# Maximum Norm Analysis of an Arbitrary Number of Nonmatching Grids Method for Nonlinears Elliptic PDES

Authors: Abida Harbi
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

We provide a maximum norm analysis of a finite element Schwarz alternating method for a nonlinear elliptic PDE on an arbitrary number of overlapping subdomains with nonmatching grids. We consider a domain which is the union of an arbitrary numberm
of overlapping subdomains where each subdomain has its own independently generated grid. The m meshes being mutually independent on the overlap regions, a triangle belonging to one triangulation does not necessarily belong to the other ones. Under the a Lipschitz assumption on the nonlinearity, we establish, on each subdomain, an optimal L∞ error estimate between the discrete Schwarz sequence and the exact solution of the PDE.


---

## Body

## 1. Introduction

The Schwarz alternating method can be used to solve elliptic boundary value problems on domains which consist of two or more overlapping subdomains. The solution is approximated by an infinite sequence of functions which result from solving a sequence of elliptic boundary value problems in each of the subdomains. The effectiveness of the Schwarz methods for various classes of nonlinear elliptic PDE problems has been demonstrated in many papers; see [1–4] and the references therein. Also the effectiveness of the Schwarz methods for these problems, especially those in fluid mechanics, has been demonstrated in many papers. See proceedings of the annual domain decomposition conference beginning with [5].In [6, 7], an optimal convergence order is obtained for nonlinear elliptic PDE, in the context of two overlapping nonmatching grids, in the sense that each subdomain has its own independent discretization by finite element method. This kind of discretization is very interesting as it can be applied to solve many practical problems which cannot be handled by global discretizations. Discretizations are earning particular attention of computational experts and engineers as they allow the choice of different mesh sizes and different orders of approximate polynomials in different subdomains according to the different properties of the solution and different requirements of the practical problems. Quite a few works on maximum norm error analysis of overlapping nonmatching grids methods for elliptic problems are known in the literature; compare and confer, for example, [8–10].This paper is a continuation of previous work [6], attempting to generalize the obtained result related to convergence order also in the context of nonmatching grids in domain which consists of a union of an arbitrary number of subdomains. It is proved that the error estimate remains true also for more than two subdomains.The proof of the main result which consists of estimating the error in the maximum norm between the continuous solution of the problem and the discrete Schwarz iterates stands on a Lipschitz continuous dependency with respect to both the boundary condition and the source term for linear elliptic equations. The optimal convergence order is then derived making use of standard finite elementL∞error estimate for linear elliptic equations.Now, we give an outline of the paper. In Section2 we state the continuous alternating Schwarz sequences and define their respective finite element counterparts in the context of nonmatching overlapping grids. Section 3 is devoted to theL∞error analysis of the method.


## 2. Preliminaries

We begin by laying down some definitions and classical results related to linear elliptic equations.


### 2.1. Linear Elliptic Equations

LetΩbe a bounded polyhedral domain ofℝ2orℝ3with sufficiently smooth boundary∂Ω. We consider the following: the bilinear form
(1)a(u,v)=∫Ω(∇u·∇v)dx;
the linear form
(2)(f,v)=∫Ωf(x)·v(x)dx;
the right hand side
(3)f,aregularfunction;
the space
(4)V(g)={v∈H1(Ω)suchthatv=gon∂Ω},
wheregis a regular function defined on∂Ω.We consider the linear elliptic equation: Findζ∈V(g)such that
(5)a(ζ,v)+c(ζ,v)=(f,v),∀v∈V(g),
wherec∈ℝc>0, such that
(6)c≥β>0.LetVhbe the space of finite elements consisting of continuous piecewise linear functionsvvanishing on∂Ω, and lets,s=1,2,…,m(h)be the basis function ofVh. The discrete counterpart of (5) consists of findingζh∈Vh(g)such that
(7)a(ζh,v)+c(ζh,v)=(f,v),∀v∈Vh(g),
where
(8)Vh(g)={v∈Vh:v=πhgon∂Ω},
andπhis an interpolation operator on∂Ω.Theorem 1 (cf. [11]).
Under the suitable regularity of the solution of problem (5), there exists a constantCindependent ofhsuch that
(9)∥ζ-ζh∥L∞(Ω)≤Ch2|logh|.Lemma 2 (cf. [4]).
Letw∈H1(Ω)∩C(Ω¯)satisfya(w,ϕ)+c(w,ϕ)≥0∀nonnegativeϕ∈H01(Ω), andw≥0on∂Ω. Thenw≥0onΩ¯.The proposition below establishes a Lipschitz continuous dependency of the solution with respect to the data.Notation. Let(f;g);(f~,g~)be a pair of data andζ=σ(f,g);ζ~=σ(f~,g~)the corresponding solutions to (5).Proposition 3 (cf. [6]).
Under conditions of the preceding lemma, one has:(10)∥ζ-ζ~∥L∞(Ω)≤max{1β∥f-f~∥L∞(Ω),∥g-g~∥L∞(∂Ω)}.Remark 4.
Lemma2 stays true in the discrete case.Indeed, assume that the discrete maximum principle (d.m.p) holds; that is, the matrix resulting from the finite element discretization is an M-Matrix. Then one has the following.Lemma 5.
Letw∈Vhsatisfya(w,s)+c(w,s)≥0,s=1,2,…,m(h)andw≥0on∂Ω. Then,w≥0onΩ¯.Proof.
The proof is a direct consequence of the discrete maximum principle.Let(f,g);(f~,g~)be a pair of data andζh=σh(f,g);ζ~h=σh(f~,g~)the corresponding solutions to (7).Proposition 6 (cf. [6]).
Let the d.m.p hold. Then, under conditions of Lemma5, one has
(11)∥ζh-ζ~h∥L∞(Ω)≤max{1β∥f-f~∥L∞(Ω),∥g-g~∥L∞(∂Ω)}.


### 2.2. The Schwarz Alternating Methods for Nonlinear PDEs

Consider the nonlinear PDE:(12)-Δu+cu=f(u)inΩ,u=0on∂Ω
or in its weak form
(13)a(u,v)+c(u,v)=(f(u),v)∀v∈H01(Ω),
wheref(·)is a nondecreasing nonlinearity. We assume thatf(·)is a Lipschitz continuous onℝ, that is
(14)|f(x)-f(y)|≤k|x-y|,∀x,y∈ℝ
such that
(15)0<kβ<1,
whereβis the constant defined in (6).We suppose that the problem (12) has a subsolutionu_and asupersolutionu¯which satisfy0<u_<u¯.Define the sector of smooth functions as follows:
(16)A={u∈C2(Ω),u_≤u≤u¯onΩ¯}.
Finally we assume that
(17)F(u)-F(v)<0,v<u∈A,
where
(18)F(u)=f(u)u.
With these assumptions cf. [12], the problem (12) has a unique solution in the sectorA.We can easily see that (12) is equivalent to
(19)-Δu=h(u)inΩ,u=0on∂Ω,
whereh(u)=f(u)-cu,and we note that the functionalh(u)satisfies (17) and (18). We introduce the following comparison lemma.Lemma 7 (cf. [3]).
Suppose that-Δu≥h(u)and-Δv≤h(v)onΩwith bothu,vpositive inΩandu≥v≥0on∂Ω.Thenu≥vonΩ¯.We decomposeΩintomoverlapping smooth subdomainsΩisuch that
(20)Ω=⋃‍i=1mΩi.
We denote by∂Ωithe boundary ofΩiand the interior boundaries by
(21)Γi=∂Ωi∩Ω.
We assume that the intersection ofΓ¯iandΓ¯j,i≠jis empty. Let
(22)Vi(wj)={v∈C2(Ωi)suchthatv=wjonΓi,v=0on∂Ωi∩∂ΩC2}.
We associate problem (13) with the following system. Findui∈∏i=1mVi(gi)solution of
(23)-Δui+cui=f(ui)inΩi,
where
(24)ui=uΩi,i=1,…,m
and (cf. [13])
(25)gi={ui-1onΓi∩Ωi-1ui-2onΓi∩(Ωi-2-Ωi-1)⋮u1onΓi∩(Ω1-Ωi-1-⋯-Ω1)umonΓi∩(Ωm-⋃‍i-1j=1Ωj)um-1onΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮ui+1onΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).In the sequel, we define a subsolution and a supersolution of (23)Definition 8.
A smooth functionu_iis a subsolution of (23) if
(26)-Δu_i+cu_i≤f(u_i)inΩi,u_i=g_ionΓi,u_i=0on∂Ω∩∂Ωi,
where
(27)g_i={u_i-1onΓi∩Ωi-1u_i-2onΓi∩(Ωi-2-Ωi-1)⋮u_1onΓi∩(Ω1-Ωi-1-⋯-Ω1)u_monΓi∩(Ωm-⋃‍i-1j=1Ωj)u_m-1onΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮u_i+1onΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).
Similarly, a supersolutionu¯iof (23) satisfies
(28)-Δu¯i+cu¯i≥f(u¯i)inΩi,ui=g¯ionΓi,u¯i=0on∂Ω∩∂Ωi.


