R E S E A R C H
Open Access
Maximum norm analysis of a
nonmatching grids method for a class of
variational inequalities with nonlinear
source terms
Abida Harbi
**Correspondence:
Laboratory of Applied Mathematics, Department of Mathematics, Badji Mokhtar-Annaba University, P.O. Box 12, Annaba, 23000, Algeria
Abstract
In this paper, we study a nonmatching grid finite element approximation of a class of elliptic variational inequalities with nonlinear source terms in the context of the Schwarz alternating domain decomposition. We show that the approximation converges optimally in the maximum norm, on each subdomain, making use of a Lipschitz continuous dependence with respect to both the boundary condition and the source term.
MSC: 65K15; 65N30; 65N15
Keywords: variational inequalities with nonlinear source terms; Schwarz alternating method; nonmatching grids; finite element;L∞-error estimate
1 Introduction
This paper deals with the error analysis in the maximum norm, in the context of the non-matching grids method, of the following variational inequalities with nonlinear source terms: findu∈Kgsuch that
a(u,v–u) +c(u,v–u)≥f(u),v–u, ∀v∈Kg, (.) wherea(u,v) =(∇u· ∇v)dxis the bilinear form defined in a bounded domainofR orR,cis a positive constant such that
c≥β> , (.)
whereβ is a positive constant,f is a nonlinear Lipschitz functional,Kg is a convex set defined by
Kg=v∈H() such thatv=gon∂andv≤ψon (.) withψ∈W,∞() such thatψ≥ on∂is the obstacle function, andg∈L∞(∂) is the boundary condition.
The concept of the nonmatching finite elements grids used in this paper consists of de-composing the whole domainin two overlapping subdomains and to discretize each subdomain by an independent finite element method. As the two discretizations are in-dependent on the overlap region, the discrete analog of problem (.) cannot be defined, and the alternating Schwarz method is then used to resolve the two discrete subproblems arising from these nonmatching finite elements grids.
We refer to [–], and the references therein for the analysis of the Schwarz alternat-ing method for elliptic obstacle problems and to the proceedalternat-ings of the annual domain decomposition conference beginning with [].
For results on maximum norm error analysis of overlapping nonmatching grids methods for elliptic problems we refer, for example, to [–].
In this paper we consider the class of variational inequalities with nonlinear source terms (.) [], where the main objective is to demonstrate that the approximation converges optimally on each subdomain making use of the characterization of the discrete solution as the upper bound of the set of discrete subsolutions [], a Lipschitz continuous depen-dence with respect to both the boundary condition and the source term, and the standard finite elementL∞error estimate for the elliptic obstacle problem [].
More precisely, ifuidenotes the true solution and (unhi) the discrete Schwarz sequence
with respect to the triangulation with mesh sizehioni, we show that
ui–unhiL∞(i)≤Ch
|logh|, (.)
whereh=max(h,h),Cis a constant independent ofnandh. This result coincides with the optimal convergence order of elliptic variational inequalities of an obstacle type prob-lem [].
2 Elliptic variational inequalities of obstacle type problem
In this section we begin by laying down some definitions and classical results related to variational inequalities, then we prove a Lipschitz continuous and discrete dependence with respect to both the boundary condition and the source term which will assume a crucial role in the proof of the main result of this paper.
Letbe a bounded polyhedral domain ofRorRwith sufficiently smooth boundary
∂. We consider the bilinear form
a(u,v) =
(∇u· ∇v)dx, (.)
the linear form
(f,v) =
f(x)·v(x)dx, (.)
the right-hand side
f ∈L∞(), (.)
the obstacle
the boundary conditiong∈L∞(∂), and the nonempty convex set
Kg=v∈H() such thatv=gon∂andv≤ψon. (.)
We consider the variational inequality (VI): findu∈Kgsuch that
a(u,v–u) +c(u,v–u)≥(f,v–u), ∀v∈Kg, (.)
wherec∈Randc> such that
c≥β> , (.)
whereβ is a positive constant. Letτhbe a triangulation ofwith mesh sizeh,Vhbe the
space of finite elements consisting of continuous piecewise linear functionsvvanishing on∂, andφs,s= , , . . . ,m(h) be the basis functions ofVh.
