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

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 number 𝑚 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.

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 Section 2 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 𝑎(𝑢,𝑣)=_Ω(∇𝑢⋅∇𝑣)𝑑𝑥; (1) the linear form 𝑓,𝑣=_Ω𝑓(𝑥)⋅𝑣(𝑥)𝑑𝑥; (2) the right hand side 𝑓,aregularfunction; (3) the space 𝑉^(𝑔)=𝑣∈𝐻¹(Ω)suchthat𝑣=𝑔on𝜕Ω, (4) where𝑔is a regular function defined on𝜕Ω.

We consider the linear elliptic equation: Find𝜁∈𝑉^(𝑔)such that 𝑎(𝜁,𝑣)+𝑐(𝜁,𝑣)=𝑓,𝑣,∀𝑣∈𝑉^(𝑔), (5) where𝑐∈ℝ𝑐>0, such that 𝑐≥𝛽>0. (6)

Let𝑉_ℎbe the space of finite elements consisting of continuous piecewise linear functions𝑣vanishing on𝜕Ω, and let𝑠,𝑠=1,2,…,𝑚(ℎ)be the basis function of𝑉_ℎ. The discrete counterpart of (5) consists of finding𝜁_ℎ∈𝑉_ℎ^(𝑔)such that 𝑎𝜁_ℎ,𝑣+𝑐𝜁_ℎ,𝑣=𝑓,𝑣,∀𝑣∈𝑉_ℎ^(𝑔), (7) where 𝑉_ℎ^(𝑔)=𝑣∈𝑉_ℎ∶𝑣=𝜋_ℎ𝑔on𝜕Ω, (8) and𝜋_ℎis 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 𝜁−𝜁_{ℎ𝐿∞(Ω)}≤𝐶ℎ²logℎ. (9)

Lemma 2 (cf. [4]). Let𝑤∈𝐻₁(Ω)∩𝐶(Ω)satisfy𝑎(𝑤,𝜙)+𝑐(𝑤,𝜙)≥0∀nonnegative𝜙∈𝐻¹₀(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);𝜁̃=𝜎(𝑓,𝑔)the corresponding solutions to (5).

Proposition 3 (cf. [6]). Under conditions of the preceding lemma, one has: 𝜁−𝜁̃_𝐿∞(Ω)≤max1𝛽𝑓−𝑓_𝐿∞(Ω),𝑔−𝑔_{𝐿∞(𝜕Ω)}. (10)

Remark 4. Lemma 2 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∈𝑉_ℎsatisfy𝑎(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 and𝜁_ℎ=𝜎_ℎ(𝑓,𝑔);𝜁̃_ℎ=𝜎_ℎ(𝑓,𝑔)the corresponding solutions to (7).

Proposition 6 (cf. [6]). Let the d.m.p hold. Then, under conditions of Lemma 5, one has 𝜁_ℎ−𝜁̃_{ℎ𝐿∞(Ω)}1≤max𝛽𝑓−𝑓_𝐿∞(Ω),𝑔−𝑔_{𝐿∞(𝜕Ω)}. (11)

2.2. The Schwarz Alternating Methods for Nonlinear PDEs

Consider the nonlinear PDE: −Δ𝑢+𝑐𝑢=𝑓(𝑢)inΩ,𝑢=0on𝜕Ω (12) or in its weak form 𝑎(𝑢,𝑣)+𝑐(𝑢,𝑣)=𝑓(𝑢),𝑣∀𝑣∈𝐻¹₀(Ω), (13) where𝑓(⋅)is a nondecreasing nonlinearity. We assume that𝑓(⋅)is a Lipschitz continuous onℝ, that is 𝑓(𝑥)−𝑓𝑦≤𝑘𝑥−𝑦,∀𝑥,𝑦∈ℝ (14) such that 𝑘0<𝛽<1, (15) where𝛽is the constant defined in (6).

