Non-conforming finite element methods for transmission eigenvalue problem

0.29 MB·

by  Yidu Yang

Non-conforming finite element methods for
transmission eigenvalue problem
6
1
Yidu Yang, Jiayu Han, Hai Bi
0
2

n School of Mathematics and Computer Science,
a
J Guizhou Normal University,Guiyang, 550001, China

6 [email protected],[email protected], [email protected]

]
A
N
Abstract : The transmission eigenvalue problem is an important and challeng-
.
h ing topic arising in the inverse scattering theory. In this paper, for the Helmholtz
t
a transmissioneigenvalueproblem,wegiveaweakformulationwhichisanonselfadjoint
m linear eigenvalue problem. Based on the weak formulation, we first discuss the non-
[ conformingfiniteelementapproximation,andprovetheerrorestimatesofthediscrete

eigenvalues obtained by the Adini element, Morley-Zienkiewicz element, modified-
1
Zienkiewicz element et. al. And we report some numerical examples to validate the
v
efficiency of our approach for solving transmission eigenvalue problem.
8
6 Keywords : transmission eigenvalue, the weak formulation, non-conforming fi-
0 niteelements, error estimates.
1
0
. 1 Introduction
1
0
6 The transmissioneigenvalue problems haveimportant physicalbackground,for
1 example, they can be used to obtain estimates for the material properties of
:
v the scattering object [7, 8, 24]. In addition, transmission eigenvalues have the-
i oretical importance in the uniqueness and reconstruction in inverse scattering
X
theory [14]. Before 2010, significant progresses of the existence of transmission
r
a eigenvaluesand applications have been made (see [8] and the survey paper [9]).
In recent years, the computation of transmission eigenvalues has attracted
the attention of many researchers. The first numerical treatment of the trans-
missioneigenvalue problemappears in [15] where three finite element methods,
including the Argyris, continuous and mixed finite element methods, are pro-
posed for the Helmholtz transmission eigenvalues, and has been further devel-
oped by [2, 11, 15, 16, 17, 19, 25, 26, 29]. In particular, [11] studied the mixed
method using the Argyris conforming elements and [29] the H2 conforming fi-
nite elementmethod, andmaderigorouserroranalysis. Moreover,basedonH2
conformingfinite elementapproximations,the iterativemethodsin[25]andthe
1
multigrid method in [17] were proposed for computing real transmission eigen-
values,andtwo-gridmethodin[29]forcomputingrealandcomplextransmission
eigenvalues. And the spectral-element method was studied in [2]. However, to
the best of our knowledge, there has no research on the non-conforming finite
element methods for the transmission eigenvalues even for arbitrary nonselfad-
joint elliptic eigenvalue problem.
Inspired by the works mentioned above, we transform the fourth order
equation of transmission eigenvalue problem into a weak formulation, which
is suitable to nonconforming elements. This formulationis a nonselfadjoint lin-
ear eigenvalue problem (see (2.11)) with a selfadjoint, continuous and coercive
sesquilinearformA(, ). Basedontheweakformulationwebuildatypeofnon-
· ·
conforming finite element discretizations with good algebraic structure, includ-
ingtheAdinielement[1],modified-Zienkiewiczelement[27],Morley-Zienkiewicz
element [23], 12-parameter triangle plate element, 15-parameter triangle plate
element et. al. (see [23]). And we prove the error estimates of the numerical
eigenvalues. The proofdifficultylies inthe non-symmetryofright-handsidesof
eigenvalue problem that involves derivatives. To overcome this difficulty, based
on Babuska-Osborn spectral approximation theory [3], the new proof method
employedinthispaperistoestablishafundamentalrelationship(4.18)anduse
it to prove the optimal error estimates of non-conforming element eigenvalues.
For fourth order equation in R3, it is difficult to implement conforming
elements in H2, whereas many non-conforming elements such as the Morley-
Zienkiewicz element have had their three dimensional versions at present (e.g.,
see [23]). Hence it is an essential and significant work to study the non-
conforming element approximation for transmission eigenvalues.
