Error Estimates on Anisotropic Q 1 Elements for Functions in Weighted Sobolev Spaces

Error Estimates on Anisotropic Q 1 Elements for Functions in Weighted Sobolev Spaces
MATHEMATICS OF COMPUTATION
Volume 00, Number 0, Pages 000–000
S 0025-5718(XX)0000-0
ERROR ESTIMATES ON ANISOTROPIC Q1 ELEMENTS FOR FUNCTIONS IN
WEIGHTED SOBOLEV SPACES
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Abstract. In this paper we prove error estimates for a piecewise Q1 average interpolation on
anisotropic rectangular elements, i.e., rectangles with sides of different orders, in two and three
dimensions.
Our error estimates are valid under the condition that neighboring elements have comparable size.
This is a very mild assumption that includes more general meshes than those allowed in previous
papers. In particular, strong anisotropic meshes arising naturally in the approximation of problems
with boundary layers fall under our hypotheses.
Moreover, we generalize the error estimates allowing on the right hand side some weighted Sobolev
norms. This extension is of interest in singularly perturbed problems.
Finally, we consider the approximation of functions vanishing on the boundary by finite element
functions with the same property, a point that was not considered in previous papers on average
interpolations for anisotropic elements.
As an application we consider the approximation of a singularly perturbed reaction-diffusion
equation and show that, as a consequence of our results, almost optimal order error estimates in the
energy norm, valid uniformly in the perturbation parameter, can be obtained.
1. Introduction
In the finite element approximation of functions which have singularities or boundary layers it
is necessary to use highly non uniform meshes such that the mesh size is much smaller near the
singularities than far from them. In the case of boundary layers these meshes contain very narrow or
anisotropic elements.
The goal of this paper is to obtain new error estimates for Q1 (piecewise bilinear in 2d or trilinear
in 3d) approximations on meshes containing anisotropic rectangular elements, i. e., rectangles with
sides of different orders. The classic error analysis is based on the so called regularity assumption
which excludes this kind of elements (see for example [8, 9]). However, it is now well known that this
assumption is not needed. Indeed, many papers have been written to prove error estimates under more
general conditions. In particular, for rectangular elements we refer to [1, 12, 18] and their references.
We will prove the error estimates for a mean average interpolation. There are two reasons to work
with this kind of approximation instead of the Lagrange interpolation. The first one is to approximate
non smooth functions for which the Lagrange interpolation is not even defined, in fact this is the reason
that motivated the introduction of average interpolations (see [10]). On the other hand, it has already
been observed that, in the three dimensional case, average interpolations have better approximation
properties than the Lagrange interpolation even for smooth functions when narrow elements are used
(see [1, 12]).
Our estimates extend previous known results in several aspects:
First, our assumptions include more general meshes than those allowed in the previous papers.
Indeed, in [12] it was required that the meshes were quasiuniform in each direction. This requirement
was relaxed in [1] but not enough to include the meshes that arise naturally in the approximation
of boundary layers, which will be included under our assumptions. To prove our error estimates we
require only that neighboring elements are of comparable size and so, our results are valid for a rather
general family of anisotropic meshes.
1991 Mathematics Subject Classification. 65N30.
Key words and phrases. Anisotropic elements, weighted norms.
The research was supported by ANPCyT under grant PICT 03-05009 and by CONICET under grant PIP 0660/98.
The first author is a member of CONICET, Argentina.
c
°1997
American Mathematical Society
1
2
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Second, we generalize the error estimates allowing weaker norms on the right hand side. These
norms are weighted Sobolev norms where the weights are related with the distance to the boundary.
The interest of working with these norms arise in the approximation of boundary layers. Indeed,
for many singular perturbed problems it is possible to prove that the solution has first and second
derivatives which are bounded, uniformly in the perturbation parameter, in appropriate weighted
Sobolev norms.
The use of weighted norms to design appropriate meshes in finite element approximations of singular
problems is a well known procedure. In particular, error estimates for functions in weighted Sobolev
spaces have been obtained in several works (see for example [4, 5, 6, 14]). In those works, the weights
considered are related with the distance to a point or an edge (in the 3d case), instead here we consider
weights related with the distance to the boundary.
Finally, we consider the approximation of functions vanishing on the boundary by finite element
functions with the same property. This is a non trivial point that was not considered in the above
mentioned references.
Our mean average interpolation is similar to that introduced in [12] but the difference is that we
define it directly on the given mesh instead of using reference elements. This is important in order to
relax the regularity assumptions on the elements.
We will prove our estimates for the domain Ω = [0, 1]d , d = 2, 3. It will be clear that the interior
estimates derived in Section 2 are valid for any domain which can be decomposed in d-rectangles.
However, the extension of our results of Section 3 for interpolations satisfying Dirichlet boundary
conditions to other domains is not straightforward and would require a further analysis.
To prove the weighted estimates we will use a result of Boas and Straube [7] which, as we show,
can be derived from the classic Hardy inequality in higher dimensions.
In Section 2 we construct the mean average interpolation and prove the error estimates for interior
elements. Section 3 deals with the approximation on boundary elements. Since the proofs of this
section are rather technical we give them in the two dimensional case. However, it is not difficult
(although very tedious!) to see that our arguments apply also in three dimensions.
Finally in Section 4, as an application of our results, we consider the finite element approximation
of the reaction diffusion equation
−ε2 ∆u + u = f
u=0
in Ω
in ∂Ω
Using that appropriate weighted norms of the solution are bounded uniformly in the perturbation
parameter ε we show that it is possible to design graded meshes independent of ε such that almost
optimal (in terms of the degrees of freedom) error estimates in the energy norm, valid uniformly in ε,
hold.
2. Error estimates for interior elements
In this section we prove error estimates for a piecewise Q1 mean average interpolation for functions
in weighted Sobolev spaces. The weights considered are powers of the distance to the boundary. This
kind of weights arise naturally in problems with boundary layers.
The approximation introduced here is a variant of that considered in [12]. The difference is that we
define it directly in the given mesh instead of using a reference one. Working in this way we are able
to remove the restrictions used in [1, 12]. In particular, our results apply for the anisotropic meshes
arising in the approximation of boundary layers.
Let T be a partition into rectangular elements of Ω = [0, 1]d , d = 2, 3. We call N the set of nodes
of T and Nin the set of interior nodes.