We suppose that the problem (12) has a subsolution𝑢and asupersolution𝑢which satisfy0<𝑢<𝑢.Define the sector of smooth functions as follows: 𝐴=𝑢∈𝐶²(Ω),𝑢≤𝑢≤𝑢onΩ. (16) Finally we assume that 𝐹(𝑢)−𝐹(𝑣)<0,𝑣<𝑢∈𝐴, (17) where 𝐹(𝑢)=𝑓(𝑢)𝑢. (18) With these assumptions cf. [12], the problem (12) has a unique solution in the sector𝐴.We can easily see that (12) is equivalent to −Δ𝑢=ℎ(𝑢)inΩ,𝑢=0on𝜕Ω, (19) 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 subdomainsΩ_𝑖such that Ω=^𝑚_𝑖=1Ω_𝑖. (20) We denote by𝜕Ω_𝑖the boundary ofΩ_𝑖and the interior boundaries by Γ_𝑖=𝜕Ω_𝑖∩Ω. (21) We assume that the intersection ofΓ_𝑖andΓ_𝑗,𝑖≠𝑗is empty. Let 𝑉^(𝑤𝑗)_𝑖=𝑣∈𝐶²Ω_𝑖suchthat𝑣=𝑤_𝑗onΓ_𝑖,𝑣=0on𝜕Ω_𝑖∩𝜕Ω. (22) We associate problem (13) with the following system. Find𝑢_𝑖∈∏^𝑚_𝑖=1𝑉^(𝑔𝑖)_𝑖solution of −Δ𝑢_𝑖+𝑐𝑢_𝑖=𝑓𝑢_𝑖inΩ_𝑖, (23) where 𝑢_𝑖=𝑢Ω_𝑖,𝑖=1,…,𝑚 (24) and (cf. [13]) 𝑔_𝑖=𝑢_𝑖−1onΓ_𝑖∩Ω_𝑖−1𝑢_𝑖−2onΓ_𝑖∩Ω_𝑖−2−Ω_𝑖−1⋮𝑢₁onΓ_𝑖∩Ω₁−Ω_𝑖−1−⋯−Ω₁𝑢_𝑚onΓ_𝑖∩Ω_𝑚−⋃^𝑖−1_𝑗=1Ω_𝑗𝑢_𝑚−1onΓ_𝑖∩Ω_𝑚−1−⋃^𝑖−1_𝑗=1Ω_𝑗−Ω_𝑚⋮𝑢_𝑖+1onΓ_𝑖∩Ω_𝑖+1−⋃^𝑖−1_𝑗=1Ω_𝑗−⋃^𝑚_𝑗=𝑖+2Ω_𝑗. (25)

In the sequel, we define a subsolution and a supersolution of (23)

Definition 8. A smooth function𝑢_𝑖is a subsolution of (23) if −Δ𝑢_𝑖+𝑐𝑢_𝑖≤𝑓𝑢_𝑖inΩ_𝑖,𝑢_𝑖=𝑔_𝑖onΓ_𝑖,𝑢_𝑖=0on𝜕Ω∩𝜕Ω_𝑖, (26) where 𝑔_𝑖=𝑢_𝑖−1onΓ_𝑖∩Ω_𝑖−1𝑢_𝑖−2onΓ_𝑖∩Ω_𝑖−2−Ω_𝑖−1⋮𝑢₁onΓ_𝑖∩Ω₁−Ω_𝑖−1−⋯−Ω₁𝑢_𝑚onΓ_𝑖∩Ω_𝑚−⋃^𝑖−1_𝑗=1Ω_𝑗𝑢_𝑚−1onΓ_𝑖∩Ω_𝑚−1−⋃^𝑖−1_𝑗=1Ω_𝑗−Ω_𝑚⋮𝑢_𝑖+1onΓ_𝑖∩Ω_𝑖+1−⋃^𝑖−1_𝑗=1Ω_𝑗−⋃^𝑚_𝑗=𝑖+2Ω_𝑗. (27)
Similarly, a supersolution𝑢_𝑖of (23) satisfies −Δ𝑢_𝑖+𝑐𝑢_𝑖≥𝑓𝑢_𝑖inΩ_𝑖,𝑢_𝑖=𝑔_𝑖onΓ_𝑖,𝑢_𝑖=0on𝜕Ω∩𝜕Ω_𝑖. (28)

2.3. The Continuous Schwarz Sequences

Let𝑢⁰_𝑖be an initialization data: 𝑢⁰_𝑖=𝑢_𝑖,𝑖=1,…,𝑚. (29)