Ournon-conformingfiniteelementdiscretizationis easytorealizeunder the
package of iFEM [12] with Matlab. We use the sparse matrix eigenvalue solver
eigs to compute the numerical eigenvalues, and numerical results indicate that
our methods are efficient for computing real and complex transmission eigen-
values as expected.
In this paper, regardingthe basic theory of finite element methods, we refer
to [3, 6, 13, 20, 23].
Throughout this paper, C denotes a positive constant independent of h,
which may not be the same constant in different places. For simplicity, we use
the symbol a.b to mean that a Cb.
≤
2 The weak formulation and non-conforming el-
ement method
Consider the Helmholtz transmission eigenvalue problem: Find k C, w,σ
∈ ∈
L2(Ω), w σ H2(Ω) such that
− ∈
∆w+k2nw =0, in Ω, (2.1)
∆σ+k2σ =0, in Ω, (2.2)
w σ =0, on ∂Ω, (2.3)
−
∂w ∂σ
=0, on ∂Ω, (2.4)
∂ν − ∂ν
where Ω Rd (d=2,3)is a bounded simply connectedinhomogeneousmedium,
⊂
ν is the unit outwardnormalto ∂Ω andthe index of refractionn(x) is positive.
2
Let Ws,p(Ω) denote the usual Sobolev space with norm , Hs(Ω) =
s,p
k·k
Ws,2(Ω), and = , H0(Ω)=L2(Ω) with the inner product (u,v) =
s,2 s 0
uvdx. Denotke·Hk2(Ω)=k·kv H2(Ω):v = ∂v =0 . Let H s(Ω) be the
0 { ∈ |∂Ω ∂ν|∂Ω } −
ΩR
“negative space”, with norm given by
(v,f)
0
v = sup | |.
s
k k− 06=f∈H0s(Ω) kfks
Itisclearthatforanyrealfunctionsv andv ,norms v +iv and v +
1 2 1 2 s,p 1 s,p
k k k k
v areequivalentinWs,p(Ω), andnorms v +iv and v + v
2 s,p 1 2 s 1 s 2 s
akrekequivalent in H s(Ω). k k− k k− k k−
−
SanindcDedeeLfifin2n(eeΩHH)il֒1be=rHtHsp011a((ΩcΩe))H×coH=m−Hp1a0(2cΩ(tΩl)y)w×(istLehe2(npΩop)r.m3w1i-kt3h(9vn,ioznr)mk[4H]k)1(va=,nzkd)vkHkH12+(=Ωkk)zv֒kk−21+H.k1z(kΩ0),
−
compactly, H→֒ H compactly. →
1
→
In this paper, we suppose that n = n(x) L (Ω) satisfying either one of
∞
∈
the following assumptions
(C1) 1+δ infn(x) n(x) supn(x)< ,
≤ Ω ≤ ≤ Ω ∞
(C2) 0<infn(x) n(x) supn(x)<1 β,
Ω ≤ ≤ Ω −
for some constant δ >0 or β >0.
From [9, 22] we know that the problem (2.1)-(2.4) can be written as an
equivalent fourth order equation for u=w σ H2(Ω):
− ∈ 0
1
(∆+k2n) (∆+k2)u=0,
n 1
−
i.e.,
1 n 1 n
∆( ∆u)= k2 ∆u k2∆( u) k4 u. (2.5)
n 1 − n 1 − n 1 − n 1
− − − −
Then the weak formulation for the transmission eigenvalue problem (2.1)-(2.4)
can be stated as follows: Find k C, u H2(Ω) such that
∈ ∈ 0
1 n 1
( ∆u,∆v) =k2( u, ( v) +k2( ( u), v)
0 0 0
n 1 ∇ ∇ n 1 ∇ n 1 ∇
− n − −
k4( u,v) , v H2(Ω). (2.6)
− n 1 0 ∀ ∈ 0
−
Introduce an auxiliary variable
ω =k2u, (2.7)
then
(ω,z) =k2(u,z) , z L2(Ω). (2.8)
0 0
∀ ∈
Thus, combining (2.6) and (2.8), we arrive at a linear weak formulation: Find
(k2,u,ω) C H2(Ω) L2(Ω) such that
∈ × 0 ×
1 1
( ∆u,∆v) =k2( ( u), v)
0 0
n 1 ∇ n 1 ∇
− n − n
+k2( u, ( v)) k2( ω,v) , v H2(Ω), (2.9)
∇ ∇ n 1 0− n 1 0 ∀ ∈ 0
− −
(ω,z) =k2(u,z) , z L2(Ω). (2.10)
0 0
∀ ∈
3
With this weak formulation, we have discussed the conforming finite element
approximations (see [29]). However, for the non-conforming element approxi-
mations,theweakformulationcannotguaranteethatthediscretebilinearform
satisfiesthe uniformH -ellipticity(seeRemark49.1in[13]). Tostudythe non-
h
conforming element approximations, next we will give a new weak formulation
referring to the weak formulation of the plate problem ( see (49.3) in [13]).
