A note of pointwise estimates on Shishkin meshes

0.13 MB·English

by  Jin Zhang

A note of pointwise estimates on Shishkin meshes
2
1 Jin Zhang∗†
0
2
January 30, 2012

n
a
J

7
Abstract
2
WeproposetheestimatesofthediscreteGreenfunctionforthestream-

] line diffusion finiteelement method (SDFEM) on Shishkinmeshes.
A
N
1 Problem
.
h
t
a We consider the singularly perturbed boundary value problem
m
[ (1.1a) Lu:=−ε∆u+b·∇u+u=f in Ω=(0,1)2,

(1.1b) u=0 on ∂Ω,
1
v
4 whereε≪1isasmallpositiveparameterandb=(b ,b )T >(0,0)T isconstant.
1 2
8 It is also assumed that f is sufficiently smooth.
6
5
. 2 The SDFEM on Shishkin meshes
1
0
2 2.1 Shishkin meshes
1
: Let N > 4 be a positive even integer. We use a piecewise uniform mesh — a
v
i so-called Shishkin mesh —with N mesh intervals in both x− and y−direction
X
which condenses in the layer regions. For this purpose we define the two mesh
r transition parameters
a
1 ε 1 ε
λ :=min ,2 lnN and λ :=min ,2 lnN .
x y
2 β 2 β
(cid:26) 1 (cid:27) (cid:26) 2 (cid:27)
Assumption 1. We assume in our analysis that ε ≤ N−1, as is generally
the case in practice. Furthermore we assume that λ = 2εβ−1lnN and λ =
x 1 y
2εβ−1lnN as otherwise N−1 is exponentially small compared with ε.
2
∗Email: [email protected]
†Address:SchoolofScience, Xi’anJiaotongUniversity,Xi’an,710049, China
1
Ω Ω
y xy
Ω Ω
s x
Figure 1: Dissection of Ω and triangulation ΩN.
The domain Ω is dissected into four parts as Ω = Ω ∪Ω ∪Ω ∪Ω (see
s x y xy
FIG. 1), where
Ω :=[0,1−λ ]×[0,1−λ ], Ω :=[1−λ ,1]×[0,1−λ ],
s x y x x y
Ω :=[0,1−λ ]×[1−λ ,1], Ω :=[1−λ ,1]×[1−λ ,1].
y x y xy x y
We introducethe setofmeshpoints{(x ,y )∈Ω:i, j =0, ··· , N}defined
i j
by
2i(1−λ )/N, for i=0, ··· , N/2,
x
x =
i
1−2(N −i)λ /N, for i=N/2+1, ··· , N
(cid:26) x
and
2j(1−λ )/N, for j =0, ··· , N/2,
y
y =
j
1−2(N −j)λ /N, for j =N/2+1, ··· , N.
(cid:26) y
Bydrawinglinesthroughthesemeshpointsparalleltothex-axisandy-axisthe
domainΩispartitionedintorectangles. ThistriangulationisdenotedbyΩN(see
FIG. 1). If D is a mesh subdomain of Ω, we write DN for the triangulation of
D. The mesh sizes h =x −x and h =y −y satisfy
x,τ i i−1 y,τ j j−1
1−λ
x
H := , for i=1, ··· , N/2,
x
N/2
h = 
x,τ λ
 hx := x , for i=N/2+1, ··· , N
N/2

2
and
1−λ
y
H := , for j =1, ··· , N/2,
y
N/2
h = 
y,τ λ
 hy := y , for j =N/2+1, ··· , N.
N/2
The mesh sizes h and h satisfy
x,τ y,τ
N−1 ≤H ,H ≤2N−1 and C εN−1lnN ≤h ,h ≤C εN−1lnN,
x y 1 x y 2
where C and C are positive constants and independent of ε and of the mesh
1 2
parameter N. The above properties are essential when inverse inequalities are
applied in our later analysis.