We define the alternating Schwarz sequences(𝑢_𝑖^𝑛+1)onΩ_𝑖,𝑖=1,…,𝑚,such that each term of each sequence𝑢_𝑖^𝑛+1∈𝑉^{(𝑔𝑖𝑛+1)},𝑛≥0solves −Δ𝑢_𝑖^𝑛+1+𝑐𝑢_𝑖^𝑛+1=𝑓𝑢_𝑖^𝑛+1inΩ_𝑖, (30) where (cf. [13]) 𝑔_𝑖^𝑛+1=𝑢^𝑛+1_𝑖−1onΓ_𝑖∩Ω_𝑖−1𝑢^𝑛+1_𝑖−2onΓ_𝑖∩Ω_𝑖−2−Ω_𝑖−1⋮𝑢₁^𝑛+1onΓ_𝑖∩Ω₁−Ω_𝑖−1−⋯−Ω₁𝑢^𝑛_𝑚onΓ_𝑖∩Ω_𝑚−⋃^𝑖−1_𝑗=1Ω_𝑗𝑢^𝑛_𝑚−1onΓ_𝑖∩Ω_𝑚−1−⋃^𝑖−1_𝑗=1Ω_𝑗−Ω_𝑚⋮𝑢^𝑛_𝑖+1onΓ_𝑖∩Ω_𝑖+1−⋃^𝑖−1_𝑗=1Ω_𝑗−⋃^𝑚_𝑗=𝑖+2Ω_𝑗. (31)