If (C1) holds, let
1
A((u,ω),(v,z))=(( µ )∆u,∆v) +(µ ∆u,∆v) +(ω,z)
1 0 1 0 0
n 1 −
−
1 ∂2u ∂2v
=(( µ )∆u,∆v) +µ dx+(ω,z) ,
1 0 1 0
n−1 − ΩZ 1≤Xi,j≤d∂xi∂xj ∂xi∂xj
1 n n
B((u,ω),(v,z))=( ( u), v) +( u, ( v)) ( ω,v) +(u,z) ,
0 0 0 0
∇ n 1 ∇ ∇ ∇ n 1 − n 1
− − −
and if (C2) holds, let
1
A((u,ω),(v,z))=(( µ )∆u,∆v) +(µ ∆u,∆v) +(ω,z)
2 0 2 0 0
1 n −
−
1 ∂2u ∂2v
=(( µ )∆u,∆v) +µ dx+(ω,z) ,
2 0 2 0
1−n − ΩZ 1≤Xi,j≤d∂xi∂xj ∂xi∂xj
1 n n
B((u,ω),(v,z))=( ( u), v) +( u, ( v)) ( ω,v) +(u,z) ,
0 0 0 0
∇ 1 n ∇ ∇ ∇ 1 n − 1 n
− − −
where µ > 0 and µ > 0 are chosen as good approximations of min( 1 ) and
1 2 n 1
min( 1 ) respectively such that 1 µ 0 and 1 µ 0. −
1 n n 1 − 1 ≥ 1 n − 2 ≥
Let λ−=k2, then (2.9)-(2.10) can b−e rewritten as: Fin−d λ C, (u,ω) H 0
∈ ∈ \{ }
such that
A((u,ω),(v,z))=λB((u,ω),(v,z)), (v,z) H. (2.11)
∀ ∈
Nextweshallseethatthediscretebilinearformof(2.11)satisfiestheuniform
H -ellipticity automatically for many non-conforming elements (see (2.22)).
h
Thus we get the following.
Theorem 2.1. The weak formulations (2.11) and (2.6) are equivalent.
Proof. If (k2,u) is an eigenpair of (2.6), then together with (2.8) we get
that (k2,u,ω) is an eigenpair of (2.9)-(2.10), thus it is an eigenpair of (2.11).
Conversely,if (k2,u,ω)satisfies (2.11), then (k2,u,ω) alsosatisfies (2.9)-(2.10);
from (2.10) we get ω = k2u, and substituting it into (2.9) we get (2.6). The
above argument indicates that (2.11) and (2.6) are equivalent. (cid:3)
For simplicity, in the next discussion we assume that (C1) holds. And the
argument is the same if (C2) holds.
It is obvious that A(, ) is a selfadjoint, continuous sesquilinear form on
H H, and · ·
×
A((v,z),(v,z)) µ v 2+ z 2 & (v,z) 2 , (2.12)
≥ 1| |2 k k0 k kH
i.e., A(, ) is coercive.
We·u·se A(, ) and A = A(, )12 as an inner product and norm on H,
· · k·k · ·
4
respectively.
Obviously, k = 0 is not an eigenvalue since A((u,ω),(u,ω)) = 0 implies
(u,ω)=0.