### 2.3. The Continuous Schwarz Sequences

Letui0be an initialization data:
(29)ui0=u_i,i=1,…,m.We define the alternating Schwarz sequences(uin+1)onΩi,i=1,…,m,such that each term of each sequenceuin+1∈V(gin+1),n≥0solves
(30)-Δuin+1+cuin+1=f(uin+1)inΩi,
where (cf. [13])
(31)gin+1={ui-1n+1onΓi∩Ωi-1ui-2n+1onΓi∩(Ωi-2-Ωi-1)⋮u1n+1onΓi∩(Ω1-Ωi-1-⋯-Ω1)umnonΓi∩(Ωm-⋃‍i-1j=1Ωj)um-1nonΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮ui+1nonΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).Theorem 9.
The Schwarz sequences(uin+1),i=1,…,mconverge to the solution of (23).Proof.
(
1
)We first show by induction, that each term of the Schwarz sequences is well defined inA.Indeed, forn=0, we begin by subdomain one, sinceu11,u¯1andu_1satisfy
(32)-Δu11+cu11=f(u11)inΩ1,u11=0on∂Ω1∩∂Ω,(33)-Δu¯1+cu¯1≥f(u¯1)inΩ1,u¯1=0on∂Ω1∩∂Ω,(34)-Δu_1+cu_1≤f(u_1)inΩ1,(35)u_1=0on∂Ω1∩∂Ω
respectively with
(36)u11=g11=g_1≤g¯1onΓ1
Lemma 7 implies that
(37)u_1≤u11≤u¯1inΩ¯1.
In subdomain 2, the condition of the interior boundaryΓ2is given by
(38)g21={u11onΓ2∩Ω1um0onΓ2∩(Ωm-Ω1)um-10onΓ2∩(Ωm-1-Ω1-Ωm)⋮u3nonΓ2∩(Ω3-Ω1-⋃‍mj=4Ωj).
so (37) implies that
(39)g_2≤g21≤g¯2onΓ2.
On the other handu21,u¯2, andu_2satisfy, respectively,
(40)-Δu21+cu21=f(u21)inΩ2,u21=0on∂Ω2∩∂Ω,-Δu¯2+cu¯2≥f(u¯2)inΩ2,u¯2=0on∂Ω2∩∂Ω,-Δu_2+cu_2≤f(u_2)inΩ2,u_2=on∂Ω2∩∂Ω.
Then Lemma 7 implies that
(41)u_2≤u21≤u¯2inΩ¯2.
The same idea is used sequentially in the remainder of the proof related to the rest of subdomainsΩi,3≤i≤m. Indeed we obtain fori=1,…,mand forn=0(42)u_i≤ui1≤u¯iinΩ¯i.
Now, let us assume that
(43)u_i≤uin≤u¯iinΩ¯i,i=1,…,m.
and prove that
(44)u_i≤uin+1≤u¯iinΩ¯i,i=1,…,m.
Indeed, in subdomain 1 (43) implies that
(45)g_1≤g1n+1≤g¯1onΓ1.
Since(46)-Δu1n+1+cu1n+1=f(u1n+1)inΩ1,u1n+1=0on∂Ω1∩∂Ω,-Δu¯1+cu¯1≥f(u¯1)inΩ1,u¯1=0on∂Ω1∩∂Ω,-Δu_1+cu_1≤f(u_1)inΩ1,u_1=0on∂Ω1∩∂Ω,
then Lemma 7 implies that
(47)u_1≤u1n+1≤u¯1inΩ¯1.
Using (47) and (43), we can write
(48)g_2≤g2n+1≤g¯2onΓ2.
We can easily obtain
(49)u_2≤u2n+1≤u¯2inΩ¯2
using (48) and by adopting the same approach used in subdomain one. The result related to rest of subdomainsΩi,3≤i≤m, is obtained sequentially by similar way; that is
(50)u_i≤uin+1≤u¯iinΩ¯i,i=1,…,m,n≥0.
(
2
)We demonstrate, by induction, that the Schwarz sequences are nondecreasing. Indeed forn=0,we have demonstrated in (50) that, fori=1,…,m,
(51)ui0≤ui1inΩ¯i
Now, let us assume that(52)uin-1≤uininΩ¯i,i=1,…,m
and prove that
(53)uin≤uin+1inΩ¯i,i=1,…,m.
Begining by subdomain one, (52) implies that
(54)g1n≤g1n+1onΓ1.
Making use of (54),
(55)-Δu1n+1+cu1n+1=f(u1n+1)inΩ1,u1n+1=0on∂Ω1∩∂Ω,-Δu1n+cu1n=f(u1n)inΩ1,u1n=0on∂Ω1∩∂Ω.
We get by using Lemma 7(56)u1n≤u1n+1inΩ¯1.
On the other hand, (52) and (56) imply
(57)g2n≤g2n+1onΓ2;
then making use of (57),
(58)-Δu2n+1+cu2n+1=f(u2n+1)inΩ2,u2n+1=0on∂Ω2∩∂Ω,-Δu2n+cu2n=f(u2n)inΩ2,u2n=0on∂Ω2∩∂Ω.
We get by using Lemma 7 the following result related to the second subdomain:
(59)u2n≤u2n+1inΩ¯2.
The same idea is sequentially applied to the rest of subdomains. Finally, we obtain
(60)uin≤uin+1inΩ¯i,i=1,…,m,n≥0.
(
3
)The sequences(uin)are bounded and nondecreasing, so they are monotone converging pointwise toui,i=1,…,m. By using an elliptic regularity argument, we can see that functionsuisatisfy the same PDE onΩicf. [3, 12].


### 2.4. The Discretization

Fori=1,…,m,letτhibe a standard regular and quasiuniform finite element triangulation inΩi;hibeing the meshsize. Themmeshes is mutually independent onΩi∩Ωj,i,j=1,…,m,wherei≠j; that is a triangle belonging to one triangulation does not necessarily belong to the other ones.We consider the following discrete spaces:(61)Vhi={v∈C(Ω¯i)∩H1(Ωi)suchthatvK∈P1∀K∈τhi},
and for everyw∈C(Γ¯i), we set
(62)Vhi(w)={v∈Vhi:v=0on∂Ωi∩∂Ω;v=πhi(w)onΓi},
whereπhidenotes the interpolation operator onΓi.The Discrete Maximum Principle (cf. [14, 15]). We assume that the respective matrices resulting from the discretiztions of problems (30) are M-matrices.Note that as the mesheshiare independent over the overlapping subdomains, it is impossible to formulate a global approximate problem which would be the direct discrete counterpart of problem (13).


### 2.5. The Discrete Schwarz Sequences

Now, we define the discrete counterparts of the continuous Schwarz sequences defined in (30). Indeed, letuh0be the discrete analog ofu0, defined in (29) and we define fori=1,…,m,the discrete sequences(uhin+1)onΩisuch thatuhin+1∈Vhi(ghin+1)solves
(63)ai(uhin+1,v)+c(uhin+1,v)=(f(uhin+1),v),∀v∈Vhi(ghin+1);n≥0,
where
(64)ai(u,v)=∫Ωi(∇u·∇v)dx,ghin+1={uhi-1n+1onΓi∩Ωi-1uhi-2n+1onΓi∩(Ωi-2-Ωi-1)⋮uh1n+1onΓi∩(Ω1-Ωi-1-⋯-Ω1)uhmnonΓi∩(Ωm-⋃‍i-1j=1Ωj)uhm-1nonΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮uhi+1nonΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+1Ωj).Notation. From now on, we will adopt the following notationsi=1,…,m:
(65)|·|i=∥·∥L∞(Γi)∥·∥i=∥·∥L∞(Ωi),πhi=πh.


## 2.1. Linear Elliptic Equations

LetΩbe a bounded polyhedral domain ofℝ2orℝ3with sufficiently smooth boundary∂Ω. We consider the following: the bilinear form
(1)a(u,v)=∫Ω(∇u·∇v)dx;
the linear form
(2)(f,v)=∫Ωf(x)·v(x)dx;
the right hand side
(3)f,aregularfunction;
the space
(4)V(g)={v∈H1(Ω)suchthatv=gon∂Ω},
wheregis a regular function defined on∂Ω.We consider the linear elliptic equation: Findζ∈V(g)such that
(5)a(ζ,v)+c(ζ,v)=(f,v),∀v∈V(g),
wherec∈ℝc>0, such that
(6)c≥β>0.LetVhbe the space of finite elements consisting of continuous piecewise linear functionsvvanishing on∂Ω, and lets,s=1,2,…,m(h)be the basis function ofVh. The discrete counterpart of (5) consists of findingζh∈Vh(g)such that
(7)a(ζh,v)+c(ζh,v)=(f,v),∀v∈Vh(g),
where
(8)Vh(g)={v∈Vh:v=πhgon∂Ω},
andπhis an interpolation operator on∂Ω.Theorem 1 (cf. [11]).
Under the suitable regularity of the solution of problem (5), there exists a constantCindependent ofhsuch that
(9)∥ζ-ζh∥L∞(Ω)≤Ch2|logh|.Lemma 2 (cf. [4]).
Letw∈H1(Ω)∩C(Ω¯)satisfya(w,ϕ)+c(w,ϕ)≥0∀nonnegativeϕ∈H01(Ω), andw≥0on∂Ω. Thenw≥0onΩ¯.The proposition below establishes a Lipschitz continuous dependency of the solution with respect to the data.Notation. Let(f;g);(f~,g~)be a pair of data andζ=σ(f,g);ζ~=σ(f~,g~)the corresponding solutions to (5).Proposition 3 (cf. [6]).
Under conditions of the preceding lemma, one has:(10)∥ζ-ζ~∥L∞(Ω)≤max{1β∥f-f~∥L∞(Ω),∥g-g~∥L∞(∂Ω)}.Remark 4.
Lemma2 stays true in the discrete case.Indeed, assume that the discrete maximum principle (d.m.p) holds; that is, the matrix resulting from the finite element discretization is an M-Matrix. Then one has the following.Lemma 5.
Letw∈Vhsatisfya(w,s)+c(w,s)≥0,s=1,2,…,m(h)andw≥0on∂Ω. Then,w≥0onΩ¯.Proof.
The proof is a direct consequence of the discrete maximum principle.Let(f,g);(f~,g~)be a pair of data andζh=σh(f,g);ζ~h=σh(f~,g~)the corresponding solutions to (7).Proposition 6 (cf. [6]).
Let the d.m.p hold. Then, under conditions of Lemma5, one has
(11)∥ζh-ζ~h∥L∞(Ω)≤max{1β∥f-f~∥L∞(Ω),∥g-g~∥L∞(∂Ω)}.


## 2.2. The Schwarz Alternating Methods for Nonlinear PDEs

Consider the nonlinear PDE:(12)-Δu+cu=f(u)inΩ,u=0on∂Ω
or in its weak form
(13)a(u,v)+c(u,v)=(f(u),v)∀v∈H01(Ω),
wheref(·)is a nondecreasing nonlinearity. We assume thatf(·)is a Lipschitz continuous onℝ, that is
(14)|f(x)-f(y)|≤k|x-y|,∀x,y∈ℝ
such that
(15)0<kβ<1,
whereβis the constant defined in (6).We suppose that the problem (12) has a subsolutionu_and asupersolutionu¯which satisfy0<u_<u¯.Define the sector of smooth functions as follows:
(16)A={u∈C2(Ω),u_≤u≤u¯onΩ¯}.
Finally we assume that
(17)F(u)-F(v)<0,v<u∈A,
where
(18)F(u)=f(u)u.
With these assumptions cf. [12], the problem (12) has a unique solution in the sectorA.We can easily see that (12) is equivalent to
(19)-Δu=h(u)inΩ,u=0on∂Ω,
whereh(u)=f(u)-cu,and we note that the functionalh(u)satisfies (17) and (18). We introduce the following comparison lemma.Lemma 7 (cf. [3]).
Suppose that-Δu≥h(u)and-Δv≤h(v)onΩwith bothu,vpositive inΩandu≥v≥0on∂Ω.Thenu≥vonΩ¯.We decomposeΩintomoverlapping smooth subdomainsΩisuch that
(20)Ω=⋃‍i=1mΩi.
We denote by∂Ωithe boundary ofΩiand the interior boundaries by
(21)Γi=∂Ωi∩Ω.
We assume that the intersection ofΓ¯iandΓ¯j,i≠jis empty. Let
(22)Vi(wj)={v∈C2(Ωi)suchthatv=wjonΓi,v=0on∂Ωi∩∂ΩC2}.
We associate problem (13) with the following system. Findui∈∏i=1mVi(gi)solution of
(23)-Δui+cui=f(ui)inΩi,
where
(24)ui=uΩi,i=1,…,m
and (cf. [13])
(25)gi={ui-1onΓi∩Ωi-1ui-2onΓi∩(Ωi-2-Ωi-1)⋮u1onΓi∩(Ω1-Ωi-1-⋯-Ω1)umonΓi∩(Ωm-⋃‍i-1j=1Ωj)um-1onΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮ui+1onΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).In the sequel, we define a subsolution and a supersolution of (23)Definition 8.
A smooth functionu_iis a subsolution of (23) if
(26)-Δu_i+cu_i≤f(u_i)inΩi,u_i=g_ionΓi,u_i=0on∂Ω∩∂Ωi,
where
(27)g_i={u_i-1onΓi∩Ωi-1u_i-2onΓi∩(Ωi-2-Ωi-1)⋮u_1onΓi∩(Ω1-Ωi-1-⋯-Ω1)u_monΓi∩(Ωm-⋃‍i-1j=1Ωj)u_m-1onΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮u_i+1onΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).
Similarly, a supersolutionu¯iof (23) satisfies
(28)-Δu¯i+cu¯i≥f(u¯i)inΩi,ui=g¯ionΓi,u¯i=0on∂Ω∩∂Ωi.