The discrete counterpart of (.) consists of findinguh∈Khgsuch that
a(uh,v–uh) +c(uh,v–uh)≥(f,v–uh), ∀v∈Khg, (.)
where
Khg={v∈Vhsuch thatv=πhgon∂andv≤rhψon}, (.)
πhis an interpolation operator on∂andrhis the usual finite element restriction operator
on.
Theorem (see []) Under conditions(.)and(.),there exists a constant C indepen-dent of h such that
ζ–ζhL∞()≤Ch|logh|. (.)
2.1 A Lipschitz continuous dependence with respect to both the boundary condition and the source term
This subsection is devoted to the establishment of a Lipschitz continuous dependence property of the solution with respect to the data whose proof is based on a monotonicity property of the solution of (.) with respect to the source term and the boundary condi-tion by which we first set out and demonstrate our result.
Proposition Let(f;g); (f˜,g)˜ be a pair of data andζ=σ(f,g);ζ˜=σ(f˜,g)˜ the correspond-ing solutions to(.).If f ≤ ˜f inand g≤ ˜g on∂thenζ≤ ˜ζ in.
Proof According to [],ζ=max{ζ}where{ζ}is the set of all the subsolutions ofζ. Hence ∀ζ∈ {ζ},ζ satisfies
with
ζ ≤g on∂.
Then the two inequalitiesf≤ ˜f inandg≤ ˜gon∂imply
a(ζ,v) +c(ζ,v)≤(f,v)≤(f˜,v), ∀v≥,
with
ζ ≤g≤ ˜g on∂.
So,ζis a subsolution ofζ˜=σ(f˜,g), that is,˜ ζ≤ ˜ζ.
Proposition Under the conditions of Proposition,we have
ζ–ζ˜L∞()≤max
β
f–f˜L∞(),g–g˜L∞(∂)
. (.)
Proof First, set
=max β
f–f˜L∞(),g–g˜L∞(∂)
. (.)
Then
˜
f ≤f +f –˜fL∞() ≤f + ()f –f˜L∞() ≤f +
c
β
f –f˜L∞()
≤f +cmax β
f –f˜L∞(),g–g˜L∞(∂)
≤f +c.
So,
˜
f ≤f +c in. (.)
On the other hand, we have
˜
ζ =g˜≤g+ on∂. (.)
Thus, making use of (.), (.), and Proposition , we obtain
˜
Sinceζ+is a solution of the following VI:
aζ+, (v+) – (ζ+)+cζ+, (v+) – (ζ+)≥f+c, (v+) – (ζ+)
with
v+,ζ+∈K(g+), we have
ζ+=σ(f+c,g+). (.)
Equations (.) and (.) imply
˜
ζ ≤ζ+ in,
thus
˜
ζ–ζ≤ in. (.)
Similarly, interchanging the roles of the couples (f;g); (f˜,g), we obtain˜
ζ–ζ˜≤ in, (.)
which completes the proof.
2.2 A Lipschitz discrete dependence with respect to both the boundary condition and the source term
Assuming that the discrete maximum principle (d.m.p.) is satisfied,i.e.the matrix resulting from the finite element discretization is anM-matrix (see [, ]), we prove the Lipschitz discrete dependence with respect to both the boundary condition and the source term by a similar study to that undertaken previously for the Lipschitz continuous dependence property.
Proposition Let(f,g); (f˜,g)˜ be a pair of data andζh=σh(f,g);ζ˜h=σh(f˜,g)˜ the
corre-sponding solutions to(.).If f ≥ ˜f inand g≥ ˜g on∂thenζh≥ ˜ζhin.
Proof The proof is similar to that of the continuous case.
The proposition below establishes a Lipschitz discrete dependence of the solution with respect to the data.
Proposition Provided that the d.m.p.is verified,then under the conditions of Proposi-tion,we have
ζh–ζ˜hL∞()≤max
β
f–f˜L∞(),g–g˜L∞(∂)
. (.)
Proof The proof is similar to that of the continuous case. Indeed, as the basis functionsφs
3 Schwarz alternating methods for VI with nonlinear source terms
We consider the following variational inequality with nonlinear source term (.): Find u∈Kg, a solution of
a(u,v–u) +c(u,v–u)≥f(u),v–u, ∀v∈Kg, (.) where
a(u,v) =
(∇u∇v)dx, (.)
f(·) is a Lipschitz continuous nondecreasing nonlinear source term onR,
f(x) –f(y)≤k|x–y|, ∀x,y∈R, (.)
withksatisfying
k<β, (.)
whereβis defined in (.).