Forthemeshelementsweshallusetwonotations: τ =[x ,x ]×[y ,y ]
ij i−1 i j−1 j
for a specific element, and τ for a generic mesh rectangle.
2.2 The streamline diffusion finite element method
Let V :=H1(Ω). On the above Shishkin mesh we define a finite element space
0
VN :={vN ∈C(Ω¯):vN| =0 and vN| is bilinear, ∀τ ∈ΩN}.
∂Ω τ
In this case, the SDFEM reads as
Find U ∈VN such that for all vN ∈VN
(2.1)
ε(∇U,∇vN)+(b·∇U +U,vN +δb·∇vN)=(f,vN +δb·∇vN).
(cid:26)
where δ =δ(x) is a user-chosen parameter (see [3]).
We set
b −b
b:= b2+b2, β := 1 /b, η := 2 /b and v :=ζT∇v
1 2 b b ζ
q (cid:18) 2(cid:19) (cid:18) 1 (cid:19)
for any vector ζ of unit length. By an easy calculation one shows that
(∇w,∇v)=(w ,v )+(w ,v ).
β β η η
We rewrite (2.1) as
ε(U ,vN)+ε(U ,vN)+(bU +U,vN +δbvN)=(f,vN +δbvN)
β β η η β β β
and, following usual practice, we set
N−1, if x∈Ω ,
δ(x):= s
0, otherwise.
(cid:26)
Fortechnicalreasonsinthelateranalysis,weincreasethecrosswinddiffusion(see
[4]) by replacing ε(U ,vN) by εˆ(U ,vN) where
η η η η
ε˜:=max(ε,N−3/2)
3
and
ε˜, x∈Ω ,
s
εˆ(x):=
ε, x∈Ω\Ω .
(cid:26) s
We now state our streamline diffusion method with artificial crosswind:
Find U ∈VN such that for all vN ∈VN
(2.2)
B(U,vN)=(f,vN +δbvN),
(cid:26) β
with
(2.3) B(U,vN):=(ε+b2δ)(U ,vN)+εˆ(U ,vN)−b(1−δ)(U,vN)+(U,vN).
β β η η β
3 The discrete Green function
Let x∗ be a mesh node in Ω. The discrete Green’s function G∈VN associated
with x∗ is defined by
B(vN,G)=vN(x∗), ∀vN ∈VN.
The weighted function ω:
(x−x∗)·β (x−x∗)·η (x−x∗)·η
ω(x):=g g g −
σ σ σ
(cid:18) β (cid:19) (cid:18) η (cid:19) (cid:18) η (cid:19)
where
2
g(r)= for r∈(−∞,∞).
1+er
and σ =kN−1lnN and σ =kε˜1/2lnN.
β η
b
|||G|||2 :=(ε+b2δ)kω−1/2G k2+εˆkω−1/2G k2+ k(ω−1)1/2Gk2+kω−1/2Gk2
ω β η 2 β
and
(3.1) |||G|||2 =B(ω−1G,G)−(ε+b2δ)((ω−1) G,G )
ω β β
−εˆ((ω−1) G,G )−bδ(ω−1G,G ).
η η β
Thus, we obtain
B(ω−1G,G)=B(E,G)+B((ω−1G)I,G)
=B(E,G)+(ω−1G)(x∗)
where E :=ω−1G−(ω−1G)I.
Lemma 1. If σ =kN−1lnN and σ =klnNε˜1/2, then for k >1 sufficiently
β η
large and independent of N and ε, we have
1
B(ω−1G,G)≥ k|G|k2.
4 ω
4
Proof. From (3.1),we estimate the following terms.
(ε+δ) ((ω−1) G,G ) ≤C(ε+δ)1/2σ−1/2·k(ω−1)1/2Gk·(ε+δ)1/2kω−1/2G k
β β β β β
(cid:12) (cid:12)≤C(ε+δ)1/2σ−1/2k|G|k2
(cid:12) (cid:12) β ω
and
εˆ ((ω−1) G,G ) ≤Cεˆ1/2σ−1·kω−1/2Gk·εˆ1/2kω−1/2G k
η η η η
(cid:12) (cid:12)≤Cεˆ1/2σ−1k|G|k2
(cid:12) (cid:12) η ω
For bδ(ω−1G,G ), we make use of integration by parts.