When n W1, (Ω), a simple calculation shows that
∞
∈
B((f,g),(v,z))
| |
1 n n
= ( ( f), v) +( f, ( v)) ( g,v) +(f,z)
0 0 0 0
| ∇ n 1 ∇ ∇ ∇ n 1 − n 1 |
− − −
. f v + f v + g v + f z
1 1 1 1 1 1 1 1
k k k k k k k k k k− k k k k k k−
.( f + g )( v + z )
1 1 1 1
k k k k− k k k k−
.k(f,g)kH1k(v,z)kH1, ∀(f,g),(v,z)∈H1. (2.13)
We can see from (2.13) that for any given (f,g) H , B((f,g),(v,z)) is a
1
continuous linear form on H. ∈
The source problem associated with (2.11) is as follows: Find (ψ,ϕ) H
∈
such that
A((ψ,ϕ),(v,z))=B((f,g),(v,z)), (v,z) H. (2.14)
∀ ∈
FromLax-Milgramtheoremweknowthat(2.14)hasoneandonlyonesolution.
Therefore, we define the corresponding solution operators T :H H by
1
→
A(T(f,g),(v,z))=B((f,g),(v,z)), (v,z) H. (2.15)
∀ ∈
Then (2.11) has the equivalent operator form:
1
T(u,ω)= (u,ω). (2.16)
λ
Theorem 2.2. Suppose n W1, (Ω), then T : H H is compact, and
∞
T :H H is compact. ∈ →
1 1
Pro→of. Let (v,z)=T(f,g) in (2.14), then from (2.12) and (2.13) we have
kT(f,g)k2H.A(T(f,g),T(f,g))=B((f,g),T(f,g)).k(f,g)kH1kT(f,g)kH1,
thus
T(f,g) H . (f,g) H , (2.17)
k k k k 1
which implies that T :H H is continuous. Because of the compact embed-
1
ding H֒ H , T :H H→is compact and T :H H is compact. (cid:3)
1 1 1
→ → →
Consider the dual problem of (2.11): Find λ C, (u ,ω ) H 0 such
∗ ∗ ∗
∈ ∈ \{ }
that
A((v,z),(u∗,ω∗))=λ∗B((v,z),(u∗,ω∗)), ∀(v,z)∈H. (2.18)
The source problem associated with (2.18) is as follows: Find (ψ ,ϕ ) H
∗ ∗
∈
such that
A((v,z),(ψ ,ϕ ))=B((v,z),(f,g)), (v,z) H. (2.19)
∗ ∗
∀ ∈
Define the corresponding solution operators T :H H by
∗ 1
→
A((v,z),T (f,g))=B((v,z),(f,g)), (v,z) H. (2.20)
∗
∀ ∈
5
Then (2.18) has the equivalent operator form:
T∗(u∗,ω∗)=λ∗−1(u∗,ω∗). (2.21)
It can be proved that T is the adjoint operator of T in the sense of inner
∗
product A(, ). In fact, from (2.15) and (2.20) we have
· ·
A(T(f,g),(v,z))=B((f,g),(v,z))=A((f,g),T (v,z)), (f,g),(v,z) H.
∗
∀ ∈
Note that since T is the adjoint operator of T, the primal and dual eigen-
∗
values are connected via λ=λ .
∗
Let π be a shape-regular mesh with size h. Let H = Sh Sh H and
h h 1
H H be a non-conforming finite element space; for exampl×e, Sh⊂ H1(Ω)
h 6⊂ ⊂ 0
is the finite element space associated with one of the Adini element, Morley-
Zienkiewiczelement, modifiedZienkiewicz element, 12-parametertriangleplate
element and 15-parameter triangle plate element et. al.
Let
1
A ((u ,ω ),(v,z))= ( µ )∆u ∆v
h h h 1 h
Z { n 1 −
κX∈πhκ −
∂2u ∂2v
h
+µ dx+(ω ,z) .