## 2.3. The Continuous Schwarz Sequences

Letui0be an initialization data:
(29)ui0=u_i,i=1,…,m.We define the alternating Schwarz sequences(uin+1)onΩi,i=1,…,m,such that each term of each sequenceuin+1∈V(gin+1),n≥0solves
(30)-Δuin+1+cuin+1=f(uin+1)inΩi,
where (cf. [13])
(31)gin+1={ui-1n+1onΓi∩Ωi-1ui-2n+1onΓi∩(Ωi-2-Ωi-1)⋮u1n+1onΓi∩(Ω1-Ωi-1-⋯-Ω1)umnonΓi∩(Ωm-⋃‍i-1j=1Ωj)um-1nonΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮ui+1nonΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+2Ωj).Theorem 9.
The Schwarz sequences(uin+1),i=1,…,mconverge to the solution of (23).Proof.
(
1
)We first show by induction, that each term of the Schwarz sequences is well defined inA.Indeed, forn=0, we begin by subdomain one, sinceu11,u¯1andu_1satisfy
(32)-Δu11+cu11=f(u11)inΩ1,u11=0on∂Ω1∩∂Ω,(33)-Δu¯1+cu¯1≥f(u¯1)inΩ1,u¯1=0on∂Ω1∩∂Ω,(34)-Δu_1+cu_1≤f(u_1)inΩ1,(35)u_1=0on∂Ω1∩∂Ω
respectively with
(36)u11=g11=g_1≤g¯1onΓ1
Lemma 7 implies that
(37)u_1≤u11≤u¯1inΩ¯1.
In subdomain 2, the condition of the interior boundaryΓ2is given by
(38)g21={u11onΓ2∩Ω1um0onΓ2∩(Ωm-Ω1)um-10onΓ2∩(Ωm-1-Ω1-Ωm)⋮u3nonΓ2∩(Ω3-Ω1-⋃‍mj=4Ωj).
so (37) implies that
(39)g_2≤g21≤g¯2onΓ2.
On the other handu21,u¯2, andu_2satisfy, respectively,
(40)-Δu21+cu21=f(u21)inΩ2,u21=0on∂Ω2∩∂Ω,-Δu¯2+cu¯2≥f(u¯2)inΩ2,u¯2=0on∂Ω2∩∂Ω,-Δu_2+cu_2≤f(u_2)inΩ2,u_2=on∂Ω2∩∂Ω.
Then Lemma 7 implies that
(41)u_2≤u21≤u¯2inΩ¯2.
The same idea is used sequentially in the remainder of the proof related to the rest of subdomainsΩi,3≤i≤m. Indeed we obtain fori=1,…,mand forn=0(42)u_i≤ui1≤u¯iinΩ¯i.
Now, let us assume that
(43)u_i≤uin≤u¯iinΩ¯i,i=1,…,m.
and prove that
(44)u_i≤uin+1≤u¯iinΩ¯i,i=1,…,m.
Indeed, in subdomain 1 (43) implies that
(45)g_1≤g1n+1≤g¯1onΓ1.
Since(46)-Δu1n+1+cu1n+1=f(u1n+1)inΩ1,u1n+1=0on∂Ω1∩∂Ω,-Δu¯1+cu¯1≥f(u¯1)inΩ1,u¯1=0on∂Ω1∩∂Ω,-Δu_1+cu_1≤f(u_1)inΩ1,u_1=0on∂Ω1∩∂Ω,
then Lemma 7 implies that
(47)u_1≤u1n+1≤u¯1inΩ¯1.
Using (47) and (43), we can write
(48)g_2≤g2n+1≤g¯2onΓ2.
We can easily obtain
(49)u_2≤u2n+1≤u¯2inΩ¯2
using (48) and by adopting the same approach used in subdomain one. The result related to rest of subdomainsΩi,3≤i≤m, is obtained sequentially by similar way; that is
(50)u_i≤uin+1≤u¯iinΩ¯i,i=1,…,m,n≥0.
(
2
)We demonstrate, by induction, that the Schwarz sequences are nondecreasing. Indeed forn=0,we have demonstrated in (50) that, fori=1,…,m,
(51)ui0≤ui1inΩ¯i
Now, let us assume that(52)uin-1≤uininΩ¯i,i=1,…,m
and prove that
(53)uin≤uin+1inΩ¯i,i=1,…,m.
Begining by subdomain one, (52) implies that
(54)g1n≤g1n+1onΓ1.
Making use of (54),
(55)-Δu1n+1+cu1n+1=f(u1n+1)inΩ1,u1n+1=0on∂Ω1∩∂Ω,-Δu1n+cu1n=f(u1n)inΩ1,u1n=0on∂Ω1∩∂Ω.
We get by using Lemma 7(56)u1n≤u1n+1inΩ¯1.
On the other hand, (52) and (56) imply
(57)g2n≤g2n+1onΓ2;
then making use of (57),
(58)-Δu2n+1+cu2n+1=f(u2n+1)inΩ2,u2n+1=0on∂Ω2∩∂Ω,-Δu2n+cu2n=f(u2n)inΩ2,u2n=0on∂Ω2∩∂Ω.
We get by using Lemma 7 the following result related to the second subdomain:
(59)u2n≤u2n+1inΩ¯2.
The same idea is sequentially applied to the rest of subdomains. Finally, we obtain
(60)uin≤uin+1inΩ¯i,i=1,…,m,n≥0.
(
3
)The sequences(uin)are bounded and nondecreasing, so they are monotone converging pointwise toui,i=1,…,m. By using an elliptic regularity argument, we can see that functionsuisatisfy the same PDE onΩicf. [3, 12].


## 2.4. The Discretization

Fori=1,…,m,letτhibe a standard regular and quasiuniform finite element triangulation inΩi;hibeing the meshsize. Themmeshes is mutually independent onΩi∩Ωj,i,j=1,…,m,wherei≠j; that is a triangle belonging to one triangulation does not necessarily belong to the other ones.We consider the following discrete spaces:(61)Vhi={v∈C(Ω¯i)∩H1(Ωi)suchthatvK∈P1∀K∈τhi},
and for everyw∈C(Γ¯i), we set
(62)Vhi(w)={v∈Vhi:v=0on∂Ωi∩∂Ω;v=πhi(w)onΓi},
whereπhidenotes the interpolation operator onΓi.The Discrete Maximum Principle (cf. [14, 15]). We assume that the respective matrices resulting from the discretiztions of problems (30) are M-matrices.Note that as the mesheshiare independent over the overlapping subdomains, it is impossible to formulate a global approximate problem which would be the direct discrete counterpart of problem (13).


## 2.5. The Discrete Schwarz Sequences

Now, we define the discrete counterparts of the continuous Schwarz sequences defined in (30). Indeed, letuh0be the discrete analog ofu0, defined in (29) and we define fori=1,…,m,the discrete sequences(uhin+1)onΩisuch thatuhin+1∈Vhi(ghin+1)solves
(63)ai(uhin+1,v)+c(uhin+1,v)=(f(uhin+1),v),∀v∈Vhi(ghin+1);n≥0,
where
(64)ai(u,v)=∫Ωi(∇u·∇v)dx,ghin+1={uhi-1n+1onΓi∩Ωi-1uhi-2n+1onΓi∩(Ωi-2-Ωi-1)⋮uh1n+1onΓi∩(Ω1-Ωi-1-⋯-Ω1)uhmnonΓi∩(Ωm-⋃‍i-1j=1Ωj)uhm-1nonΓi∩(Ωm-1-⋃‍i-1j=1Ωj-Ωm)⋮uhi+1nonΓi∩(Ωi+1-⋃‍i-1j=1Ωj-⋃‍mj=i+1Ωj).Notation. From now on, we will adopt the following notationsi=1,…,m:
(65)|·|i=∥·∥L∞(Γi)∥·∥i=∥·∥L∞(Ωi),πhi=πh.


## 3.L∞  Error Analysis

### 3.1. The Auxiliary Schwarz Problems

This section is devoted to the proof of the main result of the present paper. To that end, we begin by introducingmdiscrete auxiliary problems.We define the following problems; fori=1,…,m,whi∈Vhi(gi)solves
(66)ai(whi,v)+c(whi,v)=(f(ui),v)∀v∈Vhi(gi).
It is then clear thatwhiis the finite element approximation ofuisolutions of (23). Therefore, making use of standard maximum norm estimates for linear elliptic problems, we have
(67)∥ui-whi∥i≤Ch2|logh|,
whereCis a constant independent ofh.