β
From the definition of σ and σ and ε ≤ N−1, we take k sufficiently large
β η
and we are done.
Lemma 2. If σ = kN−1lnN, with k > 0 sufficiently large and independent
β
of N and ε. Then for each mesh point x∗ ∈Ω\Ω , we have
xy
1
(ω−1G)(x∗) ≤ k|G|k2 +CNlnN.
16 ω
(cid:12) (cid:12)
where C is independe(cid:12)nt of N, ε a(cid:12)nd x∗.
Proof. First let x∗ ∈ Ω . Let τ∗ be the unique triangle that has x∗ as its
s
north-east corner. Then
(ω−1G)(x∗) ≤CNkGk
τ∗
(cid:12) (cid:12)≤CNmax (ω−1)−1/2 ·k(ω−1)−1/2Gk
(cid:12) (cid:12) τ∗ β β τ∗
(cid:12) (cid:12)
Calculating (ω−1)−1(x) explicitly, we(cid:12)(cid:12)see that (cid:12)(cid:12)
β
(ω−1)−1(x)≤Cσ =CkN−1lnN ∀x∈τ∗
β β
Thus
1
(ω−1G)(x∗) ≤CNlnN + k|G|k2
16 ω
by means of the arith(cid:12)metic-geome(cid:12)tric mean inequality.
(cid:12) (cid:12)
Next, let x∗ ∈Ω .(The case x∗ ∈Ω is similar.)Write x∗ =(x ,y ). Then
x y i j
ω−1G(x∗) =|G(x∗)|
1
(cid:12) (cid:12)
(cid:12) (cid:12)= Gx(t,yj)dt
(cid:12)Zxi (cid:12)
(cid:12)(cid:12) 1 yj+(cid:12)(cid:12)1
≤(cid:12)CHy−1 (cid:12) |Gx(t,y)|dydt
Zxi Zyj
≤CN(εlnN ·N−1)1/2kG k
x Ωx
≤CN1/2ln1/2Nk|G|k
5
where G (t,y )= Gk,j−Gk−1,j for (t,y )∈τ .
x j hx j kj
Analysis: for the relationofboundary integraland domainintegral, we analyze
1 yj+1 N yj+1 y −y y−y
j+1 j
|G (t,y)|dydt= h f(y ) +f(y ) dy
x x j j+1
H H
Zxi Zyj k=i+1 Zyj (cid:12) y y (cid:12)
X (cid:12) (cid:12)
(cid:12) (cid:12)
where f(y )= Gk,j−Gk−1,j and f(y )= G(cid:12)k,j+1−Gk−1,j+1. (cid:12)
j hx j+1 hx
For ∆ := yj+1 f(y )yj+1−y +f(y )y−yj dy, we have
1 yj j Hy j+1 Hy
(cid:12) (cid:12)
R (cid:12) (cid:12)
|f(yj)|+(cid:12) |f(yj+1)| 1 (cid:12)
H ≥ max{|f(y ),|f(y )|}H if f(y )f(y )≥0
y j j+1 y j j+1
2 2
∆ = 
1 1f2(y )+f2(y ) 1
 j j+1 Hy ≥ max{|f(yj),|f(yj+1)|}Hy if f(yj)f(yj+1)<0
2 |f(y )−f(y )| 4
j+1 j

For ∀v ∈C2(τ), we have
|v |≤C(|v |+|v |)
x β η
|v |≤C(|v |+|v |+|v |)
xx ββ βη ηη
Similarly, we have
|v |≤C(|v |+|v |)
β x y
|v |≤C(|v |+|v |+|v |)
ββ xx xy yy
Lemma 3. Let τ ∈ΩN. Then
kω1/2DαEk ≤Ck−1/2N1/2|||G|||
Ωs ω
kω1/2DαEk ≤Ck−1ε−1/2ln−1N|||G|||
ΩN\Ωs ω
where H =max{h ,h } and |α|=1, DαG are G and G .