Theorem (see []) Problem(.)has a unique solution.
We decomposeinto two overlapping smooth subdomainsandsuch that
=∪. (.)
We denote by∂ithe boundary ofiandi=∂i∩j. We assume that the intersection
ofiandj,i=j, is empty and we associate with problem (.) the following system: Find
(u,u)∈Kg×K
g
, a solution of
ai(ui,v–ui) +c(ui,v–ui)≥(f(ui),v–ui), ∀v∈Kig,
ui/i=uj/i, i,j= , ,i=j,
(.)
such that
Kig=v∈H(i) such thatv=gon∂∩∂iandv≤ψoni
, (.)
ai(u,v) =
i
(∇u.∇v)dx, (.)
and
ui=u/i, i= , . (.)
Let
3.1 The continuous Schwarz sequences
Letu
be an initialization indefined by
u=ψ/. (.)
We, respectively, define the alternating Schwarz sequences (un+
) onsuch thatun+∈ Kg solves
b(un+,v–un+)≥(f(un+),v–un+), ∀v∈K
g
, un+/=u
n
/,
(.)
and (un+) onsuch thatun+∈K
g
solves
b(un+,v–un+)≥(f(un+),v–un+), ∀v∈K
g
, un+
/=un+/.
(.)
Theorem (see []) The two sequences(.)and(.)converge uniformly to the solution of(.).
3.2 Nonmatching grids discretization
Fori= , , letτhibe a standard regular and quasi-uniform finite element triangulation ini;hibeing its mesh size. The two meshes being mutually independent on∩in the sense that a triangle belonging to one triangulation does not necessarily belong to the other one. We consider the following discrete spaces:
Vhi=
v∈C(i)∩H(i) such thatv/T∈P,∀T∈τhi
, (.)
the convex sets
Khgi={v∈Vhisuch thatv=πhigon∂∩∂iandv≤rhiψ}, (.) whererhidenotes the restriction operator on the triangulationτ
hi. Let alsoπ
hidenote the interpolation operator oniandφis,s= , , . . . ,m(hi), be the basis functions ofVhi. The discrete maximum principle(see [, ]) We assume that the respective matrices resulting from the discretization of problems (.) and (.) are M-matrices. Note that, as the two mesheshandhare independent over the overlapping subdomains, it is im-possible to formulate a global approximate problem which would be the direct discrete counterpart of problem (.).
3.3 The discrete Schwarz sequences
Now, we define the discrete counterparts of the continuous Schwarz sequences defined in (.) and (.). Indeed, letu
h be the discrete analog ofu
defined in (.) that is, u
h=πh(u) =πh(ψ/). We, respectively, defineunh+ ∈K
g
hsuch that
b(unh+ ,v–u
n+
h )≥(f(u
n+
h ),v–u
n+
h ), ∀v∈K
g h,
un+
h /=πh(u
n h/),
andunh+ ∈Khg such that
b(unh+ ,v–u
n+
h )≥(f(u
n+
h ),v–u
n+
h ), ∀v∈K
g h,
unh+ /=πh(u
n+
h /).
(.)
4 Maximum norm error
This section is devoted to the proof of the main result of the present paper. To that end, we begin by introducing two discrete auxiliary problems.
4.1 Two auxiliary problems
We definewh∈K
g
hsuch thatwhsolves
b(wh,v–wh)≥(f(u),v–wh), ∀v∈K
g h,
wh/=πh(u/),
(.)
andwh∈K
g
h such thatwh solves
b(wh,v–wh)≥(f(u),v–wh), ∀v∈K
g h,
wh/ =πh(u/).
(.)
It is then clear thatwhandwhare the finite element approximations ofuandudefined
in (.) thus, making use of (.), we get
ui–whiL∞(i)≤Ch
|logh|, (.)
whereCis a constant independent ofh.
Notation From now on, we shall adopt the following notations:
| · |= · L∞(); | · |= · L∞(),
· = · L∞(); · = · L∞(), πh=πh=πh.
4.2 The main result
Theorem Let h=max(h,h)and letρ=βk < then there exists a constant C indepen-dent of both h and n such that
ui–unh+i i≤
( –ρ)Ch
|logh|, i= , ,n≥. (.)