1 h 0
∂x ∂x ∂x ∂x }
1 Xi,j d i j i j
≤ ≤
Denote
A ((v,z),(v,z)) v 2 + z 2 (v,z) 2.
h ≡k kh k k0 ≡k kh
For the finite element spaces mentioned above, from [13] and Lemma 5.4.3
of [23], we know that A (, ) satisfies the uniform H -ellipticity.
h h
· ·
A ((v,z),(v,z))& v 2 + z 2, (v,z) H . (2.22)
h | |2,κ k k0 ∀ ∈ h
κX∈πh
Thus (v,z) isanorminH ,andthegeneralizedPoincare-Friedrichsinequal-
h h
k k
ity holds:
k(v,z)kH1 .k(v,z)kh, ∀(v,z)∈Hh. (2.23)
The non-conforming finite element approximation of (2.11) is given by the
following: Find λ C, (u ,ω ) H 0 such that
h h h h
∈ ∈ \{ }
A ((u ,ω ),(v,z))=λ B((u ,ω ),(v,z)), (v,z) H . (2.24)
h h h h h h h
∀ ∈
Consider the approximate source problem: Find (ψ ,ϕ ) H such that
h h h
∈
A ((ψ ,ϕ ),(v,z))=B((f,g),(v,z)), (v,z) H . (2.25)
h h h h
∀ ∈
We introduce the corresponding solution operator: T :H H :
h 1 h
→
A (T (f,g),(v,z))=B((f,g),(v,z)), (v,z) H . (2.26)
h h h
∀ ∈
Then (2.24) has the operator form:
1
T (u ,ω )= (u ,ω ). (2.27)
h h h h h
λ
h
6
Thenon-conformingfiniteelementapproximationof(2.18)isgivenby: Find
λ C, (u ,ω ) H 0 such that
∗h ∈ ∗h h∗ ∈ h\{ }
A ((v,z),(u ,ω ))=λ B((v,z),(u ,ω )), (v,z) H . (2.28)
h ∗h h∗ ∗h ∗h h∗ ∀ ∈ h
Define the solution operator T :H H satisfying
h∗ 1 → h
A ((v,z),T (f,g))=B((v,z),(f,g)), (v,z) H . (2.29)
h h∗ ∀ ∈ h
And (2.28) has the following equivalent operator form
Th∗(u∗h,ωh∗)=λ∗h−1(u∗h,ωh∗). (2.30)
It can be proved that T is the adjoint operator of T in the sense of inner
h∗ h
product A (, ). In fact, from (2.26) and (2.29) we have
h
· ·
A (T (u,ω),(v,z))=B((u,ω),(v,z))=A ((u,ω),T (v,z)), (u,ω),(v,z) H .
h h h h∗ ∀ ∈ h
Hence, the primal and dual eigenvalues are connected via λ =λ .
h ∗h
Denote
d
S=( ,2].
2
DefineinterpolationoperatorI1 :H2(Ω) W3,p(Ω) Sh (p S),anddefine
h 0 ∩ → ∈
I2 :L2(Ω) Sh by
h →
(ϕ I2ϕ,z) =0, z Sh.
− h 0 ∀ ∈
And let I (ψ,ϕ)=(I1ψ,I2ϕ).
h h h
For the finite element spaces mentioned above, when ψ W3,p(Ω) with
p S, the following estimates are valid: ∈
∈
kIh1ψ−ψkh .h1+(21−p1)dkψk3,p, (2.31)
kIh1ψ−ψks .h3−s+(21−p1)dkψk3,p, s=0,1, (2.32)
and when ϕ H1(Ω)
∈ 0
I2ϕ ϕ = inf ϕ v .h ϕ , (2.33)
k h − k0 v Shk − k0 k k1
∈
(I2ϕ ϕ,v I2v)
I2ϕ ϕ = sup h − − h 0 .h2 ϕ . (2.34)
k h − k−1 v∈H01(Ω) kvk1 k k1
3 The consistency term and Strang lemma
Let(ψ,ϕ)and(ψ ,ϕ )bethesolutionsof(2.14)and(2.19),respectively. Define
∗ ∗
the consistency terms: For any (v,z) H +H,
h
∈
D ((ψ,ϕ),(v,z))=B((f,g),(v,z)) A ((ψ,ϕ),(v,z)), (3.1)
h h
−
D ((v,z),(ψ ,ϕ ))=B((v,z),(f,g)) A ((v,z),(ψ ,ϕ )). (3.2)
h∗ ∗ ∗ − h ∗ ∗
The following estimations of the consistency term play an crucial role in our
analysis.