Theorem 9. The Schwarz sequences(𝑢_𝑖^𝑛+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,𝑢₁and𝑢₁satisfy −Δ𝑢¹₁+𝑐𝑢¹₁=𝑓𝑢¹₁inΩ₁,𝑢¹₁=0on𝜕Ω₁∩𝜕Ω,−Δ𝑢₁+𝑐𝑢₁≥𝑓𝑢₁inΩ₁,𝑢₁=0on𝜕Ω₁∩𝜕Ω,−Δ𝑢₁+𝑐𝑢₁≤𝑓𝑢₁inΩ₁,𝑢₁=0on𝜕Ω₁∩𝜕Ω (32)(33)(34)(35) respectively with 𝑢¹₁=𝑔¹₁=𝑔₁≤𝑔₁onΓ₁ (36) Lemma 7 implies that 𝑢₁≤𝑢¹₁≤𝑢₁inΩ₁. (37) In subdomain 2, the condition of the interior boundaryΓ₂is given by 𝑔¹₂=𝑢¹₁onΓ₂∩Ω₁𝑢⁰_𝑚onΓ₂∩Ω_𝑚−Ω₁𝑢⁰_𝑚−1onΓ₂∩Ω_𝑚−1−Ω₁−Ω_𝑚⋮𝑢^𝑛₃onΓ₂∩Ω₃−Ω₁−⋃^𝑚_𝑗=4Ω_𝑗. (38) so (37) implies that 𝑔₂≤𝑔¹₂≤𝑔₂onΓ₂. (39) On the other hand𝑢¹₂,𝑢₂, and𝑢₂satisfy, respectively, −Δ𝑢¹₂+𝑐𝑢¹₂=𝑓𝑢¹₂inΩ₂,𝑢¹₂=0on𝜕Ω₂∩𝜕Ω,−Δ𝑢₂+𝑐𝑢₂≥𝑓𝑢₂inΩ₂,𝑢₂=0on𝜕Ω₂∩𝜕Ω,−Δ𝑢₂+𝑐𝑢₂≤𝑓𝑢₂inΩ₂,𝑢₂=on𝜕Ω₂∩𝜕Ω. (40) Then Lemma 7 implies that 𝑢₂≤𝑢¹₂≤𝑢₂inΩ₂. (41) The same idea is used sequentially in the remainder of the proof related to the rest of subdomainsΩ_𝑖,3≤𝑖≤𝑚. Indeed we obtain for𝑖=1,…,𝑚and for𝑛=0𝑢_𝑖≤𝑢¹_𝑖≤𝑢_𝑖inΩ_𝑖. (42) Now, let us assume that 𝑢_𝑖≤𝑢^𝑛_𝑖≤𝑢_𝑖inΩ_𝑖,𝑖=1,…,𝑚. (43) and prove that 𝑢_𝑖≤𝑢_𝑖^𝑛+1≤𝑢_𝑖inΩ_𝑖,𝑖=1,…,𝑚. (44) Indeed, in subdomain 1 (43) implies that 𝑔₁≤𝑔₁^𝑛+1≤𝑔₁onΓ₁. (45)
Since −Δ𝑢₁^𝑛+1+𝑐𝑢₁^𝑛+1=𝑓𝑢₁^𝑛+1inΩ₁,𝑢₁^𝑛+1=0on𝜕Ω₁∩𝜕Ω,−Δ𝑢₁+𝑐𝑢₁≥𝑓𝑢₁inΩ₁,𝑢₁=0on𝜕Ω₁∩𝜕Ω,−Δ𝑢₁+𝑐𝑢₁≤𝑓𝑢₁inΩ₁,𝑢₁=0on𝜕Ω₁∩𝜕Ω, (46) then Lemma 7 implies that 𝑢₁≤𝑢₁^𝑛+1≤𝑢₁inΩ₁. (47) Using (47) and (43), we can write 𝑔₂≤𝑔₂^𝑛+1≤𝑔₂onΓ₂. (48) We can easily obtain 𝑢₂≤𝑢₂^𝑛+1≤𝑢₂inΩ₂ (49) using (48) and by adopting the same approach used in subdomain one. The result related to rest of subdomainsΩ_𝑖,3≤𝑖≤𝑚, is obtained sequentially by similar way; that is 𝑢_𝑖≤𝑢_𝑖^𝑛+1≤𝑢_𝑖inΩ_𝑖,𝑖=1,…,𝑚,𝑛≥0. (50)
(2)We demonstrate, by induction, that the Schwarz sequences are nondecreasing. Indeed for𝑛=0,we have demonstrated in (50) that, for𝑖=1,…,𝑚, 𝑢⁰_𝑖≤𝑢¹_𝑖inΩ_𝑖 (51)
Now, let us assume that 𝑢_𝑖^𝑛−1≤𝑢^𝑛_𝑖inΩ_𝑖,𝑖=1,…,𝑚 (52) and prove that 𝑢^𝑛_𝑖≤𝑢_𝑖^𝑛+1inΩ_𝑖,𝑖=1,…,𝑚. (53)
Begining by subdomain one, (52) implies that 𝑔^𝑛₁≤𝑔₁^𝑛+1onΓ₁. (54) Making use of (54), −Δ𝑢₁^𝑛+1+𝑐𝑢₁^𝑛+1=𝑓𝑢₁^𝑛+1inΩ₁,𝑢₁^𝑛+1=0on𝜕Ω₁∩𝜕Ω,−Δ𝑢^𝑛₁+𝑐𝑢^𝑛₁=𝑓𝑢^𝑛₁inΩ₁,𝑢^𝑛₁=0on𝜕Ω₁∩𝜕Ω. (55) We get by using Lemma 7𝑢^𝑛₁≤𝑢₁^𝑛+1inΩ₁. (56) On the other hand, (52) and (56) imply 𝑔^𝑛₂≤𝑔₂^𝑛+1onΓ₂; (57) then making use of (57), −Δ𝑢₂^𝑛+1+𝑐𝑢₂^𝑛+1=𝑓𝑢₂^𝑛+1inΩ₂,𝑢₂^𝑛+1=0on𝜕Ω₂∩𝜕Ω,−Δ𝑢^𝑛₂+𝑐𝑢^𝑛₂=𝑓𝑢^𝑛₂inΩ₂,𝑢^𝑛₂=0on𝜕Ω₂∩𝜕Ω. (58) We get by using Lemma 7 the following result related to the second subdomain: 𝑢^𝑛₂≤𝑢₂^𝑛+1inΩ₂. (59) The same idea is sequentially applied to the rest of subdomains. Finally, we obtain 𝑢^𝑛_𝑖≤𝑢_𝑖^𝑛+1inΩ_𝑖,𝑖=1,…,𝑚,𝑛≥0. (60)
(3)The sequences(𝑢^𝑛_𝑖)are bounded and nondecreasing, so they are monotone converging pointwise to𝑢_𝑖,𝑖=1,…,𝑚. By using an elliptic regularity argument, we can see that functions𝑢_𝑖satisfy the same PDE onΩ_𝑖cf. [3, 12].