Given an element R ∈ T , let hR,i be the length of the side of R in the direction xi .
We assume that there exists a constant σ such that, for R, S ∈ T neighboring elements,
(2.1)
For each v ∈ N we define
hR,i
≤σ
hS,i
1 ≤ i ≤ d.
3
hv,i = min{hR,i : v is a vertex of R},
1 ≤ i ≤ d.
and hv = (hv,1 , hv,2 ) if d = 2 or hv = (hv,1 , hv,2 , hv,3 ) if d = 3. If p, q ∈ Rd we denote by p : q the
vector (p1 q1 , p2 q2 ) if d = 2 or (p1 q1 , p2 q2 , p3 q3 ) if d = 3.R Take ψ ∈ C ∞ (Rd ) with support in a ball
centered at the origin and radius r ≤ 1/σ and such that ψ = 1, and for v ∈ Nin let
µ
¶
1
v1 − x1 v2 − x2
ψv (x) =
ψ
,
hv,1 hv,2
hv,1
hv,2
if d = 2 or
µ
¶
1
v1 − x1 v2 − x2 v3 − x3
ψ
,
,
hv,1 hv,2 hv,3
hv,1
hv,2
hv,3
if d = 3. Given a function u we call P (x, y) its Taylor polynomial of degree 1 at the point x, namely,
ψv (x) =
P (x, y) = u(x) + ∇u(x) · (y − x).
Then, for v ∈ Nin we introduce the regularized average
Z
(2.2)
uv (y) =
P (x, y)ψv (x)dx.
Now, given u ∈ H01 (Ω) we define Πu as the unique piecewise (with respect to T ) Q1 function such
that, for v ∈ Nin , Πu(v) = uv (v) while Πu(v) = 0 for boundary nodes v.
Introducing the standard basis functions λv associated with the nodes v we can write
Πu(x) =
X
uv (v)λv (x).
v∈Nin
For R ∈ T and v ∈ N we define (see Figure 1 for the 2d case)
R̃ =
[
{S ∈ T : S is a neighboring element of R}
and
Rv =
[
{S ∈ T : v is a vertex of S}.
In our analysis we will also make use of the regularized average of u, namely,
Z
Qv (u) = u(x)ψv (x)dx
for v ∈ Nin .
~
R
Rv
R
v
Figure 1
4
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
We remark that, since r ≤ 1/σ, it follows from our assumption (2.1) that the support of ψv (x) is
contained in Rv .
Now we prove some weighted estimates which will be useful for our error analysis. For any set D
we call dD (x) the distance of x to the boundary of D. For a d-rectangle R = Πdi=1 (ai , bi ) we have
dR (x) = min{xi − ai , bi − xi : 1 ≤ i ≤ d}. For such R we will also consider the following function
¾
½
xi − ai bi − xi
δR (x) := min
,
:1≤i≤d .
hR,i
hR,i
We will make use of the following inequality which is known as “Hardy inequality”:
(2.3)
°
°
°
v(x) °
°
≤ Ckv 0 kL2 (0,1)
°
x(x − 1) L2 (0,1)
for v ∈ H01 (0, 1). We will also need the following generalization to higher dimensions: If D is a convex
domain and u ∈ H01 (D) then
(2.4)
° u °
° °
≤ 2k∇ukL2 (D)
° ° 2
dD L (D)
(see for example [17]).
The following lemma gives an “anisotropic” version of (2.4). It can be proved by standard scaling
arguments.
Lemma 2.1. Let R = Πdi=1 (ai , bi ) be a d-rectangle and hi = bi − ai , 1 ≤ i ≤ d. For all u ∈ H01 (R)
(2.5)
° °
°u°
° °
° δR °
≤2
L2 (R)
°
°
° ∂u °
°
hi °
° ∂xi ° 2 .
L (R)
i=1
d
X
Another consequence of (2.4) is the inequality that we prove in the following lemma. This inequality
was proved for Lipschitz domains by Boas and Straube in [7]. We give a different proof here because
we are interested in the dependence of the constant on the domain, which is not stated in [7] because
the proof given there is based on compactness arguments.
Lemma 2.2. Let R be a d-rectangle with
sides of lengths hi , 1 ≤ i ≤ d, such that 1δ ≤ hi ≤ δ, and
R
let ψ ∈ C0 (R) be a function such that R ψR = 1. Then, there exists a constant C depending only on δ
and ψ, such that, for all u ∈ H 1 (R) with R uψ = 0,
(2.6)
kukL2 (R) ≤ CkdR ∇ukL2 (R) .
R
Proof. Since v := u − ( R u)ψ has vanishing mean value, there exists F ∈ H01 (R)d such that
(2.7)
−div F = v
and such that
(2.8)
kF kH01 (R)2 ≤ CkvkL2 (R) .
Moreover, from the explicit bound for the constant given in [13] it follows that C can be taken
depending onlyR on δ.
Now, since R uψ = 0, we have from (2.7)
Z
Z
kuk2L2 (R) =
uv = −
u div F
R
R
and therefore, integrating by parts and using (2.4) for each component of F , we obtain
Z
°F °
° °
2
kukL2 (R) =
∇u · F ≤ kdR ∇ukL2 (R) ° ° 2
≤ 2kdR ∇ukL2 (R) k∇F kL2 (R)
dR L (R)
R
but,
5
kvk2L2 (R) ≤ (1 + |R|kψk2L2 (R) )kuk2L2 (R)
and so, the proof concludes by using (2.8) and the fact that the constant in that estimate depends
only on δ.
¤
As a consequence of the previous lemma we obtain the following weighted estimates.
Lemma 2.3. For v ∈ Nin there exists a constant C depending only on σ and ψ such that, for all
u ∈ H 1 (Rv ),
(2.9)
ku − Qv (u)kL2 (Rv )
°
°
°
∂u °
°
°
≤C
hv,i °δRv
° 2
∂x
i
L (Rv )
i=1
d
X
and, for all u ∈ H 2 (Rv ),
(2.10)
°
°
°
°
d
X
° ∂(u − uv ) °
°
∂2u °
°
°
°
°
≤C
.
hv,i °δRv
° ∂xj ° 2
° 2
∂x
∂x
j
i
L (Rv )
L (Rv )
i=1
Proof. Let Kv be the image of Rv by the map x → x̄ with
x̄i =
vi − x i
hv,i
1≤i≤d
and, for x̄ ∈ Kv , define ū by ū(x̄) = u(x). Then, Qv (u) = Q̄(ū) where
Z
Q̄(ū) = ū(x̄)ψ(x̄)dx̄.
Now, in view of our assumption (2.1), the d-rectangle Kv satisfies the hypothesis of
R Lemma 2.2 with
δ = 2σ. Moreover, since r ≤ σ1 , the support of ψ is contained in Kv . Therefore, since (ū− Q̄(ū))ψ = 0,
it follows from Lemma 2.2 that there exists a constant C depending only on σ and ψ such that
kū − Q̄(ū))kL2 (Kv ) ≤ CkdKv ∇ūkL2 (Kv )
and (2.9) follows by going back to the variable x.
To prove (2.10), observe that uv (y) = ū0 (ȳ) where
Z
ū0 (ȳ) = (ū(x̄) + ∇(ū)(x̄) · (ȳ − x̄))ψ(x̄)dx̄
and so, since
Z
∂(ū − ū0 )
ψ = 0,
∂ x̄i
we obtain from Lemma 2.2 that there exists a constant C depending only on σ and ψ such that
°
°
°
°
° ∂(ū − ū0 ) °
°
∂ ū °
°
°
°
°
≤ C °dKv ∇
° ∂ x̄i
° 2
∂ x̄i °L2 (Kv )
L (Kv )
and the proof concludes going back to the variable x.
¤
We can now estimate the approximation error for interior elements in terms of weighted norms. We
start with the L2 norm. From now on C will be a generic constant which depends only on σ and ψ. In
view of our hypothesis (2.1), hv,i and hR,i are equivalent up to a constant depending on σ whenever
v is a vertex of R. We will use this fact repeatedly without making it explicitly.
Theorem 2.4. There exists a constant C depending only on σ and ψ such that
(i) For all R ∈ T and u ∈ H 1 (R̃) we have
(2.11)
kΠukL2 (R) ≤ C kukL2 (R̃) .
6
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
(ii) for all R ∈ T such that R is not a boundary element and u ∈ H 1 (R̃) we have
°
°
d
X
° ∂u °
°
hR,i °
(2.12)
ku − ΠukL2 (R) ≤ C
δ
° R̃ ∂xi ° 2 .
L (R̃)
i=1
Proof. To prove (i) we write
(Πu)|R =
nR
X
uvj (vj )λvj
j=1
where {vj }n1 R are the interior nodes of R. Then,
Ã
(2.13)
kΠukL2 (R) ≤ C
d
Y
! 12
hR,i
i=1
nR
X
° °
°uv ° ∞
j L (R)
j=1
and we have to estimate kuvj kL∞ (R) for each j. To simplify notation we write v = vj (and so the
subindexes denote now the components of v). We have
(2.14)
à d
!− 12
¯Z
¯
Y
¯
¯
¯ u(x)ψv (x)dx¯ ≤ C
hR,i
kukL2 (R̃) .
¯
¯
i=1
On the other hand, since ψv = 0 on ∂ R̃, integration by parts gives
(2.15)
¯Z
¯ ¯Z
¯
Z
¯ ∂u
¯ ¯
¯
∂ψv
¯
¯
¯
¯
(x)(y
−
x
)ψ
(x)dx
(x)dx
=
u(x)ψ
(x)dx
−
u(x)(y
−
x
)
i
i
v
v
i
i
¯ ∂xi
¯ ¯
¯
∂xi
à d
!− 12
Y
kukL2 (R̃)
≤C
hR,i
i=1
where we have used that |yi − xi | ≤ Chv,i . Thus, (2.11) follows from (2.13), (2.14), (2.15) and the
definition of uv given in (2.2).
To prove (ii), choose a node of R, say v1 . Since Qv1 (u) is a constant function and R is not a
boundary element, we have ΠQv1 (u) = Qv1 (u) on R and so
(2.16)
ku − ΠukL2 (R) ≤ ku − Qv1 (u)kL2 (R) + kΠ(Qv1 (u) − u)kL2 (R)
≤ Cku − Qv1 (u)kL2 (R)
where we have used (2.11). Now, estimate (2.12) follows from (2.16) and (2.9).
¤
In what follows, we estimate the approximation error for the first derivatives for interior elements.
We will use the notation of Figure 2.
Theorem 2.5. There exists a constant C depending only on σ and ψ such that, if R ∈ T is not a
boundary element then, for all u ∈ H 2 (R̃) we have
(2.17)
°
°
° ∂
°
°
°
° ∂xj (u − Πu)°
°
°
°
∂2u °
°
°
≤C
hR,i °δR̃
° 2
∂x
∂x
i
j
L (R̃)
i=1
d
X
L2 (R)
1 ≤ j ≤ d.
Proof. We will consider the case d = 3, j = 1. Clearly, the other cases are analogous. We have