#### 3.1.1. The Main Results

Theorem 10.
Letρ=k/β<1. Then, there exists a constantCindependent of bothhandnsuch that
(68)∥ui-uhin+1∥i≤11-ρCh2|logh|;n≥0;i=1,…,m.Proof.
The proof of (68) will be carried by induction and is decomposed in three principal parts where each part is devoted to one of the following situations: situation (A), situation (B), and situation (C), respectively. The three situations are defined by
(69)(A)max1≤i≤m∥ui-uhi0∥i<Ch2|logh|,(70)(B)Ch2|logh|<∥ui-uhi0∥i,i=1,…,m,
and in the last situation we assume that a nonempty subsetJof the setM={1,…,m}exists such that
(71)(C)∀i∈M-J;∥ui-uhi0∥i<Ch2|logh|<∥uj-uhj0∥j,∀i∈M-J∥ui-uhi0∥i<Ch2|logh|∥uj-uhj0∥j∀j∈J.
To this end, we apply Proposition 6 and Theorem 1 for each subdomainΩias follows:
(72)∥ui-uhin+1∥i≤∥ui-whi∥i+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,|gi-ghin+1|i}∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,∥ui-uhin+1∥i≤+maxmaxj<i∥uj-uhjn+1∥j,maxi<j∥uj-uhjn∥j}.
The last inequality implies that
(73)∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{ρ∥ui-uhin+1∥i,maxj<i∥uj-uhjn+1∥j,∥ui-uhin+1∥i≤+maxmaxi<j∥uj-uhjn∥j}.
The interval]0,1[whichρbelongs to is divided into two subintervals as follows:
(74)]0,1[=]0,12[∪[12,1[,
and in each situation (A), (B), or (C) we deal with each subinterval separately.
Part 1. We consider situation (A). We begin the proof by the first subinterval that isρ∈]0,1/2[, so
(75)ρ1-ρ<1.
Forn=0,applying (73) to subdomain one, we get
(76)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish between two cases:
(77)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(78)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j
Case  1 implies that
(79)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then,
(80)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
Case  2 implies that
(81)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (81) byρ,we get
(82)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥j,then
(83)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(84)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is,
(85)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(86)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j,
which implies that
(87)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|,
or
(88)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (69). So for either of them there is a contradiction, and thus case 2 is impossible or case 2 is possible only if
(89)max1<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies that
(90)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2, we obtain
(91)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, we get
(92)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{max2<j≤m∥uj-uhj0∥jρ∥u2-uh21∥2,∥u1-uh11∥1,∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j}.
We have to distinguish the three following cases
(93)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(94)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=∥u1-uh11∥1
or
(95)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=max2<j≤m∥uj-uhj0∥j.
Case  1  implies that
(96)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then,
(97)∥u2-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥1∥u1-uh11∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥1max2<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We note that the two last inequalities coincide with (91) and (69), respectively. Case  2 implies that
(98)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (98) byρ, we get
(99)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1then
(100)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(101)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
Which implies that
(102)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(103)ρ1-ρCh2|logh|<∥u1-uh11∥1.
That is
(104)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(105)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they coincide with (91). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(106)∥u1-uh11∥1=ρ1-ρCh2|logh|.
Then case 2 implies
(107)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|.
With
(108)max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhj0∥j<Ch2|logh|
which coincides with (69). Case  3 implies
(109)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
By multiplying (109) byρwe get
(110)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
Soρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥jandmax2<j≤m∥uj-uhj0∥jthen w have
(111)(c)max2<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j
or
(112)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j<max2<j≤m∥uj-uhj0∥j.
Which implies
(113)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(114)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j.
That is,(115)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(116)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (c) and (d) are true because they coincide with (69). So there is contradiction and case  3 is impossible or case 3 is possible only if
(117)max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 3 implies that
(118)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|,
with
(119)∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Hence, in the three cases 1, 2, and 3 we get
(120)∥u2-uh21∥2≤11-ρCh2|logh|.
Equations (91) and (120) imply
(121)max1≤i<3∥ui-uhi1∥i≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainΩi,i=3,…,m, orderly, we get
(122)∥ui-uhi1∥i≤11-ρCh2|logh|.
Now, let us assume that
(123)∥ui-uhin∥i≤11-ρCh2|logh|,i=1,…,m
and prove that
(124)∥ui-uhin+1∥i≤11-ρCh2|logh|,i=1,…,m.
Equation (124) is obtained sequentially in the order of the numbering of subdomains. We begin by subdomain 1; indeed applying (73) to subdomain one, for iterationn+1, we get
(125)∥u1-uh1n+1∥1≤Ch2|logh|∥u1-uh1n+1∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}
We have to distinguish the two following cases(126)(1)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=ρ∥u1-uh1n+1∥1
or
(127)(2)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=max1<j≤m∥uj-uhjn∥j.
Case  1 implies
(128)∥u1-uh1n+1∥1≤Ch2|logh|+ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1.
Then
(129)∥u1-uh1n+1∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1max1<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
Case  2 implies
(130)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
By multiplying (130) byρwe get
(131)ρ∥u1-uh1n+1∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
Soρ∥u1-uh1n+1∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥jandmax1<j≤m∥uj-uhjn∥jthen
(132)(a)max1<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j
or
(133)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j<max1<j≤m∥uj-uhjn∥j.
That is
(134)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(135)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j.
Which implies
(136)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(137)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j<11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (123). So either there is a contradiction and thus case 2 is impossible or case 2 is possible only if
(138)max1<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 2 implies
(139)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j∥u1-uh1n+1∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh1n+1∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we obtain
(140)∥u1-uh1n+1∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, for iterationn+1,we get
(141)∥u1-uh2n+1∥2≤Ch2|logh|∥u1-uh2n+1∥2≤+max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+∥1,∥uj-uhjn∥j∥u1-uh2n+1∥2≤+maxDmax2<j≤m∥uj-uhjn∥j}.
We have to distinguish the three following cases
(142)(1)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=ρ∥u2-uh2n+1∥2
or
(143)(2)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=∥u1-uh1n+1∥1
or
(144)(3)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=max2<j≤m∥uj-uhjn∥j.
Case  1 implies
(145)∥u2-uh2n+1∥2≤Ch2|logh|+ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2.
Then
(146)∥u2-uh2n+1∥2≤11-ρCh2|logh|,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2∥u1-uh1n+1∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2max2<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
We note that the two last inequalities coincide with (140) and (123) respectively. Case  2 implies that
(147)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
By multiplying (147) byρwe get
(148)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρ∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
We remark thatρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρ∥u1-uh1n+1∥1and∥u1-uh1n+1∥1then
(149)(a)∥u1-uh1n+1∥1<ρCh2|logh|+ρ∥u1-uh1n+1∥1
or
(150)(b)ρCh2|logh|+ρ∥u1-uh1n+1∥1<∥u1-uh1n+1∥1.
Which implies
(151)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|
or
(152)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1.
That is
(153)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(154)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1≤11-ρCh2|logh|.
It is clear that both cases are true because they both coincide with (140). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(155)∥u1-uh1n+1∥1=ρ1-ρCh2|logh|.
Case  2 implies that
(156)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(157)max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhjn∥j<11-ρCh2|logh|
which coincides with (123). Case  3 implies
(158)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
By multiplying (158) byρwe get
(159)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
Soρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥jandmax2<j≤m∥uj-uhjn∥jthen
(160)(c)max2<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j
or
(161)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j<max2<j≤m∥uj-uhjn∥j.
Which implies
(162)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(163)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j.
That is
(164)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(165)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j≤11-ρCh2|logh|.
It is clear that both cases (c) and (d) are true because they both coincide with (123). So there is contradiction and case 3 is impossible or case 3 is possible only if
(166)max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 3 implies
(167)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(168)∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Hence in three cases 1, 2, and 3 we get
(169)∥u2-uh2n+1∥2≤11-ρCh2|logh|.
Equations (140) and (169) imply
(170)max1≤j<3∥uj-uhjn+1∥j≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainsΩi,i=3,…,m,orderly we get
(171)∥ui-uhin+1∥i≤11-ρCh2|logh|
which is the desired result. In the sequel we deal with the second subinterval[1/2,1[,indeedρ∈[1/2,1[so
(172)ρ1-ρ≥1.
Forn=0,applying (73) to subdomain one we get
(173)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish the two following cases
(174)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(175)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j.
Case  1 implies
(176)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then
(177)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1≤ρ1-ρCh2|logh|.
Case  2 implies
(178)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (178) byρwe get
(179)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥jthen
(180)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(181)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is
(182)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(183)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j.
Which implies
(184)max1<j≤m∥uj-uhj0∥j<Ch2|logh|≤ρ1-ρCh2|logh|
or
(185)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that only the case (a) is true because it coincides with (69) whereas (b) contradicts (172). Then case (a) with case 2 imply
(186)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1<Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we get
(187)∥u1-uh11∥1≤11-ρCh2|logh|.
By the same way we can obtain orderly
(188)∥ui-uhi1∥i≤11-ρCh2|logh|,i=2,…,m.
The remainder of the proof is also by induction by addopting the same idea applied in iteration one.
Part 2. We deal with situation (B). We begin by the first subinterval that isρ∈]0,1/2[;Indeed forn=0, (73) implies for subdomain 1,
(189)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases
(190)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(191)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies
(192)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So
(193)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j∥uj-uhj0∥j<Ch2|logh|.
We remark that (193) contradicts (70) so case 1 is impossible. Case  2 implies
(194)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (194) byρwe get
(195)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(196)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or
(197)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies
(198)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(199)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then
(200)Ch2|logh|≤max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(201)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (199) we get
(202)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
thus
(203)11-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
comparing this last inequalty (203) with (194), we deduce the two following possibilities
(204)∥u1-uh11∥1<11-ρCh2|logh|
or
(205)11-ρCh2|logh|<∥u1-uh11∥1
which implies
(206)∥u1-uh11∥1<11-ρCh2|logh|∥u1-uh11∥1<Ch2|logh|+max1<j∥uj-uhj0∥j
or
(207)11-ρCh2|logh|<∥u1-uh11∥111-ρCh2|logh|≤Ch2|logh|+max1<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(208)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get
(209)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥j∥u2-uh21∥2≤max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases
(210)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(211)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or
(212)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies
(213)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then
(214)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥1≤ρ∥u2-uh21∥211-ρCh2|logh|≤ρ1-ρCh2|logh|,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2max2<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
Equations (214) contradicts (70) so case 1 is impossible. Case  2 implies
(215)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (215) byρwe get
(216)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(217)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(218)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is
(219)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(220)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (208) and by addingCh2|logh|in (220) we get
(221)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (215), we distingush the two following possibilities
(222)∥u2-uh21∥2<11-ρCh2|logh|
or
(223)11-ρCh2|logh|<∥u2-uh21∥2
which imply
(224)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or
(225)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(226)∥u2-uh21∥2=11-ρCh2|logh|.
Case  3 implies
(227)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (227) byρwe get
(228)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(229)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or
(230)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is
(231)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(232)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies
(233)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(234)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (232) we get
(235)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
so
(236)11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
comparing the last inequality with (227), we consider the two following possibilties
(237)∥u2-uh21∥2<11-ρCh2|logh|
or
(238)11-ρCh2|logh|<∥u2-uh21∥2
that is(239)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
or(240)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(241)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(242)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (208) and (242) imply
(243)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(244)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one. The last step of the proof of this Part  2 is devoted to the subintervalρ∈[1/2,1[and is also by induction.Indeed for subdomain 1 andn=0,(73) implies
(245)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases(246)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(247)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies(248)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So(249)∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max1<j∥uj-uhj0∥jCh2|logh|≤ρ∥u1-uh11∥1Ch2|logh|≤ρ1-ρCh2|logh|.
We remark that (249) coincides with (70). Case  2 implies
(250)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (250) byρwe get
(251)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(252)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or(253)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies(254)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(255)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then(256)Ch2|logh|<max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(257)Ch2|logh|≤ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
It is clear that both cases (a) and (b) are true. So either there is contradiction and case 2 is impossible or case 2 is possible and we must have(258)max1<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies(259)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
So in both cases 1 and 2 we get(260)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get in this situation (B)
(261)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥uj-uhj0∥j∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(262)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(263)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or(264)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies(265)∥u1-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(266)∥u1-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2Ch2|logh|≤ρ1-ρCh2|logh|.
Case  2 implies(267)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (267) byρwe get
(268)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(269)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(270)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(271)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(272)ρ1-ρCh2|logh|<∥u1-uh11∥1.
Thus(273)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or(274)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they both coincide with (208). So either there is contradiction and case 2 is impossible or case 2 is possible and we must have
(275)∥u1-uh11∥1=ρ1-ρCh2|logh|.
With(276)Ch2|logh|<max2<j∥uj-uhj0∥j≤∥u1-uh11∥1Ch2|logh|=ρ1-ρCh2|logh|,
So case  2 implies(277)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Then(278)∥u2-uh21∥2≤11-ρCh2|logh|.
Case  3 implies(279)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (279) byρwe get
(280)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(281)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or(282)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is(283)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(284)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies(285)Ch2|logh|≤max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(286)Ch2|logh|≤ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
It is clear that both cases (c) and (d) are true. So either there is contradiction and case 3 is impossible or case 3 is possible only if.(287)max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|
Then in case 3 we obtain(288)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Thus(289)∥u2-uh21∥2≤11-ρCh2|logh|.
With(290)∥u1-uh11∥1≤max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=2,…,min order, we get
(291)∥ui-uhi1∥i≤11-ρCh2|logh|,i=1,…,m.
The remainder of the proof related to iterationsn≥2is by induction by adopting the same idea used in iteration one.
Part  3. We deal with situation (C). The proof related to this part  3 is also by induction and is so similar to the part  2 one. So we give it for the first subinterval]0,1/2[. The one for the second subinterval[1/2,1[will be omitted. Indeed forn=0andi=1we use (73) in conjunction with (71) we get
(292)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j,j∈J∥uj-uhjn∥j}
We have to distinguish the two following cases(293)(1)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(294)(2)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=max1<j,j∈J∥uj-uhj0∥j.
Case  1 implies(295)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j,j∈J∥uj-uhj0∥j≤ρ∥u1-uh11∥1max∥ujuhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We remark that (295) contradicts (71) so case 1 is impossible. Case  2 implies
(296)∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
By multiplying (296) byρwe get
(297)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥jandmax1<j,j∈J∥uj-uhj0∥j, so
(298)(a)max1<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j
or(299)(b)ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j<max1<j,j∈J∥uj-uhj0∥j.
Which implies(300)max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(301)ρ1-ρCh2|logh|<max1<j,j∈J∥uj-uhj0∥j.
Then(302)Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(303)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (301) we get
(304)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
thus(305)11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
comparing this last inequalty (305) with (296), we deduce the two following possibilities
(306)∥u1-uh11∥1<11-ρCh2|logh|
or(307)11-ρCh2|logh|<∥u1-uh11∥1,
which implies(308)∥u1-uh11∥1<11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
or(309)11-ρCh2|logh|<∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(310)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 in conjunction with (71) we get
(311)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥jmax2<j,j∈J∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(312)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(313)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=∥u1-uh11∥1
or(314)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=max2<j,j∈J∥uj-uhj0∥j.
Case  1 implies(315)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(316)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥111-ρCh2|logh|≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|max∥ujuhj0∥j<Ch2|logh|.
Equation (316) contradicts (71) so case 1 is impossible. Case  2 implies
(317)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (317) byρwe get
(318)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(319)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(320)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(321)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(322)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (310) and by addingCh2|logh|in (322) we get
(323)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (317), we distingush the two following possibilities
(324)∥u2-uh21∥2<11-ρCh2|logh|
or(325)11-ρCh2|logh|<∥u2-uh21∥2
which imply(326)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or(327)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(328)∥u2-uh21∥2=11-ρCh2|logh|.
Case 3 implies(329)∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
By multiplying (329) byρwe get
(330)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥jandmax2<j,j∈J∥uj-uhj0∥jthen
(331)(c)max2<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j
or(332)(d)ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j<max2<j,j∈J∥uj-uhj0∥j.
That is(333)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(334)ρ1-ρCh2|logh|<max2<j,j∈J∥uj-uhj0∥j.
Which implies(335)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or(336)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j,j∈J∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (334) we get
(337)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
so(338)11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
comparing the last inequality with (329), we consider the two following possibilties
(339)∥u2-uh21∥2<11-ρCh2|logh|
or(340)11-ρCh2|logh|<∥u2-uh21∥2
that is(341)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
or(342)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(343)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(344)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (310) and (344) imply
(345)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(346)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one.