2.4. The Discretization

For𝑖=1,…,𝑚,let𝜏^ℎ𝑖be a standard regular and quasiuniform finite element triangulation inΩ_𝑖;ℎ_𝑖being the meshsize. The𝑚meshes is mutually independent onΩ_𝑖∩Ω_𝑗,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: 𝑉_ℎ𝑖=𝑣∈𝐶Ω_𝑖∩𝐻¹Ω_𝑖𝑣suchthat𝐾∈𝑃₁∀𝐾∈𝜏^ℎ𝑖, (61) and for every𝑤∈𝐶(Γ_𝑖), we set 𝑉_ℎ^(𝑤)_𝑖=𝑣∈𝑉_ℎ𝑖∶𝑣=0on𝜕Ω_𝑖∩𝜕Ω;𝑣=𝜋_ℎ𝑖(𝑤)onΓ_𝑖, (62) where𝜋_ℎ𝑖denotes the interpolation operator onΓ_𝑖.

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 meshesℎ_𝑖are 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, let𝑢⁰_ℎbe the discrete analog of𝑢₀, defined in (29) and we define for𝑖=1,…,𝑚,the discrete sequences(𝑢_ℎ^𝑛+1_𝑖)onΩ_𝑖such that𝑢_ℎ^𝑛+1_𝑖∈𝑉^{(𝑔ℎ𝑖𝑛+1)}_ℎ𝑖solves 𝑎_𝑖𝑢_ℎ^𝑛+1_𝑖,𝑣+𝑐𝑢_ℎ^𝑛+1_𝑖,𝑣=𝑓𝑢_ℎ^𝑛+1_𝑖,𝑣,∀𝑣∈𝑉^{(𝑔ℎ𝑖𝑛+1)}_ℎ𝑖;𝑛≥0, (63) where 𝑎_𝑖(𝑢,𝑣)=_Ω𝑖()𝑔∇𝑢⋅∇𝑣𝑑𝑥,_{ℎ𝑛+1𝑖}=𝑢_ℎ^𝑛+1_𝑖−1onΓ_𝑖∩Ω_𝑖−1𝑢_ℎ^𝑛+1_𝑖−2onΓ_𝑖∩Ω_𝑖−2−Ω_𝑖−1⋮𝑢_ℎ^𝑛+1₁onΓ_𝑖∩Ω₁−Ω_𝑖−1−⋯−Ω₁𝑢^𝑛_ℎ𝑚onΓ_𝑖∩Ω_𝑚−⋃^𝑖−1_𝑗=1Ω_𝑗𝑢^𝑛_ℎ𝑚−1onΓ_𝑖∩Ω_𝑚−1−⋃^𝑖−1_𝑗=1Ω_𝑗−Ω_𝑚⋮𝑢^𝑛_ℎ𝑖+1onΓ_𝑖∩Ω_𝑖+1−⋃^𝑖−1_𝑗=1Ω_𝑗−⋃^𝑚_𝑗=𝑖+1Ω_𝑗. (64)

Notation. From now on, we will adopt the following notations𝑖=1,…,𝑚: |⋅|_𝑖=‖⋅‖_𝐿^∞_(Γ𝑖)‖⋅‖_𝑖=‖⋅‖_𝐿^∞_(Ω𝑖),𝜋_ℎ𝑖=𝜋_ℎ. (65)

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,𝑤_ℎ𝑖∈𝑉^(𝑔𝑖)_ℎ𝑖solves 𝑎_𝑖𝑤_ℎ𝑖,𝑣+𝑐𝑤_ℎ𝑖,𝑣=𝑓𝑢_𝑖,𝑣∀𝑣∈𝑉^(𝑔𝑖)_ℎ𝑖. (66) It is then clear that𝑤_ℎ𝑖is the finite element approximation of𝑢_𝑖solutions of (23). Therefore, making use of standard maximum norm estimates for linear elliptic problems, we have 𝑢_𝑖−𝑤_ℎ𝑖𝑖≤𝐶ℎ²logℎ, (67) 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 𝑢_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤11−𝜌𝐶ℎ²logℎ;𝑛≥0;𝑖=1,…,𝑚. (68)

