
H1(Ω). kvk
H1(Ω). Hence, for any η ∈ U
hΩ,
a(˜ u − T u, v ˜ − P
∗v) = a(˜ u − η, v − P
∗v) . k˜ u − ηk
H1(Ω)kvk
H1(Ω), (19) where we have used (18). Also
|b
2(P
∗v, λ ¯ − λ)| ˜ = |(BP
∗v, λ ¯ − λ) ˜
Γ| . kP
∗vk
H1(Ω)k λ ¯ − λk ˜
H−1/2(Γ). kvk
H1(Ω)2
−j0s0k λk ¯
Hs0(Γ), (20) where the trace theorem and Jackson estimate are used.
It follows from (17), (19) and (20) that k˜ u − T uk ˜
H1(Ω). inf
η∈UhΩ
k˜ u − ηk
H1(Ω)+ 2
−j0s0k ¯ λk
Hs0(Γ). 2
−j(s−1)k˜ uk
Hs(¤)+ 2
−j0s0k λk ¯
Hs0(Γ)
. 2
−j(s−1)k¯ uk
Hs(¤)+ 2
−j0s0k λk ¯
Hs0(Γ). Hence,
k¯ u − T uk ˜
H1(ω)= k˜ u − T uk ˜
H1(ω)≤ k˜ u − T uk ˜
H1(Ω). 2
−j(s−1)k¯ uk
Hs(¤)+ 2
−j0s0k ¯ λk
Hs0(Γ)
. (21)
Let us now estimate kˆ u − u
hk
H1(ω).
kˆ u − u
hk
H1(ω)≤ kˆ u − uk ¯
H1(ω)+ k¯ u − u
hk
H1(ω)≤ ² + k¯ u − T uk ˜
H1(ω)+ kT u ˜ − u
hk
H1(ω). (22) Note (¯ u, λ) (resp (u ¯
h, λ
h0)) is the solution of (2’) (resp (4)), we subtract these two equations to have
a(¯ u − u
h, v
h) + b
2(v
h, ¯ λ − λ
h0) = 0, ∀v
h∈ V
hΩ.
It follows then from (16) that
a(T u ˜ − u
h, v
h) = a(˜ u − u, v ¯
h) + b
2(v
h, λ
h0− λ), ˜ ∀v
h∈ V
hΩ. (23) Let P
VΩh
be the biorthogonal projector on V
hΩ[22], we have a(u
h, P
VΩh
v) = a(u
h, v), ∀(u
h, v) ∈ U
hΩ× H
1(Ω), and
kP
VΩh
vk
H1(¤). kvk
H1(¤), ∀v ∈ H
01(¤).
Hence,
kT u ˜ − u
hk
H1(ω)≤ kT u ˜ − u
hk
H1(¤0). sup
v∈H01(¤)
a(T u ˜ − u
h, v) kvk
H1(¤). sup
v∈H10(¤)
a(T u ˜ − u
h, P
VΩh
v) kP
VΩh
vk
H1(¤).
(24)
If the scale level j is sufficiently large, then P
VΩh
v ∈ ˚ V
hΩ(¤) := {v
h|v
h∈ V
hΩ, suppv
h⊂ ¤}, ∀v ∈ H
01(¤). It follows from (23) and (24) that
kT u ˜ − u
hk
H1(ω). sup
vh∈˚VhΩ(¤)
a(T u ˜ − u
h, v
h)
kv
hk
H1(¤)= sup
vh∈˚VhΩ(¤)
a(˜ u − u, v ¯
h) kv
hk
H1(¤). inf
η∈UhΩ
k˜ u − ηk
H1(¤)+ inf
η∈UhΩ
k¯ u − ηk
H1(¤). 2
−j(s−1)k¯ uk
Hs(¤).
(25)
By (21), (22) and (25), we have
kˆ u − u
hk
H1(ω). 2
−j(s−1)k¯ uk
Hs(¤)+ 2
−j0s0k λk ¯
Hs0(Γ)