Proof The proof of (.) will be carried out by induction, where the casesρ∈(,] and
ρ∈(, ) will be studied separately. Also, within each case, we will also discuss the two following situations:
(A): u–uh≤Ch
and
(B): Ch|logh|<u–uh, (.) whereu
h=πh(u
) =πh(ψ/). The basic idea of the proof is to define for each subdo-main two approximationsαhiandα˜hiin theL∞-norm ofui(a discrete subsolution and a discrete supersolution ofun
hi,n≥), such that αhi–uii≤
( –ρ)Ch
|logh|
and
˜αhi–uii≤ ( –ρ)Ch
|logh|.
Part: The first part of the proof deals with <ρ≤. So
ρ
–ρ ≤. (.)
Forn= , in domain , the discrete analogwhofudefined in (.) considered as the upper
bound of the set of discrete subsolutions [], satisfies
b
wh,ϕ
s
≤f(u),ϕs
, ∀s∈, . . . ,m(h)
,
wh=πhu on.
Since the nonlinear functional is Lipschitz and according to (.), we get
f(u) –f(wh)≤kCh
|logh|.
Then
b
wh,ϕ
s
≤f(u),ϕs
≤f(wh) +kCh
|logh|,ϕ
s
,
wh=πhu on.
Let
Wh=σh
f(wh) +kCh
|logh|,π
hu
; (.)
therefore,whis a subsolution ofWh,
wh≤Wh in. (.)
By applying (.), we get
Wh–u
h≤max
β
f(wh) +kCh
|logh|–fu
h;u–u
h
≤max β
f(wh) –f
uh+
k
β
Ch|logh|;u–uh
So
Wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u
–uh
. (.)
On the other hand, (.) generates two possibilities, that is,
(A): wh–u
h≤Wh–u
h or
(A): Wh–u
h≤wh–u
h. Case (A) in conjunction with (.) implies that
wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u
–uh
,
which lets us distinguish the following two cases:
: maxρwh–u
h+ρCh
|logh|;u
–uh
=ρwh–u
h+ρCh
|logh| (.)
and
: maxρwh–u
h+ρCh
|logh|;u
–uh
=u–uh. (.) Equation (.) implies that
wh–u
h≤ρwh–u
h+ρCh |logh|
and
u–uh≤ρwh–u
h+ρCh
|logh|.
Then
wh–u
h≤
ρ
–ρCh
|logh|
and
u–uh≤
ρ –ρCh
|logh|+ρCh|logh|
≤ ρ –ρCh
|logh|≤Ch|logh|,
which coincides with (.) and contradicts (.). So, (.) is only possible in situation (A). Equation (.) implies that
wh–u
h≤u–u
and
ρwh–u
h+ρCh
|logh|≤u
–uh.
So, by multiplying (.) byρand addingρCh|logh|, we get
ρwh–u
h+ρCh
|logh|≤ρu
–uh+ρCh
|logh|.
Then ρwh–uh+ρCh
|logh| is bounded by bothu
–uh andρu–u
h+ ρCh|logh|, so
(a): u–uh≤ρu–u
h+ρCh |logh|
or
(b): ρu–uh+ρCh
|logh| ≤u
–uh. That is,
u–uh≤ ρ –ρCh
|logh|
or
ρ
–ρCh
|logh|≤u
–uh. Thus
u–uh≤
ρ
–ρCh
|logh|≤Ch|logh|
or
ρ
–ρCh
|logh|≤u
–uh≤Ch
|logh|.
So, the two cases (a) and (b) are true because they both coincide with (.). Therefore, there is either a contradiction and thus (.) is impossible or (.) is possible only if
u–uh=
ρ
–ρCh
|logh|.
Then (.) in situation (A) implies
wh–u
h≤u–u
h=
ρ
–ρCh
|logh|,
while in situation (B) only (b) is true and leads to
wh–u
h≤u–u
h and
ρ
–ρCh
|logh|≤u
Then
wh–u
h≤
ρ
–ρCh
|logh|≤u
–uh or
ρ
–ρCh
|logh|≤w
h–u
h≤u–u
h.
We remark that both possibilities are true. There is either a contradiction and (.) is impossible or (.) is possible only if
wh–u
h=
ρ
( –ρ)Ch
|logh|.