u − Πu = (u − uv1 ) + (uv1 − Πu)
∂(u−u )
and from (2.10) we know that k ∂x1 v1 kL2 (R) is bounded by the right hand side of (2.17). Therefore,
∂(u 1 −Πu)
kL2 (R) . Since w := uv1 − Πu ∈ Q1 we have (see for example [18])
we have to estimate k v∂x
1
4
X
∂w
∂λvi
=
(w(vi ) − w(vi+4 ))
∂x1
∂x1
i=1
7
v4
v8
v
3
v
7
v2
v6
v1
x3
v
5
x2
x1
Figure 2
then,
°
°
°
°
4
X
° ∂w °
° ∂λvi °
°
°
°
°
≤
|w(v
)
−
w(v
)|
.
i
i+4 °
° ∂x1 ° 2
∂x1 °L2 (R)
L (R)
i=1
(2.18)
But, it is easy to see that
°
°
¶1
µ
° ∂λvi °
hvi ,2 hvi ,3 2
°
°
.
≤
C
° ∂x1 ° 2
hvi ,1
L (R)
(2.19)
So, we have to estimate |w(vi ) − w(vi+4 )| for 1 ≤ i ≤ 4. We have
(2.20)
w(v1 ) − w(v5 ) = uv5 (v5 ) − uv1 (v5 )
Z
Z
= P (x, v5 )ψv5 (x)dx − P (x, v5 )ψv1 (x)dx.
So, changing variables we obtain
Z
(2.21)
w(v1 ) − w(v5 ) =
[P (v5 − hv5 : y, v5 ) − P (v1 − hv1 : y, v5 )] ψ(y)dy.
We introduce the notation vi = (v1i , v2i , v3i ). Define now
¡
¢
θ = (θ1 , 0, 0) := v15 − v11 + (hv1 ,1 − hv5 ,1 )y1 , 0, 0
and
Fy (t) := P (v1 − hv1 : y + tθ, v5 ).
Then, since hv1 ,2 = hv5 ,2 , hv1 ,3 = hv5 ,3 and v21 = v25 , v31 = v35 , we have
P (v5 − hv5 : y, v5 ) − P (v1 − hv1 : y, v5 ) = Fy (1) − Fy (0)
and replacing in (2.21) we obtain
Z Z
1
w(v1 ) − w(v5 ) =
0
Z
Fy0 (t)ψ(y)dtdy =
and therefore it is enough to estimate
1
0
Z
I(t) :=
Fy0 (t)ψ(y)dy
for 0 ≤ t ≤ 1. But, from the definition of Fy and P , we have
½Z
¾
Fy0 (t)ψ(y)dy dt
8
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
¯
Z ½¯ 2
¯∂ u
¯
1
1
¯
¯
|I(t)| ≤
¯ ∂x2 (v1 − hv1 : y + tθ)¯ × |v5 − v1 + hv1 ,1 y1 − tθ1 |
1
¯
¯
¯ ∂2u
¯
¯
+¯
(v1 − hv1 : y + tθ)¯¯ × |v25 − v21 + hv1 ,2 y2 |
∂x1 ∂x2
¯
¯
¾
¯ ∂2u
¯
3
3
¯
¯
+¯
(v1 − hv1 : y + tθ)¯ × |v5 − v1 + hv1 ,3 y3 | |θ1 |ψ(y)dy.
∂x1 ∂x3
Now, for |y| ≤ 1 and 0 ≤ t ≤ 1, we have
|vi5 − vi1 + hv1 ,i yi − θ1 t| ≤ Chv1 ,i ,
|θ| = |θ1 | ≤ Chv1 ,1 ,
and therefore, since supp(ψ) ⊂ B(0, 1), we have
¯
¯
¯
Z ½¯ 2
¯∂ u
¯
¯ ∂2u
¯
2
¯
¯
¯
¯
|I(t)| ≤ C
¯ ∂x2 (v1 − hv1 : y + θt)¯ (hv1 ,1 ) + ¯ ∂x1 ∂x2 (v1 − hv1 : y + θt)¯ hv1 ,1 hv1 ,2
1
¯
¯
¾
¯ ∂2u
¯
¯
¯
+¯
(v1 − hv1 : y + θt)¯ hv1 ,1 hv1 ,3 ψ(y)dy.
∂x1 ∂x3
Now, making the change of variables z = v1 − hv1 : y + θt and calling
µ
¶
z1 − [(1 − t)v11 + tv12 ] z2 − v21
z3 − v31
φ(z) = ψ −
,−
,−
(1 − t)hv1 ,1 + thv5 ,1
hv1 ,2
hv1 ,3
we obtain
|I(t)| ≤ C
¯
Z ¯ 2
3
X
¯ ∂ u
¯
1
hv1 ,i ¯¯
(z)¯¯ φ(z)dz
hv1 ,2 hv1 ,3 i=1
∂x1 ∂xi
¢
¡
where we have used that hv1 ,1 ≥ C((1 − t)hv1 ,1 + thv5 ,1 ). But, since supp ψ ⊂ B 0, σ1 , it follows that
supp φ ⊂ R̃. Then, using the Schwarz inequality we obtain
°° °
°
3
X
°
° °
1
∂2u °
°° φ °
°
|I(t)| ≤ C
hv ,i δ
hv1 ,2 hv1 ,3 i=1 1 ° R̃ ∂x1 ∂xi ° ° δR̃ °
and from Lemma 2.1 we know that
° °
°φ°
1
° °
≤ C(hv1 ,1 hv1 ,2 hv1 ,3 ) 2 .
°δ ° 2
R̃ L (R̃)
Finally, using (2.19) we obtain
(2.22)
°
°
° ∂λv1 °
°
°
|w(v1 ) − w(v5 )| °
∂x1 °
≤C
L2 (R)
3
X
i=1
°
°
°
∂2u °
°
°
hv1 ,i °δR̃
∂x1 ∂xi °
.
L2 (R̃)
Now, to estimate |w(v2 ) − w(v6 )| we write
w(v2 ) − w(v6 ) = (uv1 (v2 ) − uv2 (v2 )) − (uv1 (v6 ) − uv6 (v6 ))
(2.23)
= (uv1 (v2 ) − uv1 (v6 )) − (uv2 (v2 ) − uv2 (v6 )) − (uv2 (v6 ) − uv6 (v6 ))
=: I − II − III.
Now we estimate I − II. We have
Z
∂u
(x)(v12 − v16 )ψv1 (x)dx
I=
∂x1
Z
and
II =
∂u
(x)(v12 − v16 )ψv2 (x)dx
∂x1
where we have used that v2 − v6 = (v12 − v16 , 0, 0). After a change of variables in both integrals we
obtain
¸
Z ·
∂u
∂u
(v1 − hv1 : y) −
(v2 − hv2 : y) (v12 − v16 )ψ(y)dy
I − II =
∂x1
∂x1
9
and so, defining θ = (0, θ2 , 0) := (0, v22 − v21 − (hv2 ,2 − hv1 ,2 )y2 , 0) and
∂u
(v1 − hv1 : y + θt)
∂x1
and taking into account that hv1 ,1 = hv2 ,1 and hv1 ,3 = hv2 ,3 we have
Z Z 1
I − II = −
Fy0 (t)(v12 − v16 )ψ(y)dtdy
0
¾
Z 1 ½Z
=−
Fy0 (t)(v12 − v16 )ψ(y)dy dt
Fy (t) =
0
1
Z
=:
I(t)dt.
0
Since
Fy0 (t) =
∂2u
(v1 − hv1 : y + θt)θ2
∂x1 ∂x2
and for y ∈ supp ψ, |y| ≤ 1, we have
¯
Z ¯
¯ ∂2u
¯
¯
|I(t)| ≤ ¯
(v1 − hv1 : y + θt)¯¯ |θ2 ||v12 − v16 ||ψ(y)|dy
∂x1 ∂x2
¯
Z ¯
¯ ∂2u
¯
¯
≤ Chv2 ,1 hv2 ,2 ¯
(v1 − hv1 : y + θt)¯¯ |ψ(y)|dy.
∂x1 ∂x2
Change now to the variable z = v1 − hv1 : y + θt and define
µ
¶
z1 − v11
z2 − [(1 − t)v21 + tv22 ] z3 − v31
φ(z) = ψ −
,−
,−
.
hv1 ,1
(1 − t)hv1 ,2 + thv2 ,2
hv1 ,3
¡
¢
Then, since supp φ ⊂ R̃ (because supp ψ ⊂ B 0, σ1 ), we can use Lemma 2.1 to obtain
¯
Z ¯
¯ ∂2u
¯
1
¯
¯ φ(z)dz
|I(t)| ≤ C
(z)
¯
hv1 ,3
∂x1 ∂x2 ¯
°
°
°
°
° φ(z) °
1 °
∂2u °
°
°
°
°
≤C
δ
hv1 ,3 ° R̃ ∂x1 ∂x2 °L2 (R̃) ° δR̃ °L2 (R̃)
°
¶1 °
µ
hv1 ,1 hv1 ,2 2 °
∂2u °
°
°
≤C
°δR̃ ∂x1 ∂x2 ° 2 .
hv1 ,3
L (R̃)
Therefore,
µ
|I − II| ≤ C
hv1 ,1 hv1 ,2
hv1 ,3
°
¶ 12 °
°
∂2u °
°δ
°
° R̃ ∂x1 ∂x2 °
.
L2 (R̃)
The term III in equation (2.23) can be bounded by the same arguments used to obtain (2.22).
Therefore we obtain
(2.24)
°
°
°
°
d
X
° ∂λv1 °
°
∂2u °
°
°
°
°
|w(v2 ) − w(v6 )| °
≤C
hv1 ,i °δR̃
° 2 .
∂x1 °L2 (R)
∂x
∂x
1
i
L (R̃)
i=1
The estimate of w(v3 ) − w(v7 ) follows by the same arguments used to estimate w(v2 ) − w(v6 ).
Then, it remains to estimate w(v4 ) − w(v8 ). We have
w(v4 ) − w(v8 ) = (uv1 (v4 ) − uv4 (v4 )) − (uv1 (v8 ) − uv8 (v8 ))
= [(uv1 (v4 ) − uv1 (v8 )) − (uv3 (v4 ) − uv3 (v8 ))]
+ [(uv3 (v4 ) − uv3 (v8 )) − (uv4 (v4 ) − uv4 (v8 ))] + [uv8 (v8 ) − uv4 (v8 )]
=: I + II + III.
Now we deal with the term I. One can check that
¸
Z ·
∂u
∂u
I=
(v1 − hv1 : y) −
(v3 − hv3 : y) (v14 − v18 )ψ(y)dy.
∂x1
∂x1
10
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Defining now
Fy (t) :=
∂u
(v3 − hv3 : y + tθ)
∂x1
where θ = (0, 0, θ3 ) := (0, 0, v31 − v33 − (hv1 ,3 − hv3 ,3 )y3 ) we have
Z Z 1
I=
Fy0 (t)(v14 − v18 )ψ(y)dydt
0
Z
1
=
0
Z
Z
Fy0 (t)(v14 − v18 )ψ(y)dtdy =:
1
I(t)dt.
0
Since
Fy0 (t) =
∂2u
(v3 − hv3 : y + tθ)θ3
∂x1 ∂x3
and |θ3 | ≤ Chv1 ,3 if |y| ≤ 1 it follows that
Z
|I(t)| ≤ hv1 ,1 hv1 ,3
¯
¯
¯ ∂2u
¯
¯
¯ ψ(y)dy
(v
−
h
:
y
+
tθ)
v3
¯ ∂x1 ∂x3 3
¯
and so, changing variables and setting
µ
¶
z1 − v13
z2 − v23
z3 − [(1 − t)v33 + tv31 ]
φ(z) = ψ −
,−
,−
,
hv3 ,1
hv3 ,2
(1 − t)hv3 ,3 + thv1 ,3
we obtain
|I(t)| ≤ C
1
hv1 ,2
¯
Z ¯
¯ ∂2u
¯
¯
¯
(z)
¯ ∂x1 ∂x3 ¯ φ(z)dz.
Now, taking into account that φ = 0 on ∂ R̃, it follows by the Schwarz inequality and Lemma 2.1 that
°
°
° °
°
°φ°
∂2u °
°δ
°
° °
R̃
°
°
hv1 ,2
∂x1 ∂x3 L2 (R̃) ° δR̃ °L2 (R̃)
°
µ
¶1 °
hv1 ,1 hv1 ,3 2 °
∂2u °
°
°
≤
°δR̃ ∂x1 ∂x3 ° 2 ,
hv1 ,2
L (R̃)
1
|I(t)| ≤ C
and therefore,
(2.25)
°
°
° ∂λv4 °
°
°
|I| °
∂x1 °
L2 (R̃)
°
°
°
∂2u °
°
°
≤ hhv1 ,3 °δR̃
.
∂x1 ∂x3 °L2 (R̃)
Finally, estimates for the terms II and III can be obtained with the arguments used for (uv1 (v2 ) −
uv1 (v6 )) − (uv2 (v2 ) − uv2 (v6 )) in (2.23) and uv5 (v5 ) − uv1 (v5 ) in (2.20) respectively. These estimates
together with the inequalities (2.22),(2.24) and (2.25) conclude the proof.
¤
3. Error estimates for boundary elements
In this section we deal with the interpolation error on boundary elements for functions satisfying a
homogeneous Dirichlet condition. For the sake of simplicity and because the proof is rather technical,
we state and prove the main Theorem in the two dimensional case. However, analogous results can
be obtained in three dimensions by using similar arguments.
We will use the notation of the previous section. Further, if R = (a1 , b1 ) × (a2 , b2 ) is a rectangle in
T , we set R1i = ai and lR,i = (ai , bi ). Also we define the function δ−,R by
¾
½
x1 − a1 x2 − a2
,
.
δ−,R (x) = min
hR,1
hR,2
We have δR (x) ≤ δ−,R (x) for all x ∈ R.
11
To estimate the error on a boundary element R we need to consider different cases according to
the position of R. So, we decompose Ω into four regions (see Figure 3):
[
Ω1 =
{R ∈ T : R ∩ ∂Ω = ∅}
[
Ω2 =
{R ∈ T : R ∩ {x : x1 = 0} = ∅ and R ∩ {x : x2 = 0} 6= ∅}
[
Ω3 =
{R ∈ T : R ∩ {x : x1 = 0} 6= ∅ and R ∩ {x : x2 = 0} = ∅}
Ω4 = R ∈ T such that (0, 0) ∈ R.
Theorem 3.1. There exists a constant C depending only on σ and ψ such that if R ∈ T for all
u ∈ H 2 (R̃) the following estimates hold,
(i) If R ⊂ Ω2 and u ≡ 0 on {x : x2 = 0}
°
°
°
° ∂
°
°
° ∂x1 (u − Πu)°
(3.1)
L2 (R)
and
L2 (R̃)