Dh((ψ,ϕ),(v,z)) .h1+(21−p1)d ψ 3,p v h, (3.3)
| | k k k k
|Dh∗((v,z),(ψ∗,ϕ∗))|.h1+(21−p1)dkψ∗k3,pkvkh. (3.4)
7
Next, we will prove the estimates (3.3) and (3.4) of the consistency term.
Itiswellknownthatthefollowing(C3)isvalidforthenon-conformingfinite
elements mentioned in Section 2 except Adini element (see Section 2.6 in [23]).
(C3) If F is the common face of element κ and κ, then
′
(v κ)ds= (v κ′)ds, v Sh; (3.5)
Z ∇ | Z ∇ | ∀ ∈
F F
if F is a face of element κ and F ∂Ω, then
∈
(v )ds=0, v Sh. (3.6)
κ
Z ∇ | ∀ ∈
F
Define the face and element average interpolation operators
1
P0f = fds, R0f =f P0f,
F meas(F)Z F − F
F
1
P0f = fdx, R0f =f P0f,
κ meas(κ)Z κ − κ
κ
where element κ π and F is an arbitrary element face of π .
h h
∈
A simple calculation shows that for arbitrary constant C ,
0
1
(f P0f)C ds=0. (3.7)
meas(F)Z − F 0
F
Theorem 3.1. Suppose that ψ,ψ W3,p(Ω) (p S), and (C3) is valid.
∗
Then for any (v,z) H +H, (3.3) and∈(3.4) hold. ∈
h
Proof. Forany∈(v,z) C (Ω) L2(Ω),bythe Green’sformulawededuce
∈ 0∞ ×
B((f,g),(v,z)) = A((ψ,ϕ),(v,z))
1
= ( ∆ψ) v+ϕzdx. (3.8)
Z −∇ n 1 ·∇
κX∈πhκ −
Since C (Ω) is dense in H1(Ω), for any (v,z) H +H the above (3.8) holds.
0∞ 0 ∈ h
8
Thus
D ((ψ,ϕ),(v,z))=B((f,g),(v,z)) A ((ψ,ϕ),(v,z))
h h
−
1 1
= ( ∆ψ) v+ϕzdx ( µ )∆ψ∆v
1
Z −∇ n 1 ·∇ − Z { n 1 −
κX∈πhκ − κX∈πhκ −
∂2u ∂2v
+µ dx (ϕ,z)
1 0
∂x ∂x ∂x ∂x } −
1 Xi,j d i j i j
≤ ≤
1
= ∆ψ v γds
− Z n 1 ∇ ·
κX∈πh∂κ −
∂2u ∂2v
+µ ∆ψ∆v dx
1
κX∈πhZκ { −1≤Xi,j≤d ∂xi∂xj ∂xi∂xj}
1
= ∆ψ v γds
− Z n 1 ∇ ·
κX∈πh∂κ −
∂2ψ ∂2v ∂2ψ ∂2v
+µ ( )dx
1κX∈πhZκ 1≤Xi6=j≤d ∂x2i ∂x2j −1≤Xi6=j≤d ∂xi∂xj ∂xi∂xj
1 ∂2ψ ∂v
= ∆ψ v γds+µ γ ds
−κX∈πh∂Zκ n−1 ∇ · 1κX∈πh∂Zκ 1≤Xi6=j≤d ∂x2i ∂xj j
∂2ψ ∂v
µ γ ds
1 i
− κX∈πh∂Zκ 1≤Xi6=j≤d∂xi∂xj ∂xj
I +I +I . (3.9)
1 2 3
≡
Note that (C3) and (3.7) are valid, and for all v H2(Ω) (3.5)-(3.6) also
hold, we deduce that (v,z) H +H, ∈ 0
h
∀ ∈
1
I = R0( ∆ψ)R0( v γ)ds. (3.10)
1 − Z F n 1 F ∇ ·
κX∈πhFX∈∂κF −
Let κˆ is a reference element, κ and κˆ be affine-equivalent. When wˆ W (κˆ)
1,ι
∈
and 1 ≤ ρ < (dd−1ι)ι, by the trace theorem we get W1,ι(κˆ) ֒→ Lρ(∂κˆ), thus we