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 (A)max_{1≤𝑖≤𝑚}𝑢_𝑖−𝑢⁰_ℎ𝑖𝑖<𝐶ℎ²,(B)logℎ𝐶ℎ²logℎ<𝑢_𝑖−𝑢⁰_ℎ𝑖𝑖,𝑖=1,…,𝑚, (69)(70) and in the last situation we assume that a nonempty subset𝐽of the set𝑀={1,…,𝑚}exists such that (C)∀𝑖∈𝑀−𝐽;𝑢_𝑖−𝑢⁰_ℎ𝑖𝑖<𝐶ℎ²logℎ<𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,∀𝑗∈𝐽. (71) To this end, we apply Proposition 6 and Theorem 1 for each subdomainΩ_𝑖as follows: 𝑢_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤𝑢_𝑖−𝑤_ℎ𝑖𝑖+𝑤_ℎ𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤𝐶ℎ²logℎ+𝑤_ℎ𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤𝐶ℎ²1logℎ+max𝛽𝑓𝑢_𝑖−𝑓𝑢_ℎ^𝑛+1_𝑖𝑖,𝑔_𝑖−𝑔_ℎ^𝑛+1_𝑖𝑖≤𝐶ℎ²1logℎ+max𝛽𝑓𝑢_𝑖−𝑓𝑢_ℎ^𝑛+1_𝑖𝑖,max_𝑗<𝑖𝑢_𝑗−𝑢_ℎ^𝑛+1_𝑗𝑗,max_𝑖<𝑗𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (72) The last inequality implies that 𝑢_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤𝐶ℎ²𝜌𝑢logℎ+max_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖,max_𝑗<𝑖𝑢_𝑗−𝑢_ℎ^𝑛+1_𝑗𝑗,max_𝑖<𝑗𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (73) The interval]0,1[which𝜌belongs to is divided into two subintervals as follows: ][=10,10,2∪12,1, (74) 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 𝜌1−𝜌<1. (75) For𝑛=0,applying (73) to subdomain one, we get 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (76) We have to distinguish between two cases: (1)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₁−𝑢¹_ℎ11 (77) or (2)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (78) Case  1 implies that 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11. (79) Then, 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (80) Case  2 implies that 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (81) By multiplying (81) by𝜌,we get 𝜌𝑢₁−𝑢¹_ℎ11≤𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (82) So𝜌‖𝑢₁−𝑢¹_ℎ1‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗,then (a)max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (83) or (b)𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (84) That is, max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (85) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗, (86) which implies that max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ, (87) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝐶ℎ²logℎ. (88) 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 max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (89) Then case 2 implies that 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (90) Hence in both cases 1 and 2, we obtain 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ. (91) Similarly applying (73) to the second subdomain, we get 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²𝜌𝑢logℎ+max₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (92) We have to distinguish the three following cases (1)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₂−𝑢¹_ℎ22 (93) or (2)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝑢₁−𝑢¹_ℎ11 (94) or (3)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (95) Case  1  implies that 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22. (96) Then, 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²,𝑢logℎ₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ21≤𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ21≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (97) We note that the two last inequalities coincide with (91) and (69), respectively. Case  2 implies that 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (98) By multiplying (98) by𝜌, we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (99) We remark that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌‖𝑢₁−𝑢¹_ℎ1‖₁and‖𝑢₁−𝑢¹_ℎ1‖₁then (a)𝑢₁−𝑢¹_ℎ11<𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11 (100) or (b)𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11<𝑢₁−𝑢¹_ℎ11. (101) Which implies that 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²logℎ (102) or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11. (103) That is 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (104) or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ. (105) 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 𝑢₁−𝑢¹_ℎ11=𝜌1−𝜌𝐶ℎ²logℎ. (106) Then case 2 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (107) With max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11=𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ (108) which coincides with (69). Case  3 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (109) By multiplying (109) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (110) So𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{2<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_{2<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗then w have (c)max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (111) or (d)𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (112) Which implies max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (113) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (114)