τ x,τ y,τ β η
Proof. Assume p∈[1,∞] and g ∈C3(τ). Then (see [2, Theorem 4])
k(g−gI) k ≤C h2 kg k +h h kg k +h2 kg k
x Lp(τ) x,τ xxx Lp(τ) x,τ y,τ xxy Lp(τ) y,τ xyy Lp(τ)
+Ch(cid:0)x,τkgxxkLp(τ), (cid:1)
k(g−gI) k ≤C h2 kg k +h h kg k +h2 kg k
y Lp(τ) x,τ xxy Lp(τ) x,τ y,τ xyy Lp(τ) y,τ yyy Lp(τ)
+Ch(cid:0)y,τkgyykLp(τ). (cid:1)
or (see [1, Comment 2.15])
k(g−gI) k ≤Ch2 kg k +Ch kg k ,
x L2(τ) y,τ xyy L2(τ) x,τ xx L2(τ)
k(g−gI) k ≤Ch2 kg k +Ch kg k .
y L2(τ) x,τ xxy L2(τ) y,τ yy L2(τ)
6
Inthe followinganalysis, Dαv denotesthe directionalderivativeofv alongβ or
η for different orders. The following analysis makes use of the former estimates
(The latter will make the analysis more shorter).
For τ ∈ΩN, we have
kω1/2E k ≤Cmaxω1/2(kE k +kE k )
β τ x τ y τ
τ
≤Cmaxω1/2{h2 k(ω−1G) k +h k(ω−1G) k +h H k(ω−1G) k
x,τ xxx τ x,τ xx τ x,τ τ xxy τ
τ
+H h k(ω−1G) k +h2 k(ω−1G) k +h k(ω−1G) k }
τ y,τ xyy τ y,τ yyy τ y,τ yy τ
≤Ch2 (k(ω−1) Gk +k(ω−1) G k )+Ch (k(ω−1) Gk +k(ω−1) G k )
x,τ xxx τ xx x τ x,τ xx τ x x τ
+Ch2 (k(ω−1) Gk +k(ω−1) G k )+Ch (k(ω−1) Gk +k(ω−1) G k )
y,τ yyy τ yy y τ y,τ yy τ y y τ
+Ch H (k(ω−1) Gk +k(ω−1) G k +k(ω−1) G k +k(ω−1) G k )
x,τ τ xxy τ xx y τ xy x τ x xy τ
+H h (k(ω−1) Gk +k(ω−1) G k +k(ω−1) G k +k(ω−1) G k )
τ y,τ xyy τ xy y τ yy x τ y xy τ
3 2
≤CH2 kDα(ω−1)DγGk +CH kDα(ω−1)DγGk
τ τ τ τ
k=2 k=1
X |α|X+|γ|=3 X |α|X+|γ|=2
|α|≥k |α|≥k
where we have used the following analysis for Dα(ω−1)DγG :
|α|=1,|γ|=2
P
k(ω−1) G k ≤Ck |Dα(ω−1)|·G k
x xy τ xy τ
|α|=1
X
≤Cmax |Dα(ω−1)|·kG k
xy τ
τ
|αX|=1
≤Ch−1max |Dα(ω−1)|·kG k
y,τ x τ
τ
|αX|=1
≤Ch−1 |Dα(ω−1)|·G .
y,τ x τ
(cid:13)|αX|=1 (cid:13)
(cid:13) (cid:13)
The same analysis can be applied to kω1/2E k .