So, in the two situations (A) and (B) and in the two cases (.) and (.) of situation (A), we get
wh–u
h≤
ρ
( –ρ)Ch
|logh|, (.)
which implies
wh– ρ
( –ρ)Ch
|logh|≤u
h≤wh+ ρ
( –ρ)Ch
|logh|.
Let us denote
αh=wh– ρ
( –ρ)Ch
|logh| (.)
and
˜
αh=wh+ ρ
( –ρ)Ch
|logh|. (.)
Then
αh≤u
h≤ ˜αh (.)
with
αh–u= wh–
ρ
( –ρ)Ch
|logh|–u
≤ wh–u+ ρ
( –ρ)Ch |logh|
≤Ch|logh|+ ρ ( –ρ)Ch
|logh|
by virtue of (.). So
αh–u≤
( –ρ)Ch
By using the same reasoning we see that (.) implies
˜αh–u≤
( –ρ)Ch
|logh|. (.)
On the other hand, (.) implies
αh–u≤uh–u≤ ˜αh–u (.)
so according to (.) and (.) we get
– ( –ρ)Ch
|logh|≤u
h–u≤
( –ρ)Ch
|logh| (.)
thus
u–uh≤ ( –ρ)Ch
|logh|. (.)
Case (A) in conjunction with (.) implies thatWh–uh is bounded by the values
wh–u
handmax{ρwh–u
h+ρCh
|logh|;u
–uh}which generates the two
situations
(c): wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u
–uh
or
(d): maxρwh–u
h+ρCh
|logh|;u
–uh
≤wh–u
h. (.) It is clear that case (c) coincides with situation (A). Let us study case (d); as in case (A),
max{ρwh–u
h+ρCh
|logh|;u
–uh}lets us distinguish the two cases (.) and
(.). Equation (.) in conjunction with (d) implies
u–uh≤ρwh–u
h+ρCh
|logh|≤w
h–u
h and (.) in conjunction with (d) implies
ρwh–u
h+ρCh
|logh|≤u
–uh≤wh–u
h. Then it is clear that in the two cases (.) and (.), we obtain
ρ
( –ρ)Ch
|logh|≤w
h–u
h (.)
with
u–uh≤wh–u
Thus,wh–uhis bounded below by both ρ
(–ρ)Ch|logh|andu–u
hso we
dis-tinguish the two following possibilities:
(e): u–uh≤
ρ
( –ρ)Ch
|logh|≤Ch|logh|
or
(f): ρ ( –ρ)Ch
|logh|≤u
–uh≤Ch
|logh|.
So, the two cases (e) and (f ) are true because they both coincide with (.). Therefore, there is either a contradiction and thus cases (.) and (.) are impossible or the two cases (.) and (.) are possible in situation (A) only if
u–uh=
ρ
( –ρ)Ch
|logh|≤w
h–u
h, while in situation (B) only the case (f ) is true and leads to
ρ
( –ρ)Ch
|logh|≤u
–uh≤wh–u
h.
In summary, in situation (A) and in the two cases (.) and (.) of situations (A) and (B), we get
ρ
( –ρ)Ch
|logh|≤w
h–u
h. (.)
Let us decompose the subdomain=,∪c,such that
ρ
( –ρ)Ch
|logh|≤w
h–u
h on, (.)
and
wh–u
h< ρ
( –ρ)Ch
|logh| onc
,. (.)
We begin with,. Ifwh–uh≥ on,then (.) impliesu
h≤αh; thus,
uh–u≤αh–u≤
( –ρ)Ch
|logh| (.)
by virtue of (.). On the other hand, (.) leads also to
– ( –ρ)Ch
|logh|≤α
h–u.
So,αh–uis bounded below by bothu
h–uand –
(–ρ)Ch
|logh|, which lets us
distin-guish the two following possibilities:
uh–u≤– ( –ρ)Ch
or
– ( –ρ)Ch
|logh|≤u
h–u.
Then
uh–u≤– ( –ρ)Ch
|logh|≤w
h–u
or
– ( –ρ)Ch
|logh|≤u
h–u≤wh–u.
So, both possibilities are true because they coincide with (.). So, there is either a con-tradiction and (.) is impossible or (.) is possible and we must have
uh–uL∞(,)=
( –ρ)Ch
|logh|. (.)