°
°
° x − R̃
∂2u °
°
° 1
11
≤ C hR,1 °
°

° hR,1 ∂x1 ∂x2 °
°
°
° ∂
°
°
°
° ∂x2 (u − Πu)°
(3.2)

°
°
°
°

° x − R̃
°
2 °
2
°
∂
u
∂
u
°
°
1
11
≤ C hR,1 °
δ−,R̃ 2 °
+ hR,2 °
°
°
°
° hR,1 ∂x1 ∂x2 °

∂x1 L2 (R̃)
L2 (R)
L2 (R̃)




°
°

2 °
°
∂
u
°
+ hR,2 °
δ
° −,R̃ ∂x2 ° 2  .
2 L (R̃)
(ii) If R ⊂ Ω3 and u ≡ 0 on {x : x1 = 0}
°
°
°
° ∂
°
°
(u
−
Πu)
°
° ∂x1
(3.3)
L2 (R)
and
(3.4)
°
°
° ∂
°
°
°
° ∂x2 (u − Πu)°

°
°
°
°
° x − R̃
°

2
2 °
°
∂
u
∂
u
°
°
2
12
°
+
h
≤ C hR,1 °
δ
°
°
R,2
° −,R̃ ∂x2 ° 2

°
°
h
∂x
∂x
R,2
1
2
1 L (R̃)
L2 (R̃)


L2 (R)
°
°
° x − R̃
∂2u °
°
° 2
12
≤ C hR,1 °
°

° hR,2 ∂x1 ∂x2 °
L2 (R̃)




°
°

2 °
°
∂ u°
+ hR,2 °
°δ−,R̃ ∂x2 ° 2  .
2 L (R̃)
(iii) If R ⊂ Ω4 and u ≡ 0 on {x : x1 = 0 or x2 = 0}
(3.5)
(
°
°
° ∂
°
°
°
(u
−
Πu)
° ∂x1
°
≤C
L2 (R)
(a)
)
°
°½
°
°
¾
2
2 °
° x1
°
°
x
∂
u
∂
u
2
°
°
hR,1 °
+ hR,2 °
° hR,1 + hR,2 ∂x1 ∂x2 ° 2
°δ−,R̃ ∂x2 ° 2
1 L (R̃)
L (R̃)
v3
v4
(b)
v3
R
v4
R
v1
v2
v1
v2
R ⊂ Ω1
(c)
v3
R ⊂ Ω2
v4
(d)
v3
R
v4
R
v1
v2
R ⊂ Ω3
v1
R ⊂ Ω4
v2
Figure 3. Relative positions of the rectangle R. The bold face line is the boundary of Ω.
12
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
and
(3.6)
°
°
° ∂
°
°
°
(u
−
Πu)
° ∂x2
°
(
≤C
)
°½
°
¾
2
° x1
°
x
∂
u
2
°
+ hR,1 °
.
° hR,1 + hR,2 ∂x1 ∂x2 ° 2
L2 (R̃)
L (R̃)
°
°
°
∂2u °
°
hR,2 °
δ
° −,R̃ ∂x2 °
2
L2 (R)
Proof of Part (i). We now use the notation of Figure 3(b). We have
Πu|R = uv3 (v3 )λv3 + uv4 (v4 )λv4 .
∂
From (2.10) we know that k ∂x
(u − uv3 )kL2 (R) is bounded by the right hand side of (3.1). So, to prove
1
∂
(3.1), it is enough to estimate k ∂x
(uv3 − Πu)kL2 (R) .
1
Since (uv3 − Πu)|R ∈ Q1 we have (see for example [18])
∂λv2
∂
(uv3 − Πu) = ((uv3 − Πu)(v2 ) − (uv3 − Πu)(v1 ))
∂x1
∂x1
(3.7)
+ ((uv3 − Πu)(v4 ) − (uv3 − Πu)(v3 ))
= (uv3 (v2 ) − uv3 (v1 ))
Taking into account that
∂u
∂x1
∂λv4
∂x1
∂λv2
∂λv4
+ (uv3 (v4 ) − uv4 (v4 ))
.
∂x1
∂x1
≡ 0 on (x1 , 0) it is easy to see that
Z
uv3 (v2 ) − uv3 (v1 ) =
(v12
−
v11 )
Z
lR̃,2
lR̃,1
Z
x2
0
∂2u
(x1 , t)ψv3 (x)dtdx1 dx2
∂x1 ∂x2
and then,
¯
¯
¯ ∂2u
¯
¯
¯
(x
,
t)
¯ ∂x1 ∂x2 1 ¯ ψv3 (x)dtdx1 dx2
lR̃,2 lR̃,1 lR̃,2
°
°
°
°
Z
° x − R̃
°
∂2u °
hv3 ,1 °
°
° 1
11
°
°
≤ Chv3 ,1
ψv (x)
°
°
° 2 dx2 .
∂x1 ∂x2 ° 2 ° 3
x
−
R̃
lR̃,2 ° hv3 ,1
1
11
L (R̃)
L (R̃)
Z
Z
Z
|uv3 (v2 ) − uv3 (v1 )| ≤ Chv3 ,1
Using the one dimensional Hardy inequality (2.3) we have
¯
¯
¯
µ
¶¯
Z
Z
¯ ∂ψ v13 − x1 v21 − x2 ¯2
¯ ψv3 (x) ¯2
C
¯
¯
¯ dx1 ≤
¯ dx1
,
¯
¯
¯
h4v3 ,1 h2v3 ,2 lR̃,1 ¯ ∂x1
hv3 ,1
hv3 ,2
lR̃,1 x1 − R̃11
(3.8)
1
≤C 3
hv3 ,1 h2v3 ,2
and then it follows that
°
°
° x − R̃
∂2u °
° 1
°
11
|uv3 (v2 ) − uv3 (v1 )| ≤ C(hv3 ,1 hv3 ,2 ) °
°
° hv3 ,1 ∂x1 ∂x2 °
1
2
L2 (R̃)
and so
(3.9)
°
°
°
°
°
° x − R̃
2
° ∂λv2 °
∂
u
°
°
1
11
°
|uv3 (v2 ) − uv3 (v1 )| °
≤ Chv3 ,2 °
°
° ∂x1 ° 2
°
h
∂x
∂x
v3 ,1
1
2°
L (R̃)
.
L2 (R̃)
On the other hand, with the same argument that we have used to obtain (2.22) in the proof of
Theorem 2.5 we can show that
Ã
!
°
°
°
°
2
2
X
° ∂λv4 °
°
°
∂
u
°
°
|(uv3 (v4 ) − uv4 (v4 ))| °
≤C
hv3 ,i °
° ∂x1 ° 2
°δR̃ ∂xi ∂xj ° 2
L (R)
L (R̃)
i=1
which together with (3.7) and (3.9) concludes the proof of (3.1).
∂
Now, to prove (3.2), using again Lemma 2.3, we have to estimate k ∂x
(uv3 − Πu)kL2 (R) . Using
2
again the expression for the derivative of a Q1 function we have
13
(3.10)
∂
∂λv3
∂λv4
∂λv4
(uv3 − Πu) = −uv3 (v1 )
+ (uv3 (v4 ) − uv4 (v4 ))
− uv3 (v2 )
∂x2
∂x2
∂x2
∂x2
∂λv4
∂λv3
+ (uv3 (v4 ) − uv3 (v2 ))
−
= −uv3 (v1 )
∂x2
∂x2
∂λv4
∂λv4
(uv4 (v4 ) − uv4 (v2 ))
− uv4 (v2 )
∂x2
∂x2
∂λv3
∂λv4
∂λv4
=: −uv3 (v1 )
+ (I − II)
− uv4 (v2 )
.
∂x2
∂x2
∂x2
Defining now
θ = (θ1 , 0) := (v14 − v13 − (hv4 ,1 − hv3 ,1 )y1 , 0)
and
Fy (t) =
∂u
(v3 − hv3 : y + θt)
∂x2
we have
Z ·
I − II = (v24 − v22 )
Z
=
(v24
−
¸
∂u
∂u
(v3 − hv3 : y) −
(v4 − hv4 : y) ψ(y)dy
∂x2
∂x2
v22 )
(Fy (0) − Fy (1))ψ(y)dy
Z Z 1
= −(v24 − v22 )
Fy0 (t)dtψ(y)dy
0
but,
∂2u
(v3 − hv3 : y + θt)θ1
∂x1 ∂x2
Fy0 (t) =
and so,
Z
I − II = −(v24 − v22 )
1
Z
0
Z
=: −(v24 − v22 )
∂2u
(v3 − hv3 : y + θt)θ1 ψ(y)dydt
∂x1 ∂x2
1
I(t)dt.
0
We will estimate I(t). Since supp ψ ⊂ B(0, 1) we have
¯
Z ¯
¯
¯ ∂2u
¯
(v3 − hv3 : y + θt)¯¯ ψ(y)dy.
|I(t)| ≤ Chv3 ,1 ¯
∂x1 ∂x2
Now, setting z = v3 − hv3 : y + θt, taking into account that Chv3 ,1 ≤ (1 − t)hv3 ,1 + thv4 ,1 (0 ≤ t ≤ 1),
and defining
µ
φ(z) = ψ
(1 − t)v13 + tv14 − z1 v23 − z2
,
(1 − t)hv3 ,1 + thv4 ,1 hv3 ,2
we obtain
|I(t)| ≤ C
1
hv3 ,2
¯
Z ¯
¯
¯ ∂2u
¯
¯
(z)
¯ ∂x1 ∂x2 ¯ φ(z)dz
and, since φ ≡ 0 on ∂ R̃ we can use Lemma 2.1 to obtain
¶
14
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
°
° °
°
°
°φ°
∂2u °
°
°
°
°
|I(t)| ≤ C
δ
hv3 ,2 ° δR̃ °L2 (R̃) ° R̃ ∂x1 ∂x2 °L2 (R̃)
Ã
!°
°
°
°
°
°
° ∂φ °
° ∂φ °
°
1
∂2u °
°
°
°
°
°
≤C
hR,1 °
+
h
δ
R,2 °
° ∂z1 ° 2
hv3 ,2
∂z2 °L2 (R̃) ° R̃ ∂x1 ∂x2 °L2 (R̃)
L (R̃)
°
¶1 °
µ
∂2u °
hR,1 2 °
°
°
≤C
°δR̃ ∂x1 ∂x2 ° 2 .
hR,2
L (R̃)
1
Therefore,
°
°
1 °
∂2u °
°
|I − II| ≤ C(hR,1 hR,2 ) 2 °
δ
° R̃ ∂x1 ∂x2 °
,
L2 (R̃)
so
°
°
°
°
°(I − II) ∂λv3 °
°
∂x2 °
(3.11)
L2 (R)
°
°
°
∂2u °
°
δ
≤ ChR,1 °
° R̃ ∂x1 ∂x2 ° 2 .
L (R̃)
∂λ
Now, to estimate the first term of formula (3.10), uv3 (v1 ) ∂xv23 , we observe that, since u(x1 , 0) ≡ 0
then one can check that
Z Z
(3.12)
uv3 (v1 ) = −
x2
t=0
∂2u
(x1 , t)(t − v21 )ψv3 (x)dtdx +
∂x22
Z
(v11 − x1 )
∂u
(x)ψv3 (x)dx
∂x1
=: A + B.
We will estimate A and B. Since v21 = 0 we have
¯
¯
¯
x1 − R̃11 t ¯¯ ∂ 2 u
hv3 ,1
|A| ≤ Chv3 ,2
(x1 , t)¯¯ ψv3 (x)
dtdx
2
¯
hv3 ,1 hv3 ,2 ∂x2
x1 − R̃11
t=0
¯
¯
Z
Z
Z
¯ ∂2u
¯
hv3 ,1
≤ Chv3 ,2
δ−,R̃ (x1 , t) ¯¯ 2 (x1 , t)¯¯ ψv3 (x)
dtdx1 dx2 .
∂x2
x1 − R̃11
lR̃,2 lR̃,1 lR̃,2
Z Z
x2
Therefore, using the Schwarz inequality and (3.8) we obtain
°
3 °
(hv3 ,2 ) 2 °
∂2u °
°
|A| ≤ C
δ−,R̃ 2 °
1
∂x °
(h ) 2 °
v3 ,1
2
L2 (R̃)
and then,
°
°
°
°
° ∂λv3 °
°
∂2u °
°
°
°
|A| °
≤ Chv3 ,2 °δ−,R̃ 2 °
.
∂x2 °L2 (R)
∂x2 °L2 (R̃)
(3.13)
In order to estimate B we note that, since
Z
∂u
∂x1 (x1 , 0)
≡ 0 then,
Z
∂u
(x)ψv3 (x)dx2 dx1
∂x
1
lR̃,1
lR̃,2
Z
Z
Z x2
∂2u
=
(v11 − x1 )
(x1 , t)ψv3 (x)dtdx2 dx1 .
lR̃,1
lR̃,2 t=0 ∂x2 ∂x1
B=
Then,
(v11
− x1 )
¯
¯
¯
¯ ∂2u
¯
¯
(x
,
t)
¯ ∂x1 ∂x2 1 ¯ ψv3 (x)dtdx1 dx2
lR̃,2 lR̃,1 lR̃,2
°
°
°
1 ° x1 − R̃11
∂2u °
°
2
≤ C(hv3 ,1 hv3 ,2 ) °
°
° hv3 ,1 ∂x1 ∂x2 ° 2
Z
Z
Z
|B| ≤ Chv3 ,1
L (R̃)
15
where we have used the Schwarz inequality and the same argument used to obtain (3.9). Consequently
we obtain
°
°
°
°
° x − R̃
°
2
° ∂λv3 °
∂
u
°
°
1
11
°
|B| °
≤
Ch
°
°
v3 ,1
° ∂x2 ° 2
°
° 2
h
∂x
∂x
v
,1
1
2
3
L (R)
L (R̃)
which together with (3.12) and (3.13) implies