## 3.1. The Auxiliary Schwarz Problems

This section is devoted to the proof of the main result of the present paper. To that end, we begin by introducingmdiscrete auxiliary problems.We define the following problems; fori=1,…,m,whi∈Vhi(gi)solves
(66)ai(whi,v)+c(whi,v)=(f(ui),v)∀v∈Vhi(gi).
It is then clear thatwhiis the finite element approximation ofuisolutions of (23). Therefore, making use of standard maximum norm estimates for linear elliptic problems, we have
(67)∥ui-whi∥i≤Ch2|logh|,
whereCis a constant independent ofh.


### 3.1.1. The Main Results

Theorem 10.
Letρ=k/β<1. Then, there exists a constantCindependent of bothhandnsuch that
(68)∥ui-uhin+1∥i≤11-ρCh2|logh|;n≥0;i=1,…,m.Proof.
The proof of (68) will be carried by induction and is decomposed in three principal parts where each part is devoted to one of the following situations: situation (A), situation (B), and situation (C), respectively. The three situations are defined by
(69)(A)max1≤i≤m∥ui-uhi0∥i<Ch2|logh|,(70)(B)Ch2|logh|<∥ui-uhi0∥i,i=1,…,m,
and in the last situation we assume that a nonempty subsetJof the setM={1,…,m}exists such that
(71)(C)∀i∈M-J;∥ui-uhi0∥i<Ch2|logh|<∥uj-uhj0∥j,∀i∈M-J∥ui-uhi0∥i<Ch2|logh|∥uj-uhj0∥j∀j∈J.
To this end, we apply Proposition 6 and Theorem 1 for each subdomainΩias follows:
(72)∥ui-uhin+1∥i≤∥ui-whi∥i+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,|gi-ghin+1|i}∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,∥ui-uhin+1∥i≤+maxmaxj<i∥uj-uhjn+1∥j,maxi<j∥uj-uhjn∥j}.
The last inequality implies that
(73)∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{ρ∥ui-uhin+1∥i,maxj<i∥uj-uhjn+1∥j,∥ui-uhin+1∥i≤+maxmaxi<j∥uj-uhjn∥j}.
The interval]0,1[whichρbelongs to is divided into two subintervals as follows:
(74)]0,1[=]0,12[∪[12,1[,
and in each situation (A), (B), or (C) we deal with each subinterval separately.
Part 1. We consider situation (A). We begin the proof by the first subinterval that isρ∈]0,1/2[, so
(75)ρ1-ρ<1.
Forn=0,applying (73) to subdomain one, we get
(76)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish between two cases:
(77)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(78)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j
Case  1 implies that
(79)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then,
(80)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
Case  2 implies that
(81)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (81) byρ,we get
(82)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥j,then
(83)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(84)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is,
(85)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(86)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j,
which implies that
(87)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|,
or
(88)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (69). So for either of them there is a contradiction, and thus case 2 is impossible or case 2 is possible only if
(89)max1<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies that
(90)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2, we obtain
(91)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, we get
(92)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{max2<j≤m∥uj-uhj0∥jρ∥u2-uh21∥2,∥u1-uh11∥1,∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j}.
We have to distinguish the three following cases
(93)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(94)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=∥u1-uh11∥1
or
(95)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=max2<j≤m∥uj-uhj0∥j.
Case  1  implies that
(96)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then,
(97)∥u2-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥1∥u1-uh11∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥1max2<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We note that the two last inequalities coincide with (91) and (69), respectively. Case  2 implies that
(98)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (98) byρ, we get
(99)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1then
(100)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(101)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
Which implies that
(102)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(103)ρ1-ρCh2|logh|<∥u1-uh11∥1.
That is
(104)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(105)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they coincide with (91). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(106)∥u1-uh11∥1=ρ1-ρCh2|logh|.
Then case 2 implies
(107)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|.
With
(108)max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhj0∥j<Ch2|logh|
which coincides with (69). Case  3 implies
(109)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
By multiplying (109) byρwe get
(110)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
Soρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥jandmax2<j≤m∥uj-uhj0∥jthen w have
(111)(c)max2<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j
or
(112)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j<max2<j≤m∥uj-uhj0∥j.
Which implies
(113)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(114)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j.
That is,(115)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(116)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (c) and (d) are true because they coincide with (69). So there is contradiction and case  3 is impossible or case 3 is possible only if
(117)max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 3 implies that
(118)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|,
with
(119)∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Hence, in the three cases 1, 2, and 3 we get
(120)∥u2-uh21∥2≤11-ρCh2|logh|.
Equations (91) and (120) imply
(121)max1≤i<3∥ui-uhi1∥i≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainΩi,i=3,…,m, orderly, we get
(122)∥ui-uhi1∥i≤11-ρCh2|logh|.
Now, let us assume that
(123)∥ui-uhin∥i≤11-ρCh2|logh|,i=1,…,m
and prove that
(124)∥ui-uhin+1∥i≤11-ρCh2|logh|,i=1,…,m.
Equation (124) is obtained sequentially in the order of the numbering of subdomains. We begin by subdomain 1; indeed applying (73) to subdomain one, for iterationn+1, we get
(125)∥u1-uh1n+1∥1≤Ch2|logh|∥u1-uh1n+1∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}
We have to distinguish the two following cases(126)(1)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=ρ∥u1-uh1n+1∥1
or
(127)(2)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=max1<j≤m∥uj-uhjn∥j.
Case  1 implies
(128)∥u1-uh1n+1∥1≤Ch2|logh|+ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1.
Then
(129)∥u1-uh1n+1∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1max1<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
Case  2 implies
(130)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
By multiplying (130) byρwe get
(131)ρ∥u1-uh1n+1∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
Soρ∥u1-uh1n+1∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥jandmax1<j≤m∥uj-uhjn∥jthen
(132)(a)max1<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j
or
(133)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j<max1<j≤m∥uj-uhjn∥j.
That is
(134)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(135)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j.
Which implies
(136)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(137)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j<11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (123). So either there is a contradiction and thus case 2 is impossible or case 2 is possible only if
(138)max1<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 2 implies
(139)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j∥u1-uh1n+1∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh1n+1∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we obtain
(140)∥u1-uh1n+1∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, for iterationn+1,we get
(141)∥u1-uh2n+1∥2≤Ch2|logh|∥u1-uh2n+1∥2≤+max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+∥1,∥uj-uhjn∥j∥u1-uh2n+1∥2≤+maxDmax2<j≤m∥uj-uhjn∥j}.
We have to distinguish the three following cases
(142)(1)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=ρ∥u2-uh2n+1∥2
or
(143)(2)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=∥u1-uh1n+1∥1
or
(144)(3)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=max2<j≤m∥uj-uhjn∥j.
Case  1 implies
(145)∥u2-uh2n+1∥2≤Ch2|logh|+ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2.
Then
(146)∥u2-uh2n+1∥2≤11-ρCh2|logh|,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2∥u1-uh1n+1∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2max2<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
We note that the two last inequalities coincide with (140) and (123) respectively. Case  2 implies that
(147)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
By multiplying (147) byρwe get
(148)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρ∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
We remark thatρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρ∥u1-uh1n+1∥1and∥u1-uh1n+1∥1then
(149)(a)∥u1-uh1n+1∥1<ρCh2|logh|+ρ∥u1-uh1n+1∥1
or
(150)(b)ρCh2|logh|+ρ∥u1-uh1n+1∥1<∥u1-uh1n+1∥1.
Which implies
(151)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|
or
(152)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1.
That is
(153)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(154)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1≤11-ρCh2|logh|.
It is clear that both cases are true because they both coincide with (140). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(155)∥u1-uh1n+1∥1=ρ1-ρCh2|logh|.
Case  2 implies that
(156)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(157)max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhjn∥j<11-ρCh2|logh|
which coincides with (123). Case  3 implies
(158)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
By multiplying (158) byρwe get
(159)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
Soρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥jandmax2<j≤m∥uj-uhjn∥jthen
(160)(c)max2<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j
or
(161)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j<max2<j≤m∥uj-uhjn∥j.
Which implies
(162)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(163)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j.
That is
(164)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(165)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j≤11-ρCh2|logh|.
It is clear that both cases (c) and (d) are true because they both coincide with (123). So there is contradiction and case 3 is impossible or case 3 is possible only if
(166)max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 3 implies
(167)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(168)∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Hence in three cases 1, 2, and 3 we get
(169)∥u2-uh2n+1∥2≤11-ρCh2|logh|.
Equations (140) and (169) imply
(170)max1≤j<3∥uj-uhjn+1∥j≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainsΩi,i=3,…,m,orderly we get
(171)∥ui-uhin+1∥i≤11-ρCh2|logh|
which is the desired result. In the sequel we deal with the second subinterval[1/2,1[,indeedρ∈[1/2,1[so
(172)ρ1-ρ≥1.
Forn=0,applying (73) to subdomain one we get
(173)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish the two following cases
(174)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(175)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j.
Case  1 implies
(176)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then
(177)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1≤ρ1-ρCh2|logh|.
Case  2 implies
(178)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (178) byρwe get
(179)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥jthen
(180)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(181)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is
(182)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(183)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j.
Which implies
(184)max1<j≤m∥uj-uhj0∥j<Ch2|logh|≤ρ1-ρCh2|logh|
or
(185)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that only the case (a) is true because it coincides with (69) whereas (b) contradicts (172). Then case (a) with case 2 imply
(186)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1<Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we get
(187)∥u1-uh11∥1≤11-ρCh2|logh|.
By the same way we can obtain orderly
(188)∥ui-uhi1∥i≤11-ρCh2|logh|,i=2,…,m.
The remainder of the proof is also by induction by addopting the same idea applied in iteration one.