The casewh–uh< on,is studied in a similar manner and leads to the same result
(.). Equation (.) is studied in the same way as that for case (A) and leads to
uh–uL∞(c
,)≤
( –ρ)Ch
|logh|. (.)
Equations (.) and (.) imply
u
h–u≤ ( –ρ)Ch
|logh|. (.)
Finally, in the two cases (A) and (A) and in the two situations (A) and (B), we get
u–uh≤ ( –ρ)Ch
|logh|. (.)
Forn= in domain , the discrete analogwhofu, defined in (.) and considered as the
upper bound of the set of discrete subsolutions [], satisfies
b
wh,ϕ
s
≤f(u),ϕs
, ∀s∈, . . . ,m(h)
,
wh=πhu on.
The nonlinear functional is Lipschitz and according to (.)
f(u) –f(wh)≤kCh
|logh|.
Then
b
wh,ϕ
s
≤f(u),ϕs
≤f(wh) +kCh
|logh|,ϕ
s
,
Let
Wh=σh
f(wh) +kCh
|logh|,π
hu
; (.)
therefore,wh is a subsolution ofWh, so
wh≤Wh in. (.)
By applying (.), we get
Wh–u
h≤max
β
f(wh) +kCh
|logh|–fu
h;u–u
h
≤max β
f(wh) –f
uh+
k
β
Ch|logh|;u–uh
.
So
Wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u –uh
. (.)
On the other hand, (.) generates two possibilities, that is,
(B): wh–u
h≤Wh–u
h or
(B): Wh–u
h≤wh–u
h. Case (B) in conjunction with (.) implies that
wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u –uh
,
which lets us distinguish the following two cases:
: maxρwh–u
h+ρCh
|logh|;u –uh
=ρwh–u
h+ρCh
|logh| (.)
and
: maxρwh–u
h+ρCh
|logh|;u
–uh
=u–uh. (.) Equation (.) implies that
wh–u
h≤ρwh–u
h+ρCh |logh|
and
u–uh≤ρwh–u
h+ρCh
Then
wh–u
h≤
ρ
–ρCh
|logh|
and
u–uh≤
ρ
–ρCh
|logh|+ρCh|logh|
≤ ρ –ρCh
|logh|< –ρCh
|logh|,
which coincides with (.). Equation (.) implies that
wh–u
h≤u–u
h (.)
and
ρwh–u
h+ρCh
|logh|≤u
–uh.
So, by multiplying (.) byρand addingρCh|logh|we get
ρwh–u
h+ρCh
|logh|≤ρu
–uh+ρCh
|logh|;
thenρwh –uh+ρCh
|logh| is bounded above byu
–uh andρu–uh+
ρCh|logh|. So, either
(a): u–uh≤ρu–u
h+ρCh |logh|
or
(b): ρu–uh+ρCh
|logh| ≤u
–uh, that is,
u–uh≤
ρ
–ρCh
|logh|< –ρCh
|logh|
or
ρ
–ρCh
|logh|≤u
–uh≤ –ρCh
|logh|.
So, the two cases (a) and (b) are true because they both coincide with (.). Therefore, there is either a contradiction and thus (.) is impossible or (.) is possible only if
u–uh=
ρ
–ρCh
|logh|
thus
wh–u
h≤u–u
h=
ρ
–ρCh
In summary, in the two cases (.) and (.) of situation (B), we get
wh –u
h≤
ρ
( –ρ)Ch |logh|
so
wh– ρ
( –ρ)Ch
|logh|≤u
h≤wh+ ρ
( –ρ)Ch
|logh|.
Let us denote
αh=wh– ρ
( –ρ)Ch
|logh| (.)
and
˜
αh=wh+ ρ
( –ρ)Ch
|logh|; (.)
then
αh≤u
h≤ ˜αh. (.)
By using a same reasoning as adopted in subdomainfor (.) and (.), we get
αh–u≤
( –ρ)Ch
|logh| (.)
and
˜αh–u≤
( –ρ)Ch
|logh|. (.)
Equation (.) implies
αh–u≤u
h–u≤ ˜αh–u
and according to (.) and (.), we obtain
– ( –ρ)Ch
|logh|≤u
h–u≤
( –ρ)Ch
|logh|, (.)
that is,
u–uh≤ ( –ρ)Ch
|logh|.