°
°
°
°
°
°

°
° x − R̃
2 °
2
°
°
°
∂
u
∂λ
∂
u
°
°
v
1
11
3
°uv3 (v1 )
°
°
°
≤
C
h
δ
+
h
°
°
v3 ,1
°
 v3 ,2 ° −,R̃ ∂x22 °L2 (R̃)
° hv3 ,1 ∂x1 ∂x2 °
∂x2 °L2 (R)


L2 (R̃)

.
∂λ
Clearly an analogous estimate follows for kuv4 (v2 ) ∂xv24 kL2 (R) , and then, in view of (3.10) and (3.11)
we conclude the proof of inequality (3.2).
The proof of Part (ii) is, of course, analogous to that of Part (i).
Proof of Part (iii): We will use the notation of the Figure 3(d). Then
Πu|R = uv4 (v4 )λv4 .
In this case the error can be split as
(u − Πu)|R = (u − uv4 ) + (uv4 − Πu)
and it is enough to bound uv4 − Πu, which is piecewise Q1 . Then we have
∂
∂λv4
(uv4 − Πu) = ((uv4 − Πu)(v4 ) − (uv4 − Πu)(v3 ))
∂x1
∂x1
(3.14)
+ ((uv4 − Πu)(v2 ) − (uv4 − Πu)(v1 ))
= −uv4 (v3 )
∂λv2
∂x1
∂λv4
∂λv2
+ (uv4 (v2 ) − uv4 (v1 ))
.
∂x1
∂x1
First we estimate |uv4 (v2 ) − uv4 (v1 )|. Using that
∂u
∂x1 (x1 , 0)
≡ 0 we have
Z
uv4 (v2 ) − uv4 (v1 ) =
(P (x, v2 ) − P (x, v1 ))ψv4 (x)dx
Z
∂u
= (v12 − v11 )
(x)ψv4 (x)dx
∂x1
Z Z x2
∂2u
= (v12 − v11 )
(x1 , t)ψv4 (x)dtdx.
t=0 ∂x1 ∂x2
It follows that
¯
¯
¯ ∂2u
¯
¯
¯
|uv4 (v2 ) − uv4 (v1 )| ≤ Chv4 ,1
¯ ∂x1 ∂x2 (x1 , t)¯ ψv4 (x)dtdx1 dx2
lR̃,2 lR̃,1 lR̃,2
¯
¯
Z
Z
Z
¯
hv ,1
x1 ¯¯ ∂ 2 u
(x1 , t)¯¯ ψv4 (x) 4 dx1 dtdx2 ,
≤ Chv4 ,1
¯
x1
lR̃,2 lR̃,2 lR̃,1 hv4 ,1 ∂x1 ∂x2
Z
Z
Z
and an argument similar to that used to obtain (3.9) gives
°
°
1 ° x1
∂2u °
°
|uv4 (v2 ) − uv4 (v1 )| ≤ C(hv4 ,1 hv4 ,2 ) 2 °
° hv ,1 ∂x1 ∂x2 °
4
.
L2 (R̃)
Therefore,
(3.15)
°
°
°
°
° ∂λv2 °
°
∂2u °
° ≤ Chv4 ,2 ° x1
°
|uv4 (v2 ) − uv4 (v1 )| °
° ∂x1 °
° hv ,1 ∂x1 ∂x2 ° 2 .
4
L (R̃)
Now we consider the other term in (3.14). We have to estimate |uv4 (v3 )|. Using that u(0, x2 ) ≡ 0
and v3 = (0, v23 ) we obtain
16
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Z Z
x1
∂2u
uv4 (v3 ) = −
t 2 (t, x2 )ψv4 (x)dtdx +
t=0 ∂x1
=: A + B
Z
(v23 − x2 )
∂u
(x)ψv4 (x)dx
∂x2
and we have to estimate A and B. We have
¯
¯
¯
x2 ¯¯ ∂ 2 u
hv ,2
|A| ≤ hv4 ,1
(t, x2 )¯¯ ψv4 (x) 4 dtdx
2
¯
h
h
∂x
x2
t=0 v4 ,1 v4 ,2
1
¯ 2
¯
Z
Z
Z
¯∂ u
¯
hv ,2
¯
δ−,R̃ (t, x2 ) ¯ 2 (t, x2 )¯¯ ψv4 (x) 4 dtdx2 dx1 .
≤ Chv4 ,1
∂x
x2
lR̃,1 lR̃,2 lR̃,1
1
Z Z
x1
t
But again, by an argument similar to that used in the proof of (3.9), we obtain
°
3 °
(hv4 ,1 ) 2 °
∂2u °
°
°
.
|A| ≤ C
δ
1
−,R̃
∂x21 °L2 (R̃)
(hv4 ,2 ) 2 °
Therefore,
°
°
°
°
°
° ∂λv4 °
∂2u °
°
°
°
|A| °
≤ Chv4 ,1 °δ−,R̃ 2 °
.
∂x1 °L2 (R)
∂x1 °L2 (R̃)
(3.16)
∂u
∂x2 (0, x2 )
On the other hand, using now that
≡ 0, we have
Z
∂u
(v23 − x2 )
(x)ψv4 (x)dx
∂x2
Z x1
Z
Z
∂2u
=
(v23 − x2 )
(t, x2 )ψv4 (x)dtdx2 dx1
t=0 ∂x1 ∂x2
lR̃,1 lR̃,2
B=
and then
Z
|B| ≤ Chv4 ,2
lR̃,1
Z
Z
lR̃,2
lR̃,1
x2
∂2u
hv ,2
(t, x2 )ψv4 (x) 4 dtdx2 dx1 .
hv4 ,2 ∂x1 ∂x2
x2
Hence
(3.17)
°
°
° ∂λv4 °
°
°
|B| °
∂x1 °
L2 (R)
°
°
° x2
∂2u °
°
°
≤ Chv4 ,2 °
hv ,2 ∂x1 ∂x2 °
4
.
L2 (R̃)
Now, inequality (3.5) follows from (3.14), (3.15), (3.16) and (3.17).
Since (3.6) is analogous to (3.5) the proof is concluded.
¤
4. Application to a reaction-diffusion problem
As an example of application of our results we consider in this section the singular perturbation
model problem
(4.1)
−ε2 ∆u + u = f
u=0
in (0, 2) × (0, 2)
on ∂{(0, 2) × (0, 2)}.
Compatibility conditions are assumed in order to have the regularity results proved in [15] and
[16]. As we will show, appropriate graded anisotropic meshes can be defined in order to obtain almost
optimal order error estimates in the energy norm valid uniformly in the parameter ε. These estimates
follow from our results of Sections 2 and 3.
The meshes that we construct are very different from the Shishkin type meshes that have been
used in other papers for this problem (see for example [2, 16]). In particular, our almost optimal error
estimate in the energy norm is obtained with meshes independent of ε.
17
Given a partition Th of (0, 2)× (0, 2) into rectangles, we call uh the Q1 finite element approximation
of the solution of problem (4.1). Since uh is the orthogonal projection in the scalar product associated
with the energy norm
n
o 12
kvkε = ε2 k∇vk2L2 ((0,2)2 ) + kvk2L2 ((0,2)2 )
we know that, for any vh in the finite element space,
ku − uh kε ≤ ku − vh kε .
In particular, if Π is the average interpolation operator associated with the partition Th introduced
in Section 2, we have
(4.2)
ku − uh kε ≤ ku − Πukε .
Therefore, we will construct the meshes in order to have a good estimate for the right hand side of