Part 2. We deal with situation (B). We begin by the first subinterval that isρ∈]0,1/2[;Indeed forn=0, (73) implies for subdomain 1,
(189)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases
(190)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(191)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies
(192)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So
(193)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j∥uj-uhj0∥j<Ch2|logh|.
We remark that (193) contradicts (70) so case 1 is impossible. Case  2 implies
(194)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (194) byρwe get
(195)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(196)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or
(197)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies
(198)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(199)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then
(200)Ch2|logh|≤max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(201)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (199) we get
(202)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
thus
(203)11-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
comparing this last inequalty (203) with (194), we deduce the two following possibilities
(204)∥u1-uh11∥1<11-ρCh2|logh|
or
(205)11-ρCh2|logh|<∥u1-uh11∥1
which implies
(206)∥u1-uh11∥1<11-ρCh2|logh|∥u1-uh11∥1<Ch2|logh|+max1<j∥uj-uhj0∥j
or
(207)11-ρCh2|logh|<∥u1-uh11∥111-ρCh2|logh|≤Ch2|logh|+max1<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(208)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get
(209)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥j∥u2-uh21∥2≤max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases
(210)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(211)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or
(212)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies
(213)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then
(214)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥1≤ρ∥u2-uh21∥211-ρCh2|logh|≤ρ1-ρCh2|logh|,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2max2<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
Equations (214) contradicts (70) so case 1 is impossible. Case  2 implies
(215)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (215) byρwe get
(216)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(217)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(218)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is
(219)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(220)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (208) and by addingCh2|logh|in (220) we get
(221)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (215), we distingush the two following possibilities
(222)∥u2-uh21∥2<11-ρCh2|logh|
or
(223)11-ρCh2|logh|<∥u2-uh21∥2
which imply
(224)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or
(225)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(226)∥u2-uh21∥2=11-ρCh2|logh|.
Case  3 implies
(227)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (227) byρwe get
(228)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(229)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or
(230)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is
(231)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(232)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies
(233)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(234)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (232) we get
(235)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
so
(236)11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
comparing the last inequality with (227), we consider the two following possibilties
(237)∥u2-uh21∥2<11-ρCh2|logh|
or
(238)11-ρCh2|logh|<∥u2-uh21∥2
that is(239)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
or(240)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(241)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(242)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (208) and (242) imply
(243)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(244)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one. The last step of the proof of this Part  2 is devoted to the subintervalρ∈[1/2,1[and is also by induction.Indeed for subdomain 1 andn=0,(73) implies
(245)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases(246)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(247)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies(248)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So(249)∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max1<j∥uj-uhj0∥jCh2|logh|≤ρ∥u1-uh11∥1Ch2|logh|≤ρ1-ρCh2|logh|.
We remark that (249) coincides with (70). Case  2 implies
(250)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (250) byρwe get
(251)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(252)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or(253)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies(254)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(255)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then(256)Ch2|logh|<max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(257)Ch2|logh|≤ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
It is clear that both cases (a) and (b) are true. So either there is contradiction and case 2 is impossible or case 2 is possible and we must have(258)max1<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies(259)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
So in both cases 1 and 2 we get(260)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get in this situation (B)
(261)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥uj-uhj0∥j∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(262)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(263)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or(264)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies(265)∥u1-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(266)∥u1-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2Ch2|logh|≤ρ1-ρCh2|logh|.
Case  2 implies(267)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (267) byρwe get
(268)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(269)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(270)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(271)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(272)ρ1-ρCh2|logh|<∥u1-uh11∥1.
Thus(273)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or(274)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they both coincide with (208). So either there is contradiction and case 2 is impossible or case 2 is possible and we must have
(275)∥u1-uh11∥1=ρ1-ρCh2|logh|.
With(276)Ch2|logh|<max2<j∥uj-uhj0∥j≤∥u1-uh11∥1Ch2|logh|=ρ1-ρCh2|logh|,
So case  2 implies(277)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Then(278)∥u2-uh21∥2≤11-ρCh2|logh|.
Case  3 implies(279)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (279) byρwe get
(280)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(281)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or(282)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is(283)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(284)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies(285)Ch2|logh|≤max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(286)Ch2|logh|≤ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
It is clear that both cases (c) and (d) are true. So either there is contradiction and case 3 is impossible or case 3 is possible only if.(287)max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|
Then in case 3 we obtain(288)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Thus(289)∥u2-uh21∥2≤11-ρCh2|logh|.
With(290)∥u1-uh11∥1≤max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=2,…,min order, we get
(291)∥ui-uhi1∥i≤11-ρCh2|logh|,i=1,…,m.
The remainder of the proof related to iterationsn≥2is by induction by adopting the same idea used in iteration one.
Part  3. We deal with situation (C). The proof related to this part  3 is also by induction and is so similar to the part  2 one. So we give it for the first subinterval]0,1/2[. The one for the second subinterval[1/2,1[will be omitted. Indeed forn=0andi=1we use (73) in conjunction with (71) we get
(292)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j,j∈J∥uj-uhjn∥j}
We have to distinguish the two following cases(293)(1)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(294)(2)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=max1<j,j∈J∥uj-uhj0∥j.
Case  1 implies(295)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j,j∈J∥uj-uhj0∥j≤ρ∥u1-uh11∥1max∥ujuhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We remark that (295) contradicts (71) so case 1 is impossible. Case  2 implies
(296)∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
By multiplying (296) byρwe get
(297)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥jandmax1<j,j∈J∥uj-uhj0∥j, so
(298)(a)max1<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j
or(299)(b)ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j<max1<j,j∈J∥uj-uhj0∥j.
Which implies(300)max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(301)ρ1-ρCh2|logh|<max1<j,j∈J∥uj-uhj0∥j.
Then(302)Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(303)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (301) we get
(304)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
thus(305)11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
comparing this last inequalty (305) with (296), we deduce the two following possibilities
(306)∥u1-uh11∥1<11-ρCh2|logh|
or(307)11-ρCh2|logh|<∥u1-uh11∥1,
which implies(308)∥u1-uh11∥1<11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
or(309)11-ρCh2|logh|<∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(310)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 in conjunction with (71) we get
(311)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥jmax2<j,j∈J∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(312)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(313)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=∥u1-uh11∥1
or(314)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=max2<j,j∈J∥uj-uhj0∥j.
Case  1 implies(315)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(316)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥111-ρCh2|logh|≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|max∥ujuhj0∥j<Ch2|logh|.
Equation (316) contradicts (71) so case 1 is impossible. Case  2 implies
(317)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (317) byρwe get
(318)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(319)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(320)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(321)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(322)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (310) and by addingCh2|logh|in (322) we get
(323)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (317), we distingush the two following possibilities
(324)∥u2-uh21∥2<11-ρCh2|logh|
or(325)11-ρCh2|logh|<∥u2-uh21∥2
which imply(326)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or(327)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(328)∥u2-uh21∥2=11-ρCh2|logh|.
Case 3 implies(329)∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
By multiplying (329) byρwe get
(330)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥jandmax2<j,j∈J∥uj-uhj0∥jthen
(331)(c)max2<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j
or(332)(d)ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j<max2<j,j∈J∥uj-uhj0∥j.
That is(333)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(334)ρ1-ρCh2|logh|<max2<j,j∈J∥uj-uhj0∥j.
Which implies(335)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or(336)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j,j∈J∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (334) we get
(337)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
so(338)11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
comparing the last inequality with (329), we consider the two following possibilties
(339)∥u2-uh21∥2<11-ρCh2|logh|
or(340)11-ρCh2|logh|<∥u2-uh21∥2
that is(341)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
or(342)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(343)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(344)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (310) and (344) imply
(345)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(346)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one.