deduce the follow−ing trace inequality:
∂κ
Z |w|ρds=Z |wˆ|ρ|∂κˆ|dsˆ.hdκ−1kwˆkρ0,ρ,∂κˆ .hdκ−1kwˆkρ1,ι,κˆ
∂κ ∂κˆ
.hdκ−1(kwˆkρ0,ι,κˆ+|wˆ|ρ1,ι,κˆ).hdκ−1(hκ−ριdkwkρ0,ι,κ+hκρ−ριd|w|ρ1,ι,κ)
.hdκ−ριd−1kwkρ0,ι,κ+hκρ+d−ριd−1|w|ρ1,ι,κ, ∀κ∈πh. (3.11)
Since p ∈ S, W1,p(κˆ) ֒→ Lρ(∂κˆ) with ρ ∈ (d−1,(dd−1p)p). Choose ρ1′ = 1− 1ρ,
then ρ′ < dd−21 and W1,2(κˆ)֒→Lρ′(∂κˆ). And thus, by−the Ho¨lder inequality, the
−
9
trace inequality (3.11) and the interpolation error estimate we deduce that
1
|I1|. kRF0(n 1∆ψ)k0,ρ,FkRF0(∇v·γ)k0,ρ′,F
κX∈πhFX∈∂κ −
1
. kRκ0(n 1∆ψ)k0,ρ,FkRκ0(∇v·γ)k0,ρ′,F
κX∈πhFX∈∂κ −
. (hκd−ρpd−1kRκ0(n 1 1∆ψ)kρ0,p,κ+hκρ+d−ρpd−1|Rκ0(n 1 1∆ψ)|ρ1,p,κ)ρ1
κX∈πh − −
×(hdκ−ρ2′d−1kRκ0(∇v·γ)kρ0,′κ+hκρ′+d−ρ2′d−1|Rκ0(∇v·γ)|ρ1,′κ)ρ1′
. (hκρ+d−ρpd−1)ρ1 ψ 3,p,κ (hρκ′+d−ρ2′d−1)ρ1′ v 2,κ
k k × k k
κX∈πh
.h1+(21−p1)d ψ 3,p v h, (v,z) Hh+H. (3.12)
k k k k ∀ ∈
Similarly we deduce
I2 .h1+(21−p1)d ψ 3,p v h, (v,z) Hh+H, (3.13)
| | k k k k ∀ ∈
I3 .h1+(21−p1)d ψ 3,p v h, (v,z) Hh+H. (3.14)
| | k k k k ∀ ∈
Substituting (3.12), (3.13) and (3.14) into (3.9) we get (3.3).
Using the same argument as above, we can prove (3.4). (cid:3)
Next, we shall analyze Adini rectangle element approximation. We suppose
that Ω R2, and the boundary of Ω and the edges of elements are parallel to
⊂
the coordinate axis. Although (C3) is not valid, thanks to [13], we can prove
(3.3) and (3.4) still hold.
Theorem 3.2. Suppose that Sh is Adini element space, then for any
(v,z) H (3.3) and (3.4) are valid.
h
Pr∈oof. We shall analyze the terms I , I and I on the right-hand side of
1 2 3
(3.9). Noticing that the edges of elements are parallel to the coordinate axis,
using the proof method of Theorem 50.1 in [13], we can deduce that for any
(v,z) H ,
h
∈
I1 + I2 .h1+(21−p1)d ψ 3,p v h.
| | | | k k k k
And from line 11 on page 304 in [13], we see
I =0.
3
Substituting the aboveestimatesinto(3.9)weget(3.3). Similarlywecanprove
(3.4). (cid:3)
The following lemma is a generalizationof Strang Lemma (1972).
Lemma 3.1. Let (ψ,ϕ) be the solution of (2.14) and (ψ ,ϕ ) be the
h h
10