η τ
For τ ∈Ω , we have
s
kω1/2E k ≤Ck−5/2N−2 (σ−5/2+σ−3/2σ−1+σ−1/2σ−2)max(ω−1)1/2+σ−3maxω−1/2 kGk
β τ β β η β η τ β η τ τ
+Ck−3/2H2 (σ−3/2+σ−(cid:2)1/2σ−1)max(ω−1)1/2+σ−2maxω−1/2 ·NkGk (cid:3)
τ β β η τ β η τ τ
+Ck−3/2H (cid:2)(σ−3/2+σ−1/2σ−1)max(ω−1)1/2+σ−2maxω−1/2(cid:3)kGk
τ β β η τ β η τ τ
+Cmaxω1/2(cid:2)H kDα(ω−1)k · kDγGk (cid:3)
τ L∞(τ) τ
τ
|αX|=1 |γX|=1
≤Ck−1/2N1/2|||G|||
ω
wherewehaveusedtheestimatesofω−1,standardinverseestimatesandHo¨lder
7
inequalities. Similarly, we have
kω1/2E k ≤Ck−1/2N1/2|||G||| .
η Ωs ω
For τ ∈ΩN \ΩN, we have
s
kω1/2E k ≤Ck−5/2H2 (σ−5/2+σ−3/2σ−1+σ−1/2σ−2)max(ω−1)1/2+σ−3maxω−1/2 kGk
β τ τ β β η β η τ β η τ τ
+Ck−2ε−1/2H2 σ−2+(cid:2)σ−1σ−1+σ−2 maxω−1/2·ε1/2 kDγGk (cid:3)
τ β β η η τ
τ
(cid:2) (cid:3) |γX|=1
+Ck−3/2H (σ−3/2+σ−1/2σ−1)max(ω−1)1/2+σ−2maxω−1/2 kGk
τ β β η τ β η τ τ
+Ck−1ε−1/2(cid:2)H σ−1+σ−1 maxω−1/2·ε1/2 kDγGk (cid:3)
τ β η τ τ
(cid:2) (cid:3) |γX|=1
≤Ck−1ε−1/2ln−1N|||G|||
ω
Similarly, we have
kω1/2E k ≤Ck−1ε−1/2ln−1N|||G||| .
η Ω\Ωs ω
Lemma 4. Let τ ∈ΩN. Let E =ω−1G−(ω−1G)I where (ω−1G)I denote the
s
bilinear function that interpolates to ω−1G at the vertices of τ. Then
kω1/2Ek ≤Ck−1N−1/2|||G|||
Ωs ω
kω1/2Ek ≤Ck−1ε1/2|||G|||
Ω\Ωs ω
where H =max{h ,h }.
τ x,τ y,τ
Proof. We make use of the following standard interpolation error bounds
ku−uIk ≤h2 ku k +h2 ku k
Lp(τ) x,τ xx Lp(τ) y,τ yy Lp(τ)
where p∈[1,∞] and u∈C(τ)∩W2,p(τ).
Then, we have
kEk ≤h2 k(ω−1G) k +h2 k(ω−1G) k
τ x,τ xx τ y,τ yy τ
≤CH2 kDα(ω−1)·Gk +CH2k(|(ω−1) |+|(ω−1) |)·(|G |+|G |)k
τ τ τ β η β η τ
|αX|=2
≤CH2 k(ω−1) G k +k(ω−1) G k +k(ω−1) G k +k(ω−1) G k
τ β β τ β η τ η β τ η η τ
+CHτ2k(cid:8) |Dα(ω−1)|·Gkτ. (cid:9)
|αX|=2
From the above inequality, we have
kω1/2Ek ≤Ck−1N−1/2|||G|||
Ωs ω
8
and
kω1/2Ek ≤Ck−1ε−1/2N−1|||G||| .