(4.2).
We will obtain our estimates in Ω = (0, 1) × (0, 1). Clearly, analogous arguments can be applied
for the rest of the domain. The constant C will be always independent of ε.
In order to bound the part of the error which contains the first derivatives we will make use of the
estimates obtained in the previous sections together with the fact that the solution of (4.1) satisfies
some weighted a priori estimates which are valid uniformly in the parameter ε. We state these a priori
estimates in the next two lemmas but postpone the proofs until the end of the section.
Lemma 4.1. There exists a constant C such that if α ≥
(4.3)
°
°
° α ∂u °
°x1
°
≤C
° ∂x1 ° 2 3
L ((0, )×(0, 3 ))
2
and
2
2
°
°
° α ∂2u °
°
ε°
≤ C,
x
° 1 ∂x2 ° 2 3
1 L ((0, )×(0, 3 ))
2
(4.5)
°
° α
ε°
°x1
≤C
1
2
2
then
°
°
° α ∂2u °
°
ε°
≤ C,
x
° 2 ∂x2 ° 2 3
2 L ((0, )×(0, 3 ))
2
°
∂u °
°
∂x1 ∂x2 °
then
°
°
° α ∂u °
°x2
°
≤ C.
° ∂x2 ° 2 3
L ((0, )×(0, 3 ))
Lemma 4.2. There exists a constant C such that if α ≥
(4.4)
1
2
2
and
L2 ((0, 32 )×(0, 23 ))
2
°
°
° α ∂2u °
°
ε°
x
° 2 ∂x1 ∂x2 °
≤ C.
L2 ((0, 23 )×(0, 32 ))
To estimate the error in the L2 norm we will use a priori estimates in the following norms. For
v : R → R, where R is the rectangle R = l1 × l2 , define
(4.6)
°
°
°
°
kvk∞×1,R := °kv(x1 , ·)kL1 (l2 ) °
L∞ (l1 )
and
°
°
°
°
kvk1×∞,R := °kv(·, x2 )kL1 (l1 ) °
L∞ (l2 )
.
Then we have the following lemma, which also will be proved at the end of the section.
Lemma 4.3. There exists a constant C such that
°
°
° ∂u °
°
°
≤ C and
° ∂x1 °
1×∞,(0, 3 )×(0, 3 )
2
2
°
°
° ∂u °
°
°
° ∂x2 °
≤ C.
∞×1,(0, 32 )×(0, 32 )
Let us now define the graded meshes. Given a parameter h > 0 and α ∈ (0, 1) we introduce the
1
α
1−α , ξ
partition {ξi }N
i+1 = ξi + hξi for i = 2, · · · , N − 2,
i=0 of the interval [0, 1] given by ξ0 = 0, ξ1 = h
α
where N is such that ξN −1 < 1 and ξN −1 + hξN −1 ≥ 1, and ξN = 1. We assume that the last interval
(ξN −1 , 1) is not too small in comparison with the previous one (ξN −2 , ξN −1 ) (if this is not the case we
just eliminate the node ξN −1 ).
We define the partitions Th,α such that they are symmetric with respect to the lines x1 = 1 and
x2 = 1 and in the subdomain Ω = (0, 1) × (0, 1) are given by
{R ⊂ Ω : R = (ξi−1 , ξi ) × (ξj−1 , ξj ) for 1 ≤ i, j ≤ N }.
18
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Observe that the family of meshes Th,α satisfies our local regularity condition (2.1) with σ = 2α ,
that is, if S, T ∈ Th,α are neighboring elements then
hT,i
≤ 2α .
hS,i
For these meshes we have the following error estimates. We set Ω̃ = ∪{R̃ : R ⊂ Ω} where we are
using the notations of the previous sections.
Theorem 4.4. If u ∈ H 2 (Ω) and u ≡ 0 on {x : x1 = 0 or x2 = 0} then there exists a constant C
such that
)
(°
°
°
°
° α ∂u °
° α ∂u °
°
°
°
°
+ x
ku − ΠukL2 (Ω) ≤ Ch °x1
∂x1 °L2 (Ω̃) ° 2 ∂x2 °L2 (Ω̃)
)
(°
(4.7)
°
°
°
° ∂u °
° ∂u °
1
°
°
+ Ch 2−2α °
+°
,
° ∂x2 °
° ∂x1 °
∞×1,Ω̃
1×∞,Ω̃
(4.8)
ð
!
°
°
°
°
°
2
° ∂(u − Πu) °
° α ∂2u °
° α
°
∂
u
α
°
°
°
°
≤ Ch °
+°
,
°
°x1 ∂x2 ° 2
°(x1 + x2 ) ∂x1 ∂x2 ° 2
° 2
∂x1
1 L (Ω̃)
L (Ω)
L (Ω̃)
and
(4.9)
°
°
° ∂(u − Πu) °
°
°
°
°
∂x2
L2 (Ω)
!
°
°
° α ∂2u °
°
.
+°
°x2 ∂x2 ° 2
2 L (Ω̃)
L2 (Ω̃)
ð
°
° α
∂2u °
α
°
°
≤ Ch °(x1 + x2 )
∂x1 ∂x2 °
Proof. We will estimate the error on each element according to its position. So, we decompose the
domain Ω into four parts, Ωi , i = 1, . . . , 4 defined as
Ω1 = [ξ1 , ξN ]2
Ω2 = [ξ1 , ξN ] × [0, ξ1 ]
Ω3 = [0, ξ1 ] × [ξ1 , ξN ]
Ω4 = [0, ξ1 ]2 ,
and we set Ω̃i = ∪{R̃ : R ⊂ Ωi }, i = 1, . . . , 4.
In order to prove (4.7) we split the error as follows
(4.10)
ku − Πuk2L2 (Ω) =
4
X
ku − Πuk2L2 (Ωi ) =: S1 + S2 + S3 + S4 .
i=1
First we estimate S1 . If R̃ ∩ {x : x1 = 0 or x2 = 0} = ∅ we have that, for each S ⊂ R̃, hS,1 ≤ hxα
1 and
for
all
(x
,
x
)
∈
S
and
then,
Theorem
2.4
gives
hS,2 ≤ hxα
1
2
2
(
ku −
Πuk2L2 (R)
≤C
≤C
h2R,1
X
S⊂R̃
≤C
X
S⊂S̃
(
¯
Z ¯
Z
¯ ∂u ¯2
¯
¯ dx + h2R,2
¯ ∂x1 ¯
R̃
R̃
)
¯
¯
¯ ∂u ¯2
¯
¯
¯ ∂x2 ¯ dx
¯
Z ¯
Z
¯ ∂u ¯2
2
2
¯
¯
hS,1
¯ ∂x1 ¯ dx + hS,2
S
S
)
¯
¯
¯ ∂u ¯2
¯ dx
¯
¯ ∂x2 ¯
¯2
¯
¯2 )
¯
Z
Z
¯
¯
¯
¯
2α ¯ ∂u ¯
2α ¯ ∂u ¯
2
2
x2 ¯
x1 ¯
dx + h
dx
h
¯
∂x1
∂x2 ¯
(
S
S
(
=C
¯
¯2
¯
¯2 )
Z
Z
¯
¯
¯
¯
2α ¯ ∂u ¯
2α ¯ ∂u ¯
2
2
x1 ¯
x2 ¯
h
dx + h
dx .
¯
∂x1
∂x2 ¯
R̃
R̃
Now, suppose that R ⊂ Ω1 , R̃ ∩ {x : x2 = 0} 6= ∅ and R̃ ∩ {x : x1 = 0} = ∅. Then R̃12 = 0 and,
1
1−α . Therefore, using Theorem 2.4 we
if S ⊂ R̃, we have hS,1 ≤ hxα