Case (B) in conjunction with (.) implies thatWh–uhis bounded by the values
wh –uh andmax{ρwh –uh+ρCh
|logh|;u
–uh}, which generates two situations,
(c): wh–u
h≤max
ρwh–u
h+ρCh
|logh|;u
–uh
or
(d): maxρwh–u
h+ρCh
|logh|;u
–uh
≤wh–u
h.
It is clear that case (c) coincides with case (B). Let us study case (d); as in case (B)
max{ρwh –uh+ρCh
|logh|;u
–uh}lets us distinguish the two cases (.) and (.). Equation (.) in conjunction with (d) implies
u–uh≤ρwh–u
h+ρCh
|logh|≤w
h–u
h and (.) in conjunction with (d) implies
ρwh–u
h+ρCh
|logh|≤u
–uh≤wh–u
h. Then the two cases (.) and (.) imply
ρ
( –ρ)Ch
|logh|≤w
h–u
h and
u–uh≤wh–u
h. wh–uhis bounded below by
ρ
(–ρ)Ch
|logh|andu
–uh so we distinguish the
two following possibilities:
(e): u–uh≤
ρ
( –ρ)Ch
|logh|< ( –ρ)Ch
|logh|
or
(f): ρ ( –ρ)Ch
|logh|≤u
–uh≤ ( –ρ)Ch
|logh|.
So, the two cases (e) and (f ) are true because they both coincide with (.). Therefore, there is either a contradiction and thus cases (.) and (.) are impossible or the two cases (.) and (.) are possible only if
u–uh=
ρ
( –ρ)Ch
|logh|≤w
h–u
h. So, in the two cases (.) and (.) of situation (B), we get
ρ
( –ρ)Ch
|logh|≤w
h–u
h.
The remainder of the proof related to situation (B) rests on the same arguments used in subdomain for situation (A) that is, on a decomposition of=,∪c,and showing that
u–uhL∞(,)≤
( –ρ)Ch
and
u–uhL∞(c,)≤ ( –ρ)Ch
|logh|.
Finally, in the two situations (B) and (B) we get
u–uh≤ ( –ρ)Ch
|logh|. (.)
Now, let us assume that
u–unh≤ ( –ρ)Ch
|logh|,
u–unh≤ ( –ρ)Ch
|logh|,
(.)
and prove that
u–unh+ ≤ ( –ρ)Ch
|logh|,
u–unh+ ≤ ( –ρ)Ch
|logh|.
(.)
By using the definition ofWhin (.) and by applying (.), we get Wh–u
n+
h ≤max
β
f(wh) +kCh
|logh|–fun+
h ;u–u
n h
≤max β
f(wh) –f
unh+ +
k
β
Ch|logh|;u–unh
so
Wh–u
n+
h ≤max
ρwh–u
n+
h +ρCh
|logh|;u
–unh
. (.)
On the other hand, (.) generates two possibilities, that is
(C): wh–u
n+
h ≤Wh–u
n+
h or
(C): Wh–u
n+
h ≤wh–u
n+
h . Case (C) in conjunction with (.) implies that
wh–u
n+
h ≤max
ρwh–u
n+
h +ρCh
|logh|;u
–unh
,
which lets us distinguish the following two cases:
: maxρwh–u
n+
h +ρCh
|logh|;u
–unh
=ρwh–u
n+
h +ρCh
and
: maxρwh–u
n+
h +ρCh
|logh|;u
–unh
=u–unh. (.)
Equation (.) implies that
wh–u
n+
h ≤ρwh–u
n+
h +ρCh |logh|
and
u–unh≤ρwh–u
n+
h +ρCh
|logh|.
Then
wh–u
n+
h ≤
ρ
–ρCh
|logh|
and
u–unh≤
ρ
–ρCh
|logh|< –ρCh
|logh|,
which coincides with (.). Equation (.) implies that
wh–u
n+
h ≤u–u
n
h (.)
and
ρwh–u
n+
h +ρCh
|logh|≤u
–unh.
So, by multiplying (.) byρand addingρCh|logh|we get
ρwh–u
n+
h +ρCh
|logh|≤ρu
–unh+ρCh
|logh|;
thenρwh–u
n+
h +ρCh
|logh| is bounded by bothu
–unh andρu–u
n h+ ρCh|logh|. So
(a): u–unh≤ρu–u
n
h+ρCh |logh|
or
(b): ρu–unh+ρCh
|logh| ≤u
–unh. Thus
u–unh≤
ρ
–ρCh
|logh|< –ρCh
or
ρ
–ρCh
|logh|≤u
–unh≤ –ρCh
|logh|.