## 3.1.1. The Main Results

Theorem 10.
Letρ=k/β<1. Then, there exists a constantCindependent of bothhandnsuch that
(68)∥ui-uhin+1∥i≤11-ρCh2|logh|;n≥0;i=1,…,m.Proof.
The proof of (68) will be carried by induction and is decomposed in three principal parts where each part is devoted to one of the following situations: situation (A), situation (B), and situation (C), respectively. The three situations are defined by
(69)(A)max1≤i≤m∥ui-uhi0∥i<Ch2|logh|,(70)(B)Ch2|logh|<∥ui-uhi0∥i,i=1,…,m,
and in the last situation we assume that a nonempty subsetJof the setM={1,…,m}exists such that
(71)(C)∀i∈M-J;∥ui-uhi0∥i<Ch2|logh|<∥uj-uhj0∥j,∀i∈M-J∥ui-uhi0∥i<Ch2|logh|∥uj-uhj0∥j∀j∈J.
To this end, we apply Proposition 6 and Theorem 1 for each subdomainΩias follows:
(72)∥ui-uhin+1∥i≤∥ui-whi∥i+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|+∥whi-uhin+1∥i∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,|gi-ghin+1|i}∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{(1β)∥f(ui)-f(uhin+1)∥i,∥ui-uhin+1∥i≤+maxmaxj<i∥uj-uhjn+1∥j,maxi<j∥uj-uhjn∥j}.
The last inequality implies that
(73)∥ui-uhin+1∥i≤Ch2|logh|∥ui-uhin+1∥i≤+max{ρ∥ui-uhin+1∥i,maxj<i∥uj-uhjn+1∥j,∥ui-uhin+1∥i≤+maxmaxi<j∥uj-uhjn∥j}.
The interval]0,1[whichρbelongs to is divided into two subintervals as follows:
(74)]0,1[=]0,12[∪[12,1[,
and in each situation (A), (B), or (C) we deal with each subinterval separately.
Part 1. We consider situation (A). We begin the proof by the first subinterval that isρ∈]0,1/2[, so
(75)ρ1-ρ<1.
Forn=0,applying (73) to subdomain one, we get
(76)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish between two cases:
(77)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(78)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j
Case  1 implies that
(79)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then,
(80)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
Case  2 implies that
(81)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (81) byρ,we get
(82)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥j,then
(83)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(84)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is,
(85)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(86)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j,
which implies that
(87)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|,
or
(88)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (69). So for either of them there is a contradiction, and thus case 2 is impossible or case 2 is possible only if
(89)max1<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies that
(90)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2, we obtain
(91)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, we get
(92)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{max2<j≤m∥uj-uhj0∥jρ∥u2-uh21∥2,∥u1-uh11∥1,∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j}.
We have to distinguish the three following cases
(93)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(94)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=∥u1-uh11∥1
or
(95)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j}=max2<j≤m∥uj-uhj0∥j.
Case  1  implies that
(96)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then,
(97)∥u2-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥1∥u1-uh11∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhj0∥j≤ρ∥u2-uh21∥1max2<j≤m∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We note that the two last inequalities coincide with (91) and (69), respectively. Case  2 implies that
(98)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (98) byρ, we get
(99)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1then
(100)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(101)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
Which implies that
(102)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(103)ρ1-ρCh2|logh|<∥u1-uh11∥1.
That is
(104)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(105)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they coincide with (91). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(106)∥u1-uh11∥1=ρ1-ρCh2|logh|.
Then case 2 implies
(107)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|.
With
(108)max2<j≤m∥uj-uhj0∥j≤∥u1-uh11∥1max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhj0∥j<Ch2|logh|
which coincides with (69). Case  3 implies
(109)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
By multiplying (109) byρwe get
(110)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j≤m∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j.
Soρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥jandmax2<j≤m∥uj-uhj0∥jthen w have
(111)(c)max2<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j
or
(112)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhj0∥j<max2<j≤m∥uj-uhj0∥j.
Which implies
(113)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(114)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j.
That is,(115)max2<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(116)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that both cases (c) and (d) are true because they coincide with (69). So there is contradiction and case  3 is impossible or case 3 is possible only if
(117)max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 3 implies that
(118)∥u2-uh21∥2≤Ch2|logh|+max2<j≤m∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh21∥2=11-ρCh2|logh|,
with
(119)∥u1-uh11∥1≤max2<j≤m∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Hence, in the three cases 1, 2, and 3 we get
(120)∥u2-uh21∥2≤11-ρCh2|logh|.
Equations (91) and (120) imply
(121)max1≤i<3∥ui-uhi1∥i≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainΩi,i=3,…,m, orderly, we get
(122)∥ui-uhi1∥i≤11-ρCh2|logh|.
Now, let us assume that
(123)∥ui-uhin∥i≤11-ρCh2|logh|,i=1,…,m
and prove that
(124)∥ui-uhin+1∥i≤11-ρCh2|logh|,i=1,…,m.
Equation (124) is obtained sequentially in the order of the numbering of subdomains. We begin by subdomain 1; indeed applying (73) to subdomain one, for iterationn+1, we get
(125)∥u1-uh1n+1∥1≤Ch2|logh|∥u1-uh1n+1∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}
We have to distinguish the two following cases(126)(1)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=ρ∥u1-uh1n+1∥1
or
(127)(2)max{ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j}=max1<j≤m∥uj-uhjn∥j.
Case  1 implies
(128)∥u1-uh1n+1∥1≤Ch2|logh|+ρ∥u1-uh1n+1∥1,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1.
Then
(129)∥u1-uh1n+1∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhjn∥j≤ρ∥u1-uh1n+1∥1max1<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
Case  2 implies
(130)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
By multiplying (130) byρwe get
(131)ρ∥u1-uh1n+1∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j,ρ∥u1-uh1n+1∥1≤max1<j≤m∥uj-uhjn∥j.
Soρ∥u1-uh1n+1∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥jandmax1<j≤m∥uj-uhjn∥jthen
(132)(a)max1<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j
or
(133)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhjn∥j<max1<j≤m∥uj-uhjn∥j.
That is
(134)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(135)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j.
Which implies
(136)max1<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(137)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhjn∥j<11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because both coincide with (123). So either there is a contradiction and thus case 2 is impossible or case 2 is possible only if
(138)max1<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 2 implies
(139)∥u1-uh1n+1∥1≤Ch2|logh|+max1<j≤m∥uj-uhjn∥j∥u1-uh1n+1∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh1n+1∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we obtain
(140)∥u1-uh1n+1∥1≤11-ρCh2|logh|.
Similarly applying (73) to the second subdomain, for iterationn+1,we get
(141)∥u1-uh2n+1∥2≤Ch2|logh|∥u1-uh2n+1∥2≤+max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+∥1,∥uj-uhjn∥j∥u1-uh2n+1∥2≤+maxDmax2<j≤m∥uj-uhjn∥j}.
We have to distinguish the three following cases
(142)(1)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=ρ∥u2-uh2n+1∥2
or
(143)(2)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=∥u1-uh1n+1∥1
or
(144)(3)max{ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j}=max2<j≤m∥uj-uhjn∥j.
Case  1 implies
(145)∥u2-uh2n+1∥2≤Ch2|logh|+ρ∥u2-uh2n+1∥2,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2.
Then
(146)∥u2-uh2n+1∥2≤11-ρCh2|logh|,∥u1-uh1n+1∥1≤ρ∥u2-uh2n+1∥2∥u1-uh1n+1∥1≤ρ1-ρCh2|logh|<11-ρCh2|logh|,max2<j≤m∥uj-uhjn∥j≤ρ∥u2-uh2n+1∥2max2<j≤m∥uj-uhjn∥j≤ρ1-ρCh2|logh|<11-ρCh2|logh|.
We note that the two last inequalities coincide with (140) and (123) respectively. Case  2 implies that
(147)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
By multiplying (147) byρwe get
(148)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρ∥u1-uh1n+1∥1,ρ∥u2-uh2n+1∥2≤∥u1-uh1n+1∥1,max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1.
We remark thatρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρ∥u1-uh1n+1∥1and∥u1-uh1n+1∥1then
(149)(a)∥u1-uh1n+1∥1<ρCh2|logh|+ρ∥u1-uh1n+1∥1
or
(150)(b)ρCh2|logh|+ρ∥u1-uh1n+1∥1<∥u1-uh1n+1∥1.
Which implies
(151)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|
or
(152)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1.
That is
(153)∥u1-uh1n+1∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(154)ρ1-ρCh2|logh|<∥u1-uh1n+1∥1≤11-ρCh2|logh|.
It is clear that both cases are true because they both coincide with (140). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have
(155)∥u1-uh1n+1∥1=ρ1-ρCh2|logh|.
Case  2 implies that
(156)∥u2-uh2n+1∥2≤Ch2|logh|+∥u1-uh1n+1∥1∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(157)max2<j≤m∥uj-uhjn∥j≤∥u1-uh1n+1∥1max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|max2<j≤m∥uj-uhjn∥j<11-ρCh2|logh|
which coincides with (123). Case  3 implies
(158)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
By multiplying (158) byρwe get
(159)ρ∥u2-uh2n+1∥2≤ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j,ρ∥u2-uh2n+1∥2≤max2<j≤m∥uj-uhjn∥j,∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j.
Soρ∥u2-uh2n+1∥2is bounded by bothρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥jandmax2<j≤m∥uj-uhjn∥jthen
(160)(c)max2<j≤m∥uj-uhjn∥j<ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j
or
(161)(d)ρCh2|logh|+ρmax2<j≤m∥uj-uhjn∥j<max2<j≤m∥uj-uhjn∥j.
Which implies
(162)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|
or
(163)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j.
That is
(164)max2<j≤m∥uj-uhjn∥j<ρ1-ρCh2|logh|<11-ρCh2|logh|
or
(165)ρ1-ρCh2|logh|<max2<j≤m∥uj-uhjn∥j≤11-ρCh2|logh|.
It is clear that both cases (c) and (d) are true because they both coincide with (123). So there is contradiction and case 3 is impossible or case 3 is possible only if
(166)max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Then case 3 implies
(167)∥u2-uh2n+1∥2≤Ch2|logh|+max2<j≤m∥uj-uhjn∥j∥u2-uh2n+1∥2=Ch2|logh|+ρ1-ρCh2|logh|∥u2-uh2n+1∥2=11-ρCh2|logh|.
With
(168)∥u1-uh1n+1∥1≤max2<j≤m∥uj-uhjn∥j=ρ1-ρCh2|logh|.
Hence in three cases 1, 2, and 3 we get
(169)∥u2-uh2n+1∥2≤11-ρCh2|logh|.
Equations (140) and (169) imply
(170)max1≤j<3∥uj-uhjn+1∥j≤11-ρCh2|logh|.
Similarly applying the same idea for the rest of subdomainsΩi,i=3,…,m,orderly we get
(171)∥ui-uhin+1∥i≤11-ρCh2|logh|
which is the desired result. In the sequel we deal with the second subinterval[1/2,1[,indeedρ∈[1/2,1[so
(172)ρ1-ρ≥1.
Forn=0,applying (73) to subdomain one we get
(173)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}.
We have to distinguish the two following cases
(174)(1)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(175)(2)max{ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j}=max1<j≤m∥uj-uhj0∥j.
Case  1 implies
(176)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
Then
(177)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j≤m∥uj-uhj0∥j≤ρ∥u1-uh11∥1≤ρ1-ρCh2|logh|.
Case  2 implies
(178)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
By multiplying (178) byρwe get
(179)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j≤m∥uj-uhj0∥j.
Soρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥jandmax1<j≤m∥uj-uhj0∥jthen
(180)(a)max1<j≤m∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j
or
(181)(b)ρCh2|logh|+ρmax1<j≤m∥uj-uhj0∥j<max1<j≤m∥uj-uhj0∥j.
That is
(182)max1<j≤m∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(183)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j.
Which implies
(184)max1<j≤m∥uj-uhj0∥j<Ch2|logh|≤ρ1-ρCh2|logh|
or
(185)ρ1-ρCh2|logh|<max1<j≤m∥uj-uhj0∥j<Ch2|logh|.
It is clear that only the case (a) is true because it coincides with (69) whereas (b) contradicts (172). Then case (a) with case 2 imply
(186)∥u1-uh11∥1≤Ch2|logh|+max1<j≤m∥uj-uhj0∥j∥u1-uh11∥1<Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
Hence in both cases 1 and 2 we get
(187)∥u1-uh11∥1≤11-ρCh2|logh|.
By the same way we can obtain orderly
(188)∥ui-uhi1∥i≤11-ρCh2|logh|,i=2,…,m.
The remainder of the proof is also by induction by addopting the same idea applied in iteration one.
Part 2. We deal with situation (B). We begin by the first subinterval that isρ∈]0,1/2[;Indeed forn=0, (73) implies for subdomain 1,
(189)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases
(190)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or
(191)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies
(192)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So
(193)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1max1<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|max1<j∥uj-uhj0∥j<Ch2|logh|.
We remark that (193) contradicts (70) so case 1 is impossible. Case  2 implies