Ω\Ωs ω
Following the techniques of (see[5, Lemma 4.4]), we have
Γ(x)
(ω1/2E)(x)= (ω1/2E) ds
η
x
Z
where x ∈ Ω \ Ω , Γ(x) ∈ Γ satisfies (x −Γ(x)) · β = 0 and the following
s
condition:
For ∀y ∈Γ, (x−y)·β =0,
|x−Γ(x)|=min|x−y|.
y
From the above representation of ω1/2E, we have
1−λy Γ(x) 2
kω1/2Ek2 = (ω1/2E) ds dΩ
Ωx Zλ0 Zl(xlu,Γ(xlu))"Zx η #
1 Γ(x) 2 λ0 Γ(x) 2
+ (ω1/2E) ds dΩ+ (ω1/2E) ds dΩ
η η
Z1−λxZl(xu,Γ(xu))"Zx # Z0 Zl(xld,Γ(xld))"Zx #
≤Cλ2 k(ω1/2) Ek2 +kω1/2E k2
x η Ωx η Ωx
n o
≤Cε2ln2N σ−2kω1/2Ek2 +kω1/2E k2
η Ωx η Ωx
≤Ck−2ε2ln2nN{N3/2ln−2N ·ε−1N−2+ε−o1ln−2N}|||G|||2
ω
≤Ck−2ε|||G|||2
ω
where λ = b1λ and
0 b2 x
• x ∈{(1−λ ,y):λ ≤y ≤1−λ };
lu x 0 y
• x ∈{(1−λ ,y):0≤y ≤λ };
ld x 0
• x ∈{(x,1−λ ):1−λ ≤x≤1}.
u y x
Lemma 5. If σ = kN−1lnN and σ = kε˜1/2lnN , where k > 1 sufficiently
β η
large and independent of N and ε. Then
1
B((ω−1G)I −ω−1G,G)≤ k|G|k2.
16 ω
Proof. Cauchy-Schwarzsinequality gives
|B(E,G)|≤(ε+b2δ)1/2kω1/2E k·(ε+b2δ)1/2kω−1/2G k+εˆ1/2kω1/2E k·εˆ1/2kω−1/2G k
β β η η
+Ckω1/2Ek·kω−1/2G k+kω1/2Ek·kω−1/2Gk
β
From Lemma 3 and Lemma 4, we are done.
9
Theorem 3.1. Assumethat σ =kN−1lnN and σ =kε˜1/2lnN, where k >0
β η
issufficientlylargeandindependentofεandN. Letx∗ ∈Ω\Ω . Thenforeach
xy
nonnegative integer υ, there exists a positive constant C =C(υ) and K =K(υ)
such that
kGk ≤CN−υ,
W1,∞(Ωs\Ω′0)
ε|G| +kGk ≤CN−υ
W1,∞((Ωx∪Ωy)\Ω′0) L∞((Ωx∪Ωy)\Ω′0)
and
ε|G| +kGk ≤Cε−1/2N−υ.
W1,∞(Ωxy\Ω′0) L∞(Ωxy\Ω′0)
Proof. On Ω , we apply an inverse estimate.
s
On Ω\Ω the application of an inverse estimate does not yield a satisfactory
s
result, so we use a different technique.
Let x∈Ω \Ω′ be arbitrary. Starting from x we choose a polygonal curve
x 0
Γ ⊂ (Ω\Ω )\Ω′ that joints x with some point on outflow boundaries. If
xy 0
(x−x∗)·η <0, we can choose Γ as a line parallel to β. If (x−x∗)·η >0, the
situation is a little complicated. We can choose Γ as follows:
In Ω \Ω′, we choose the direction of Γ along η or the negative direction of
x 0
x-axis so that Γ∩Ω = Φ. In (Ω ∪Ω )\Ω′, we choose the direction of Γ
xy s y 0
along η or the positive direction of y−axis.
Let TN be the set of mesh rectangle τ in (Ω\Ω )\Ω′ that Γ intersects. Note
xy 0
that the length ofthe segmentof Γ that lies in eachτ is at mostCεN−1lnN if
τ ∈Ω or τ ∈Ω .
x y
Then, by the fundamental theorem of calculus and inverse estimates in dif-
ferent domain, we can obtain the results.
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