1 for (x1 , x2 ) ∈ S and hS,2 ≤ Ch
obtain
19
¯
¯
¯
¯
Z
¯ ∂u ¯2
¯ ∂u ¯
2α
2
2α
¯
¯
¯
¯ dx
δR̃ (x) ¯
dx + ChR,2
δR̃ (x) ¯
∂x1 ¯
∂x2 ¯
R̃
R̃
(
¯2
¯2 )
¯
Z ¯
Z
¯
¯
¯
¯
∂u
∂u
2−2α
¯
¯ dx
¯
¯
h2S,1
x2α
2 ¯
¯ ∂x1 ¯ dx + ChS,2
¯
∂x
2
S
S
Z
ku −
Πuk2L2 (R)
≤Ch2R,1
X
≤C
S⊂R̃
X
≤C
(
Z
h
2
S
S⊂R̃
x2α
1
¯
¯
¯
¯2 )
Z
¯ ∂u ¯2
¯
¯
2α ¯ ∂u ¯
¯
¯ dx + Ch2
x2 ¯
dx
¯ ∂x1 ¯
∂x2 ¯
S
¯
¯
¯2
¯2 )
Z
Z
¯
¯
¯
¯
2
2α ¯ ∂u ¯
2
2α ¯ ∂u ¯
dx + h
x2 ¯
dx .
h
x1 ¯
¯
∂x1
∂x2 ¯
(
=C
R̃
R̃
Now, if 0 ∈ R̃, that is R̃ ∩ {x : x1 = 0} 6= ∅ and R̃ ∩ {x : x2 = 0} 6= ∅, then, R̃11 = R̃12 = 0 and
1
1
hR,1 ≤ Ch 1−α , hR,2 ≤ Ch 1−α . Then, from Theorem 2.4 we have
¯
¯
¯
¯
Z
¯ ∂u ¯2
¯ ∂u ¯
2α
2
2α
¯
¯
¯
¯ dx
δR̃ (x) ¯
dx + ChR,2
δR̃ (x) ¯
∂x1 ¯
∂x2 ¯
R̃
R̃
¯
¯
¯2
¯2
Z
Z
¯
¯
¯
¯
2−2α
2−2α
2α ¯ ∂u ¯
2α ¯ ∂u ¯
x1 ¯
x2 ¯
≤ChR,1
dx + ChR,2
dx
¯
∂x1
∂x2 ¯
R̃
R̃
Z
ku −
Πuk2L2 (R)
≤Ch2R,1
(
≤C
Z
h
2
R̃
x2α
1
¯
¯
¯2 )
¯
Z
¯ ∂u ¯2
¯
¯
2α ¯ ∂u ¯
¯
¯ dx + h2
x2 ¯
dx .
¯ ∂x1 ¯
∂x2 ¯
R̃
A similar estimate can be obtained for ku − ΠukL2 (R) when R̃ ∩ {x : x1 = 0} 6= ∅ and R̃ ∩ {x : x2 =
0} = ∅. Therefore, we have
( Z
¯2
¯2 )
¯
¯
Z
X
¯
¯
¯
¯
2
2α ¯ ∂u ¯
2
2α ¯ ∂u ¯
S1 ≤C
h
x1 ¯
dx + h
x2 ¯
dx
¯
∂x1
∂x2 ¯
R̃
R̃
R⊂Ω1
(4.11)
¯
¯2
¯
¯
Z
Z
¯
¯ ∂u ¯2
¯
2
2α ¯ ∂u ¯
¯
¯
≤Ch2
x2α
dx
+
Ch
x
dx.
1 ¯
2 ¯
¯
∂x1
∂x2 ¯
Ω̃
Ω̃
Now, we estimate S2 . From Theorem 2.4 we know that kΠukL2 (R) ≤ CkukL2 (R̃) for all R ∈ Th,α and
therefore
X
X
(4.12)
S2 =
ku − Πuk2L2 (R) ≤ C
kuk2L2 (R̃) ≤ Ckuk2L2 (Ω̃2 ) .
R⊂Ω2
R⊂Ω2
1
So, we have to estimate kukL2 (Ω̃2 ) . We have Ω̃2 = lΩ̃2 ,1 × lΩ̃2 ,2 with |lΩ̃2 ,1 | ≤ C and |lΩ̃2 ,2 | ≤ Ch 1−α .
Using that u(x1 , 0) ≡ 0 we have
Z
kuk2L2 (Ω̃2 )
u2 (x)dx
=
lΩ̃
2 ,1
lΩ̃
2 ,2
¾2
∂u
=
(x1 , t)dt dx2 dx1
∂x2
lΩ̃ ,1 lΩ̃ ,2
0
2
2
°
°°
°2
Z
°
°° ∂u
°
°
°°
(x1 , ·)°
≤C
dx2
°
°°
°
°
°
∂x
1
2
lΩ̃ ,2
L (l
)
Z
(4.13)
Z
Z
½Z
x2
Ω̃2 ,2
2
1
≤Ch 1−α
L∞ (lΩ̃
2 ,1
°
°
° ∂u °2
°
°
° ∂x2 °
∞×1,Ω̃2
and so, it follows from (4.12) and (4.13) that
(4.14)
S2 ≤ Ch
1
1−α
°
°
° ∂u °2
°
°
° ∂x2 °
∞×1,Ω̃2
.
)
20
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
Analogously we can prove that
(4.15)
S3 ≤ Ch
1
1−α
°
°
° ∂u °2
°
°
° ∂x1 °
,
1×∞,Ω̃3
(4.16)
S4 ≤ Ch
2
1−α
°
°
° ∂u °2
°
°
° ∂x2 °
and
S4 ≤ Ch
∞×1,Ω̃4
2
1−α
°
°
° ∂u °2
°
°
° ∂x1 °
1×∞,Ω̃4
and inserting inequalities (4.11), (4.14), (4.15) and (4.16) in (4.10) we obtain (4.7) (note that Ω̃4 ⊂ Ω̃2
and Ω̃4 ⊂ Ω̃3 ).
Let us now prove (4.8). Inequality (4.9) follows in a similar way. Again we use the decomposition
of Ω into the four subsets Ωi , i = 1, . . . , 4 defined above. Then we have
°2
4 °
X
° ∂
°
°
°
=
° ∂x1 (u − Πu)° 2
°
°2
° ∂
°
°
°
° ∂x1 (u − Πu)° 2
(4.17)
L (Ω)
=: S1 + S2 + S3 + S4
L (Ωi )
i=1
and we have to estimate Si , i = 1, . . . , 4.
For S1 , Theorem 2.5 gives
°2
X °
° ∂
°
°
°
S1 =
° ∂x1 (u − Πu)° 2
L (R)
R⊂Ω1
(
¯
¯
¯2 )
¯ 2
Z
Z
X
¯ ∂2u
¯
¯ ∂ u ¯2
2
2α
2
2α
¯
¯
¯
δR̃ (x) ¯
(x)¯¯ dx
≤
hR,1
δR̃ (x) ¯ 2 (x)¯ dx + hR,2
∂x1
∂x1 ∂x2
R̃
R̃
R⊂Ω1
X
=:
IR .
R⊂Ω1
Now, if R̃ ∩ {x : x1 = 0 or x2 = 0} = ∅ we have
(
X
|IR | ≤ C
h2T,1
T ⊂R̃
Z
Z ¯ 2 ¯2
¯∂ u¯
¯ dx + h2T,2
¯
¯ 2¯
T
T ∂x1
)
¯
¯
¯ ∂ 2 u ¯2
¯
¯
¯ ∂x1 ∂x2 ¯ dx
but, for T ⊂ Ω1 , we have that
(4.18)
hT,1 ≤ Chxα
1 ,
hT,2 ≤ Chxα
2
∀(x1 , x2 ) ∈ T,
and therefore,
(
|IR | ≤ C
h
Z
2
R̃
x2α
1
)
¯ 2 ¯2
¯
¯
Z
¯∂ u¯
¯ ∂ 2 u ¯2
2α
2
¯
¯ dx .
¯
x2 ¯¯
¯ ∂x2 ¯ dx + h
∂x1 ∂x2 ¯
R̃
1
On the other hand, if R̃ ∩ {x : x2 = 0} 6= ∅ and R̃ ∩ {x : x1 = 0} = ∅, there are some elements T ⊂ R̃
1
that not verify condition (4.18). For a such elements T we have hT,2 ≤ h 1−α while the condition on
hT,1 in (4.18) remains valid. So we obtain
( Z
¯ 2 ¯2
¯2 )
¯
Z
2
X
¯
¯
¯
¯
2−2α
2α ¯ ∂ u ¯
2
2α ¯ ∂ u ¯
x1 ¯ 2 ¯ dx + hT,2
h
|IR | ≤ C
x2 ¯
dx
∂x1
∂x1 ∂x2 ¯
T
T
T ⊂R̃
( Z
¯ 2 ¯2
¯2 )
¯
Z
2
X
¯
¯
¯
¯
2α ¯ ∂ u ¯
2
2α ¯ ∂ u ¯
2
x1 ¯ 2 ¯ dx + h
h
≤C
x2 ¯
dx .
∂x1
∂x1 ∂x2 ¯
T
T
T ⊂R̃
1
1
Now, if (0, 0) ∈ R̃ we have hT,1 ≤ Ch 1−α and hT,2 ≤ Ch 1−α for all T ⊂ R̃ and therefore,
21
(
)
¯ 2 ¯2
¯
¯
Z
¯∂ u¯
¯ ∂ 2 u ¯2
2−2α
2α
¯
¯
¯ dx
|IR | ≤ C
x2 ¯¯
¯ ∂x2 ¯ dx + hT,2
∂x1 ∂x2 ¯
T
T
1
T ⊂R̃
( Z
¯ 2 ¯2
¯
¯2 )
Z
2
X
¯
¯
¯
¯
2
2α ¯ ∂ u ¯
2
2α ¯ ∂ u ¯
h
≤C
x1 ¯ 2 ¯ dx + h
x2 ¯
dx .
∂x1
∂x1 ∂x2 ¯
T
T
X
Z
h2−2α
T,1
x2α
1
T ⊂R̃
If R̃ ∩ {x : x1 = 0} 6= ∅ and R̃ ∩ {x : x2 = 0} = ∅ we can estimate IR analogously and so we obtain
(
(4.19)
S1 ≤ C
¯
¯ 2 ¯2
¯2 )
Z
2
¯
¯
¯
¯
∂
u
∂
u
2
¯
¯
¯
¯ dx .
x2α
x2α
2 ¯
1 ¯
2 ¯ dx + h
¯
∂x
∂x
∂x
1
2
Ω̃
Ω̃
1
Z
h2
Let us now estimate S2 . From Theorem 3.1(i) we have