So, the two cases (a) and (b) are true because they both coincide with (.). Therefore, there is either a contradiction and thus (.) is impossible or (.) is possible only if
u–unh=
ρ
–ρCh
|logh|.
Then (.) implies
wh–u
n+
h ≤u–u
n h=
ρ
–ρCh
|logh|.
Thus, in situation (C) and in the two cases (.) and (.), we get
wh–u
n+
h ≤
ρ
( –ρ)Ch |logh|
so
αh≤u
n+
h ≤ ˜αh (.)
and
αh–u≤u
n+
h –u≤ ˜αh–u.
So, according to (.) and (.), we get
– ( –ρ)Ch
|logh|≤un+
h –u≤
( –ρ)Ch
|logh|.
Thus
u–unh+ ≤ ( –ρ)Ch
|logh|.
Case (C) in conjunction with (.) implies thatWh–unh+ is bounded by the values
wh–unh+andmax{ρwh–u
n+
h +ρCh
|logh|;u
–unh}, which generates two situations,
(c): wh–u
n+
h ≤max
ρwh–u
n+
h +ρCh
|logh|;u
–unh
or
(d): maxρwh–u
n+
h +ρCh
|logh|;u
–unh
≤wh–u
n+
h .
It is clear that case (c) coincides with case (C). Let us study case (d); as in case (C),
max{ρwh–u
n+
h +ρCh
|logh|;u
and (.). Equation (.) in conjunction with (d) implies
u–unh≤ρwh–u
n+
h +ρCh
|logh|≤w
h–u
n+
h and (.) in conjunction with (d) implies
ρwh–u
n+
h +ρCh
|logh|≤u
–unh≤wh–u
n+
h . Then in the two cases (.) and (.), we get
ρ
( –ρ)Ch
|logh|≤w
h–u
n+
h and
u–unh≤wh–u
n+
h .
Hence,wh–unh+ is bounded below by both ρ
(–ρ)Ch
|logh|andu
–unh so we
distinguish the two following possibilities:
(e): u–unh≤
ρ
( –ρ)Ch
|logh|< ( –ρ)Ch
|logh|
or
(f): ρ ( –ρ)Ch
|logh|≤u
–unh≤ ( –ρ)Ch
|logh|.
So, the two cases (e) and (f ) are true because they both coincide with (.). Therefore, there is either a contradiction and the two cases (.) and (.) are impossible or the two cases (.) and (.) are possible only if
u–unh=
ρ
( –ρ)Ch
|logh|≤w
h–u
n+
h ;
thus, in the two cases (.) and (.) of situation (C), we get
ρ
( –ρ)Ch
|logh|≤w
h–u
n+
h .
The remainder of the proof related to situation (C) rests on the same arguments used in subdomainfor situation (A) at iterationn= , that is, on a decomposition of=
,∪c,and on showing that
u–uhn+L∞(,)≤
( –ρ)Ch
|logh|
and
u–unh+ L∞(c,)≤ ( –ρ)Ch
Finally, in the two situations (C) and (C) we get the desired result,
u–un+
h ≤ ( –ρ)Ch
|logh|. (.)
Estimate (.) in domain can be proved similarly using estimate (.). Part: This second part of the proof is devoted to<ρ< . So
ρ
–ρ > . (.)
Forn= , in domain , like as in part , (.) generates two different situations (A) and (A), which we study separately. According to (.), situation (A) in conjunction with (.) implies
wh–u
h≤ρwh–u
h+ρCh |logh|
and
u–uh≤ρwh–u
h+ρCh
|logh|.
Then
wh–u
h≤
ρ
–ρCh
|logh|
and
u–uh≤
ρ
–ρCh
|logh|.
So, we can write for (.)
u–uh≤Ch|logh|< ρ –ρCh
|logh|
and for (.)
Ch|logh|<u–uh≤
ρ
–ρCh
|logh|.
Equation (.) implies that
wh–u
h≤u–u
h (.)
and
ρwh–u
h+ρCh
|logh|≤u
–uh.
So, by multiplying (.) byρand addingρCh|logh|we get
ρwh–u
h+ρCh
|logh|≤ρu
–uh+ρCh