(194)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (194) byρwe get
(195)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(196)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or
(197)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies
(198)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(199)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then
(200)Ch2|logh|≤max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(201)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (199) we get
(202)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
thus
(203)11-ρCh2|logh|<Ch2|logh|+max1<j∥uj-uhj0∥j
comparing this last inequalty (203) with (194), we deduce the two following possibilities
(204)∥u1-uh11∥1<11-ρCh2|logh|
or
(205)11-ρCh2|logh|<∥u1-uh11∥1
which implies
(206)∥u1-uh11∥1<11-ρCh2|logh|∥u1-uh11∥1<Ch2|logh|+max1<j∥uj-uhj0∥j
or
(207)11-ρCh2|logh|<∥u1-uh11∥111-ρCh2|logh|≤Ch2|logh|+max1<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(208)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get
(209)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥j∥u2-uh21∥2≤max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases
(210)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or
(211)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or
(212)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies
(213)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then
(214)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥1≤ρ∥u2-uh21∥211-ρCh2|logh|≤ρ1-ρCh2|logh|,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2max2<j∥uj-uhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
Equations (214) contradicts (70) so case 1 is impossible. Case  2 implies
(215)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (215) byρwe get
(216)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(217)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or
(218)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is
(219)∥u1-uh11∥1<ρ1-ρCh2|logh|
or
(220)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (208) and by addingCh2|logh|in (220) we get
(221)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (215), we distingush the two following possibilities
(222)∥u2-uh21∥2<11-ρCh2|logh|
or
(223)11-ρCh2|logh|<∥u2-uh21∥2
which imply
(224)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or
(225)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if
(226)∥u2-uh21∥2=11-ρCh2|logh|.
Case  3 implies
(227)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (227) byρwe get
(228)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(229)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or
(230)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is
(231)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or
(232)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies
(233)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or
(234)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (232) we get
(235)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
so
(236)11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
comparing the last inequality with (227), we consider the two following possibilties
(237)∥u2-uh21∥2<11-ρCh2|logh|
or
(238)11-ρCh2|logh|<∥u2-uh21∥2
that is(239)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j∥uj-uhj0∥j
or(240)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(241)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(242)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (208) and (242) imply
(243)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(244)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one. The last step of the proof of this Part  2 is devoted to the subintervalρ∈[1/2,1[and is also by induction.Indeed for subdomain 1 andn=0,(73) implies
(245)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}.
We have to distinguish the two following cases(246)(1)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(247)(2)max{ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j}=max1<j∥uj-uhj0∥j.
Case  1 implies(248)∥u1-uh11∥1≤Ch2|logh|+ρ∥u1-uh11∥1,max1<j∥uj-uhj0∥j≤ρ∥u1-uh11∥1.
So(249)∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max1<j∥uj-uhj0∥jCh2|logh|≤ρ∥u1-uh11∥1Ch2|logh|≤ρ1-ρCh2|logh|.
We remark that (249) coincides with (70). Case  2 implies
(250)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
By multiplying (250) byρwe get
(251)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j∥uj-uhj0∥jandmax1<j∥uj-uhj0∥j, so
(252)(a)max1<j∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j∥uj-uhj0∥j
or(253)(b)ρCh2|logh|+ρmax1<j∥uj-uhj0∥j<max1<j∥uj-uhj0∥j.
Which implies(254)max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(255)ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
Then(256)Ch2|logh|<max1<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(257)Ch2|logh|≤ρ1-ρCh2|logh|<max1<j∥uj-uhj0∥j.
It is clear that both cases (a) and (b) are true. So either there is contradiction and case 2 is impossible or case 2 is possible and we must have(258)max1<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
Then case 2 implies(259)∥u1-uh11∥1≤Ch2|logh|+max1<j∥uj-uhj0∥j∥u1-uh11∥1=Ch2|logh|+ρ1-ρCh2|logh|∥u1-uh11∥1=11-ρCh2|logh|.
So in both cases 1 and 2 we get(260)∥u1-uh11∥1≤11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 we get in this situation (B)
(261)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥uj-uhj0∥j∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(262)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(263)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=∥u1-uh11∥1
or(264)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j∥uj-uhj0∥j}=max2<j∥uj-uhj0∥j.
Case  1 implies(265)∥u1-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(266)∥u1-uh21∥2≤11-ρCh2|logh|,∥u1-uh11∥1≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|∥u1-uh11∥1≤11-ρCh2|logh|,Ch2|logh|<max2<j∥uj-uhj0∥j≤ρ∥u2-uh21∥2Ch2|logh|≤ρ1-ρCh2|logh|.
Case  2 implies(267)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (267) byρwe get
(268)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(269)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(270)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(271)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(272)ρ1-ρCh2|logh|<∥u1-uh11∥1.
Thus(273)∥u1-uh11∥1<ρ1-ρCh2|logh|<11-ρCh2|logh|
or(274)ρ1-ρCh2|logh|<∥u1-uh11∥1≤11-ρCh2|logh|.
It is clear that both cases (a) and (b) are true because they both coincide with (208). So either there is contradiction and case 2 is impossible or case 2 is possible and we must have
(275)∥u1-uh11∥1=ρ1-ρCh2|logh|.
With(276)Ch2|logh|<max2<j∥uj-uhj0∥j≤∥u1-uh11∥1Ch2|logh|=ρ1-ρCh2|logh|,
So case  2 implies(277)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Then(278)∥u2-uh21∥2≤11-ρCh2|logh|.
Case  3 implies(279)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
By multiplying (279) byρwe get
(280)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j∥uj-uhj0∥jandmax2<j∥uj-uhj0∥jthen
(281)(c)max2<j∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j∥uj-uhj0∥j
or(282)(d)ρCh2|logh|+ρmax2<j∥uj-uhj0∥j<max2<j∥uj-uhj0∥j.
That is(283)max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(284)ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
Which implies(285)Ch2|logh|≤max2<j∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(286)Ch2|logh|≤ρ1-ρCh2|logh|<max2<j∥uj-uhj0∥j.
It is clear that both cases (c) and (d) are true. So either there is contradiction and case 3 is impossible or case 3 is possible only if.(287)max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|
Then in case 3 we obtain(288)∥u2-uh21∥2≤Ch2|logh|+max2<j∥uj-uhj0∥j∥u2-uh21∥2=Ch2|logh|+ρ1-ρCh2|logh|
Thus(289)∥u2-uh21∥2≤11-ρCh2|logh|.
With(290)∥u1-uh11∥1≤max2<j∥uj-uhj0∥j=ρ1-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=2,…,min order, we get
(291)∥ui-uhi1∥i≤11-ρCh2|logh|,i=1,…,m.
The remainder of the proof related to iterationsn≥2is by induction by adopting the same idea used in iteration one.
Part  3. We deal with situation (C). The proof related to this part  3 is also by induction and is so similar to the part  2 one. So we give it for the first subinterval]0,1/2[. The one for the second subinterval[1/2,1[will be omitted. Indeed forn=0andi=1we use (73) in conjunction with (71) we get
(292)∥u1-uh11∥1≤Ch2|logh|∥u1-uh11∥1≤+max{ρ∥u1-uh1n+1∥1,max1<j,j∈J∥uj-uhjn∥j}
We have to distinguish the two following cases(293)(1)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=ρ∥u1-uh11∥1
or(294)(2)max{ρ∥u1-uh11∥1,max1<j,j∈J∥uj-uhj0∥j}=max1<j,j∈J∥uj-uhj0∥j.
Case  1 implies(295)∥u1-uh11∥1≤11-ρCh2|logh|,max1<j,j∈J∥uj-uhj0∥j≤ρ∥u1-uh11∥1max∥ujuhj0∥j≤ρ1-ρCh2|logh|<Ch2|logh|.
We remark that (295) contradicts (71) so case 1 is impossible. Case  2 implies
(296)∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
By multiplying (296) byρwe get
(297)ρ∥u1-uh11∥1≤ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j,ρ∥u1-uh11∥1≤max1<j,j∈J∥uj-uhj0∥j.
It is clear thatρ∥u1-uh11∥1is bounded by bothρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥jandmax1<j,j∈J∥uj-uhj0∥j, so
(298)(a)max1<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j
or(299)(b)ρCh2|logh|+ρmax1<j,j∈J∥uj-uhj0∥j<max1<j,j∈J∥uj-uhj0∥j.
Which implies(300)max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(301)ρ1-ρCh2|logh|<max1<j,j∈J∥uj-uhj0∥j.
Then(302)Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(303)ρ1-ρCh2|logh|<Ch2|logh|≤max1<j,j∈J∥uj-uhj0∥j.
We can see that only case (b) is possible and by addingCh2|logh|in (301) we get
(304)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
thus(305)11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
comparing this last inequalty (305) with (296), we deduce the two following possibilities
(306)∥u1-uh11∥1<11-ρCh2|logh|
or(307)11-ρCh2|logh|<∥u1-uh11∥1,
which implies(308)∥u1-uh11∥1<11-ρCh2|logh|<Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j
or(309)11-ρCh2|logh|<∥u1-uh11∥1≤Ch2|logh|+max1<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(310)∥u1-uh11∥1=11-ρCh2|logh|.
Similarly applying (73) to subdomain 2 in conjunction with (71) we get
(311)∥u2-uh21∥2≤Ch2|logh|∥u2-uh21∥2≤+max{ρ∥u2-uh21∥2,∥u1-uh11∥1,∥uj-uhj0∥jmax2<j,j∈J∥uj-uhj0∥j}.
So we have to distinguish between the three following cases(312)(1)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=ρ∥u2-uh21∥2
or(313)(2)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=∥u1-uh11∥1
or(314)(3)max{ρ∥u2-uh21∥2,∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j}=max2<j,j∈J∥uj-uhj0∥j.
Case  1 implies(315)∥u2-uh21∥2≤Ch2|logh|+ρ∥u2-uh21∥2,∥u1-uh11∥1≤ρ∥u2-uh21∥2,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2.
Then(316)∥u2-uh21∥2≤11-ρCh2|logh|,11-ρCh2|logh|=∥u1-uh11∥111-ρCh2|logh|≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|,max2<j,j∈J∥uj-uhj0∥j≤ρ∥u2-uh21∥2≤ρ1-ρCh2|logh|max∥ujuhj0∥j<Ch2|logh|.
Equation (316) contradicts (71) so case 1 is impossible. Case  2 implies
(317)∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
By multiplying (317) byρwe get
(318)ρ∥u2-uh21∥2≤ρCh2|logh|+ρ∥u1-uh11∥1,ρ∥u2-uh21∥2≤∥u1-uh11∥1,max2<j,j∈J∥uj-uhj0∥j≤∥u1-uh11∥1.
It is clear thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρ∥u1-uh11∥1and∥u1-uh11∥1so
(319)(a)∥u1-uh11∥1<ρCh2|logh|+ρ∥u1-uh11∥1
or(320)(b)ρCh2|logh|+ρ∥u1-uh11∥1<∥u1-uh11∥1.
That is(321)∥u1-uh11∥1<ρ1-ρCh2|logh|
or(322)ρ1-ρCh2|logh|<∥u1-uh11∥1.
It is clear that only case (b) is true because it coincides with (310) and by addingCh2|logh|in (322) we get
(323)11-ρCh2|logh|=Ch2|logh|+ρ1-ρCh2|logh|11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
comparing this last inequalty with (317), we distingush the two following possibilities
(324)∥u2-uh21∥2<11-ρCh2|logh|
or(325)11-ρCh2|logh|<∥u2-uh21∥2
which imply(326)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+∥u1-uh11∥1
or(327)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+∥u1-uh11∥1.
It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if(328)∥u2-uh21∥2=11-ρCh2|logh|.
Case 3 implies(329)∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
By multiplying (329) byρwe get
(330)ρ∥u2-uh21∥2≤ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j,ρ∥u2-uh21∥2≤max2<j,j∈J∥uj-uhj0∥j,∥u1-uh11∥1≤max2<j,j∈J∥uj-uhj0∥j.
We remark thatρ∥u2-uh21∥2is bounded by bothρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥jandmax2<j,j∈J∥uj-uhj0∥jthen
(331)(c)max2<j,j∈J∥uj-uhj0∥j<ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j
or(332)(d)ρCh2|logh|+ρmax2<j,j∈J∥uj-uhj0∥j<max2<j,j∈J∥uj-uhj0∥j.
That is(333)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|
or(334)ρ1-ρCh2|logh|<max2<j,j∈J∥uj-uhj0∥j.
Which implies(335)max2<j,j∈J∥uj-uhj0∥j<ρ1-ρCh2|logh|<Ch2|logh|
or(336)ρ1-ρCh2|logh|<Ch2|logh|≤max2<j,j∈J∥uj-uhj0∥j.
It is clear that only case (d) is possible. By addingCh2|logh|in (334) we get
(337)Ch2|logh|+ρ1-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
so(338)11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
comparing the last inequality with (329), we consider the two following possibilties
(339)∥u2-uh21∥2<11-ρCh2|logh|
or(340)11-ρCh2|logh|<∥u2-uh21∥2
that is(341)∥u2-uh21∥2<11-ρCh2|logh|<Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j
or(342)11-ρCh2|logh|<∥u2-uh21∥2≤Ch2|logh|+max2<j,j∈J∥uj-uhj0∥j.
It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if(343)∥u2-uh21∥2=11-ρCh2|logh|.
So case 1 is impossible and in both cases 2 and 3 we obtain(344)∥u2-uh21∥2=11-ρCh2|logh|.
Equations (310) and (344) imply
(345)max1≤i<3∥ui-uhi1∥i=11-ρCh2|logh|.
We adopt the same approach for the other subdomainsΩi,i=3,…,min order, we get
(346)∥ui-uhi1∥i=11-ρCh2|logh|,i=1,…,m.
The remainder of the proof is by induction and is by adopting the same approach used in iteration one.



---

*Source: 893182-2013-11-28.xml*