!2α ¯
¯ 2 ¯2
¯2 
Z Ã
Z
2
X 
¯
¯
¯
¯
∂ u
x1 − R̃11
¯ ∂ u ¯ dx
S2 ≤
h2
δ 2α (x) ¯¯ 2 ¯¯ dx + h2R,2
¯ ∂x1 ∂x2 ¯
 R,1 R̃ −,R̃

∂x1
hR,1
R̃
R∈Ω2
(4.20)
X
=:
IR .
R∈Ω2
Now, if R ⊂ Ω2 is such that R̃ ∩ {x : x1 = 0} = ∅ then, we have


¯2 
Z ¯ 2 ¯2
Z ¯
X
2
X
¯
¯
¯
¯
¯ ∂ u ¯ dx +
¯ ∂ u ¯ dx
|IR | ≤ C
h2R,1
h2R,2
¯ ∂x2 ¯
¯
¯


T
T ∂x1 ∂x2
1
T ⊂R̃
T ⊂R̃
but, in this case, for T ⊂ R̃,
hT,2 ≤ ChT,1 ≤ Chxα
1
∀x = (x1 , x2 ) ∈ T
and therefore,
(
(4.21)
|IR | ≤ C
Z
h
2
R̃
x2α
1
¯ 2 ¯2
¯
¯2 )
Z
2
¯∂ u¯
¯
¯
2α ¯ ∂ u ¯
¯ dx + h2
¯
x1 ¯
dx .
¯ ∂x2 ¯
∂x1 ∂x2 ¯
R̃
1
On the other hand, if R ⊂ Ω2 is such that R̃ ∩ {x : x1 = 0} 6= ∅, R̃11 = 0 and so, it follows from (4.20)
that (note that hR,2 ≤ hR,1 )


¯ 2 ¯2
¯
¯2 
Z
Z
X
2
X
¯
¯
¯
¯
∂
u
∂
u
¯
¯ dx +
¯
¯ dx
|IR | ≤ C
h2−2α
x2α
h2−2α
x2α
1 ¯
1 ¯
R,1
R,1


∂x21 ¯
∂x1 ∂x2 ¯
T
T
T ⊂R̃
T ⊂R̃
1
but in this case, for T ⊂ R̃, hT,1 ≤ Ch 1−α and then
Z
(4.22)
|IR | ≤ h
2
R̃
x2α
1
¯ 2 ¯2
¯2
¯
Z
2
¯∂ u¯
¯
¯
2α ¯ ∂ u ¯
¯
¯ dx + h2
x1 ¯
dx.
¯ ∂x2 ¯
∂x1 ∂x2 ¯
R̃
1
Therefore, inserting inequalities (4.21) and (4.22) in (4.20) we obtain
( Z
¯ 2 ¯2
¯2 )
¯
Z
2
¯
¯
¯
¯
2α ¯ ∂ u ¯
2
2α ¯ ∂ u ¯
2
x1 ¯ 2 ¯ dx + h
(4.23)
S2 ≤ C h
x1 ¯
dx .
∂x1
∂x1 ∂x2 ¯
Ω̃2
Ω̃2
Let us now estimate S3 . Using Theorem 3.1(ii) we have


!2α ¯
¯ 2 ¯2
¯2 
Z
Z Ã
2
X 
¯
¯
¯
¯
∂
u
∂
u
x
−
R̃
2
12
¯
¯
S3 ≤ C
h2
δ 2α (x) ¯¯ 2 ¯¯ dx + h2R,2
¯ ∂x1 ∂x2 ¯ dx
 R,1 R̃ −,R̃
∂x1
hR,2
R̃
R∈Ω3
X
=:
IR .
R∈Ω3
22
RICARDO G. DURÁN AND ARIEL L. LOMBARDI
If R ⊂ Ω3 is such that R̃ ∩ {x : x2 = 0} = ∅ then, for T ⊂ R̃,
1
hT,2 ≤ Chxα
2
hT,1 ≤ Ch 1−α
and so
(
X
|IR | ≤ C
Z
h2−2α
T,1
T ⊂R̃
(4.24)
≤
X
(
T
Z
2
h
T
T ⊂R̃
x2α
1
x2α
1
∀(x1 , x2 ) ∈ T,
¯ 2 ¯2
Z
¯∂ u¯
¯
¯ dx + h2T,2
¯ ∂x2 ¯
T
1
)
¯
¯
¯ ∂ 2 u ¯2
¯
¯ dx
¯ ∂x1 ∂x2 ¯
¯ 2 ¯2
¯
¯2 )
Z
2
¯∂ u¯
¯
¯
2α ¯ ∂ u ¯
¯
¯ dx + h2
x2 ¯
dx .
¯ ∂x2 ¯
∂x1 ∂x2 ¯
T
1
If R̃ ∩ {x : x2 = 0} 6= ∅ then R̃12 = 0 and so (4.24) can be obtained also for this case using similar
arguments. Therefore, we have
(
(4.25)
S3 ≤ C
Z
h
2
Ω̃3
x2α
1
)
¯
¯
¯ 2 ¯2
Z
¯∂ u¯
¯ ∂ 2 u ¯2
2
2α
¯
¯ dx .
¯
x2 ¯¯
¯ ∂x2 ¯ dx + h
∂x1 ∂x2 ¯
Ω̃3
1
Finally, to estimate S4 , note that Ω4 contains only one element R. Now, using Theorem 3.1(iii)
1
and the fact that for this element hR,1 = hR,2 = h 1−α we obtain
(4.26)
(°
°
2 °2
°
2 ° α ∂ u°
S4 ≤ Ch °x1 2 °
∂x
2
1
L (R̃)
°
°
° α ∂ 2 u °2
°
°
+ °x1
∂x1 ∂x2 ° 2
L (R̃)
°
°
° α ∂ 2 u °2
°
°
+ °x2
∂x1 ∂x2 ° 2
)
.
L (R̃)
Collecting the inequalities (4.19), (4.23), (4.25) and (4.26) we obtain (4.8) concluding the proof. ¤
As a consequence of Theorem 4.4 and the a priori estimates for the solution of problem (4.1) we
obtain the following error estimates for the finite element approximations obtained using the family
of meshes Th,α . To simplify notation we omit the subscript α in the approximate solution.
Corollary 4.5. Let u be the solution of (4.1) and uh its Q1 finite element approximation obtained
using the mesh Th,α with 12 ≤ α < 1 . If N is the number of nodes of Th,α then, there exists a constant
C independent of ε and N such that
(4.27)
ku − uh kε ≤ C
1
1
√ log N.
1−α N
Proof. From (4.2), Lemmas 4.1, 4.2 and 4.3, and Theorem 4.4 (and its extension to the rest of
(0, 2) × (0, 2)) it follows that if h is small enough (h < 12 is sufficient) and α ≥ 21 , then
ku − uh kε ≤ Ch.
So we have to estimate h in terms of N . If we denote with M the number of nodes in each direction
in the subdomain Ω, we have N ∼ M 2 and we will estimate M . Let f (ξ) = ξ + hξ α . Then, ξ0 = 0,
1
ξ1 = h 1−α and ξi+1 = f (ξi ), i = 1, . . . , Mf − 1 where Mf (= M ) is the first number i such that ξi ≥ 1.
Since α < 1 we have that
f (ξ) > ξ + hξ =: g(ξ),
Mg
{ηi }i=0
∀ξ ∈ (0, 1).
Now, consider the sequence
given by η1 = ξ1 , and ηi+1 = g(ηi ), i = 2, . . . Mg where Mg is
defined analogously to Mf . Then, it is easy to see that Mf < Mg and therefore, it is enough to
1
estimate Mg . But, Mg = [m] where m solves (1 + h)m−1 ξ1 = 1. Since ξ1 = h 1−α , for 0 ≤ h ≤ 1, we
obtain
1
1 1
1
1 1
log ≤ m − 1 ≤ C
log .
(4.28)
1−αh
h
1−αh
h
Now, from inequalities (4.28) we easily arrive at
1 1
h≤C
log M
1−αM
for all h small enough.
¤
23
The Lemmas 4.1, 4.2 and 4.3 are straightforward consequences of the following estimates
¯ k
¯
n
o
¯∂ u
¯
x
2−x1
−k − ε1
¯
¯
(4.29)
(x
,
x
)
+ ε−k e− ε
¯ ∂xk 1 2 ¯ ≤ C 1 + ε e
¯ k1
¯
n
o
¯∂ u
¯
x
2−x
¯
¯ ≤ C 1 + ε−k e− ε2 + ε−k e− ε 2
(4.30)
(x
,
x
)
1
2
¯ ∂xk
¯
2
provided that 0 ≤ k ≤ 4 and (x1 , x2 ) ∈ [0, 2] × [0, 2], which are proved in [16]. As an example we prove
r
the first inequality in (4.5). Observe that, for r = 0, 1, 2, ∂∂xur (x1 , x2 ) ≡ 0 when x2 = 0 or x2 = 2 for
i
i = 1 and when x1 = 0 or x1 = 2 for i = 2. Then we have
Z
2
0
Z
3
2
0
(4.31)
¯
¯
Z 32 Z 2
¯ ∂ 2 u ¯2
∂u ∂ 3 u
¯
¯
dx
dx
=
−
x2α
dx2 dx1
x2α
1
2
1 ¯
1
¯
∂x1 ∂x2
∂x1 ∂x1 ∂x22
0
0
)
¯3
µ
¶ 2
Z 2(
Z 32
2 ¯2
∂
∂ u
2α ∂u ∂ u ¯
2α ∂u
=−
x1
−
x1
dx1 dx2
∂x1 ∂x22 ¯x1 =0
∂x1 ∂x22
0
0 ∂x1
Z 2
Z 2 Z 32
3
∂u 3
∂2u 3
∂u ∂ 2 u
= −( )2α
( , x2 ) 2 ( , x2 )dx2 +
2αx2α−1
dx1 dx2
1
2
∂x2 2
∂x1 ∂x22
0 ∂x1 2
0
0
Z 2 Z 23
∂2u ∂2u
x2α
dx1 dx2
+
1
∂x21 ∂x22
0
0
=: I + II + III.
Now, since
¯
¯
¯ ∂u 3
¯
¯
¯
¯ ∂x1 ( 2 , x2 )¯ ≤ C
¯
¯ 2
¯
¯∂ u 3
−2
¯
¯
¯ ∂x2 ( 2 , x2 )¯ ≤ C(1 + ε )
2
¯
¯
¯
¯ ∂u
x
−1 − ε1
¯
¯
) (0 ≤ x1 ≤ 3/2)
¯ ∂x1 (x1 , x2 )¯ ≤ C(1 + ε e
¯ 2
¯
¯∂ u
¯
x
−2 − ε1
¯
¯
) (0 ≤ x1 ≤ 3/2)
¯ ∂x2 (x1 , x2 )¯ ≤ C(1 + ε e
¯ 21
¯
¯∂ u
¯
−2
¯
¯
¯ ∂x2 (x1 , x2 )¯ ≤ C(1 + ε )
2
we easily obtain
(4.32)
(4.33)
(4.34)
|I|
≤ C(1 + ε−2 )
|II| ≤ C(ε−2 + ε2α−3 )
|III| ≤ C(ε−2 + ε2α−3 ).
Now, using inequalities (4.32), (4.33) and (4.34) in (4.31) we conclude the proof.
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RICARDO G. DURÁN AND ARIEL L. LOMBARDI
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119.
Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,
1428 Buenos Aires, Argentina.
E-mail address: rduran@dm.uba.ar
Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,
1428 Buenos Aires, Argentina.
E-mail address: aldoc7@dm.uba.ar
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