That is, max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ (115) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝐶ℎ²logℎ. (116) 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 max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (117) Then case 3 implies that 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ, (118) with 𝑢₁−𝑢¹_ℎ11≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (119) Hence, in the three cases 1, 2, and 3 we get 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²logℎ. (120) Equations (91) and (120) imply max_1≤𝑖<3𝑢_𝑖−𝑢¹_ℎ𝑖𝑖≤11−𝜌𝐶ℎ²logℎ. (121) Similarly applying the same idea for the rest of subdomainΩ_𝑖,𝑖=3,…,𝑚, orderly, we get 𝑢_𝑖−𝑢¹_ℎ𝑖𝑖≤11−𝜌𝐶ℎ²logℎ. (122) Now, let us assume that 𝑢_𝑖−𝑢^𝑛_ℎ𝑖𝑖≤11−𝜌𝐶ℎ²logℎ,𝑖=1,…,𝑚 (123) and prove that 𝑢_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤11−𝜌𝐶ℎ²logℎ,𝑖=1,…,𝑚. (124)
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 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢_ℎ^𝑛+1₁₁,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗 (125)
We have to distinguish the two following cases (1)max𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁ (126) or (2)max𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (127) Case  1 implies 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝐶ℎ²logℎ+𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁. (128) Then 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤11−𝜌𝐶ℎ²logℎ,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ. (129) Case  2 implies 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (130) By multiplying (130) by𝜌we get 𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (131) So𝜌‖𝑢₁−𝑢_ℎ^𝑛+1₁‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢^𝑛_ℎ𝑗‖_𝑗andmax_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢^𝑛_ℎ𝑗‖_𝑗then (a)max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗 (132) or (b)𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (133) That is max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (134) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (135) Which implies max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (136) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<11−𝜌𝐶ℎ²logℎ. (137) 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 max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (138) Then case 2 implies 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (139) Hence in both cases 1 and 2 we obtain 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤11−𝜌𝐶ℎ²logℎ. (140) Similarly applying (73) to the second subdomain, for iteration𝑛+1,we get 𝑢₁−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²𝜌𝑢logℎ+max₂−𝑢_ℎ^𝑛+1₂₂,𝑢₁−𝑢_ℎ^𝑛+₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (141) We have to distinguish the three following cases (1)max𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂,𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂ (142) or (2)max𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂,𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝑢₁−𝑢_ℎ^𝑛+1₁₁ (143) or (3)max𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂,𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (144) Case  1 implies 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²logℎ+𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂,𝑢₁−𝑢_ℎ^𝑛+1₁₁≤𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂. (145) Then 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤11−𝜌𝐶ℎ²,𝑢logℎ₁−𝑢_ℎ^𝑛+1₁₁≤𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ. (146) We note that the two last inequalities coincide with (140) and (123) respectively. Case  2 implies that 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²logℎ+𝑢₁−𝑢_ℎ^𝑛+1₁₁,𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝑢₁−𝑢_ℎ^𝑛+1₁₁. (147) By multiplying (147) by𝜌we get 𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁,𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝑢₁−𝑢_ℎ^𝑛+1₁₁,max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝑢₁−𝑢_ℎ^𝑛+1₁₁. (148) We remark that𝜌‖𝑢₂−𝑢_ℎ^𝑛+1₂‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌‖𝑢₁−𝑢_ℎ^𝑛+1₁‖₁and‖𝑢₁−𝑢_ℎ^𝑛+1₁‖₁then (a)𝑢₁−𝑢_ℎ^𝑛+1₁₁<𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁ (149) or (b)𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢_ℎ^𝑛+1₁₁<𝑢₁−𝑢_ℎ^𝑛+1₁₁. (150) Which implies 𝑢₁−𝑢_ℎ^𝑛+1₁₁<𝜌1−𝜌𝐶ℎ²logℎ (151) or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢_ℎ^𝑛+1₁₁. (152) That is 𝑢₁−𝑢_ℎ^𝑛+1₁₁<𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (153) or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢_ℎ^𝑛+1₁₁≤11−𝜌𝐶ℎ²logℎ. (154)
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 𝑢₁−𝑢_ℎ^𝑛+1₁₁=𝜌1−𝜌𝐶ℎ²logℎ. (155) Case  2 implies that 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²logℎ+𝑢₁−𝑢_ℎ^𝑛+1₁₁=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (156) With max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤𝑢₁−𝑢_ℎ^𝑛+1₁₁=𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (157) which coincides with (123). Case  3 implies 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²logℎ+max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝑢₁−𝑢_ℎ^𝑛+1₁₁≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (158) By multiplying (158) by𝜌we get 𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝜌𝑢₂−𝑢_ℎ^𝑛+1₂₂≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗,𝑢₁−𝑢_ℎ^𝑛+1₁₁≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (159) So𝜌‖𝑢₂−𝑢_ℎ^𝑛+1₂‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{2<𝑗≤𝑚}‖𝑢_𝑗−𝑢^𝑛_ℎ𝑗‖_𝑗andmax_{2<𝑗≤𝑚}‖𝑢_𝑗−𝑢^𝑛_ℎ𝑗‖_𝑗then (c)max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗 (160) or (d)𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (161) Which implies max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (162) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗. (163) That is max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (164) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗≤11−𝜌𝐶ℎ²logℎ. (165) 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 max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (166) Then case 3 implies 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤𝐶ℎ²logℎ+max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (167) With 𝑢₁−𝑢_ℎ^𝑛+1₁₁≤max_{2<𝑗≤𝑚}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (168) Hence in three cases 1, 2, and 3 we get 𝑢₂−𝑢_ℎ^𝑛+1₂₂≤11−𝜌𝐶ℎ²logℎ. (169) Equations (140) and (169) imply max_1≤𝑗<3𝑢_𝑗−𝑢_ℎ^𝑛+1_𝑗𝑗≤11−𝜌𝐶ℎ²logℎ. (170) Similarly applying the same idea for the rest of subdomainsΩ_𝑖,𝑖=3,…,𝑚,orderly we get 𝑢_𝑖−𝑢_ℎ^𝑛+1_𝑖𝑖≤11−𝜌𝐶ℎ²logℎ (171) which is the desired result. In the sequel we deal with the second subinterval[1/2,1[,indeed𝜌∈[1/2,1[so 𝜌1−𝜌≥1. (172) For𝑛=0,applying (73) to subdomain one we get 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (173) We have to distinguish the two following cases (1)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₁−𝑢¹_ℎ11 (174) or (2)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (175) Case  1 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11. (176) Then 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ,max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11≤𝜌1−𝜌𝐶ℎ²logℎ. (177) Case  2 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (178) By multiplying (178) by𝜌we get 𝜌𝑢₁−𝑢¹_ℎ11≤𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (179) So𝜌‖𝑢₁−𝑢¹_ℎ1‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_{1<𝑗≤𝑚}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗then (a)max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (180) or (b)𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (181) That is max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (182) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (183) Which implies max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎj𝑗<𝐶ℎ²≤𝜌logℎ1−𝜌𝐶ℎ²logℎ (184) or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝐶ℎ²logℎ. (185) 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 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗≤𝑚}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (186) Hence in both cases 1 and 2 we get 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ. (187) By the same way we can obtain orderly 𝑢_𝑖−𝑢¹_ℎ𝑖𝑖≤11−𝜌𝐶ℎ²logℎ,𝑖=2,…,𝑚. (188) 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, 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (189) We have to distinguish the two following cases (1)max𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₁−𝑢¹_ℎ11 (190) or (2)max𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (191) Case  1 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11. (192) So 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (193) We remark that (193) contradicts (70) so case 1 is impossible. Case  2 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (194) By multiplying (194) by𝜌we get 𝜌𝑢₁−𝑢¹_ℎ11≤𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (195) It is clear that𝜌‖𝑢₁−𝑢¹_ℎ1‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_1<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_1<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗, so (a)max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (196) or (b)𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (197) Which implies max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (198) or 𝜌1−𝜌𝐶ℎ²logℎ<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (199) Then 𝐶ℎ²logℎ≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (200) or 𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (201) We can see that only case (b) is possible and by adding𝐶ℎ²|logℎ|in (199) we get 𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (202) thus 11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (203) comparing this last inequalty (203) with (194), we deduce the two following possibilities 𝑢₁−𝑢¹_ℎ11<11−𝜌𝐶ℎ²logℎ (204) or 11−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11 (205) which implies 𝑢₁−𝑢¹_ℎ11<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (206) or 11−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (207) 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 𝑢₁−𝑢¹_ℎ11=11−𝜌𝐶ℎ²logℎ. (208) Similarly applying (73) to subdomain 2 we get 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²𝜌𝑢logℎ+max₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (209) So we have to distinguish between the three following cases (1)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₂−𝑢¹_ℎ22 (210) or (2)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝑢₁−𝑢¹_ℎ11 (211) or (3)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (212) Case  1 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22. (213) Then 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²,1logℎ1−𝜌𝐶ℎ²logℎ=𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²logℎ,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (214) Equations (214) contradicts (70) so case 1 is impossible. Case  2 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (215) By multiplying (215) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (216) It is clear that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌‖𝑢₁−𝑢¹_ℎ1‖₁and‖𝑢₁−𝑢¹_ℎ1‖₁so (a)𝑢₁−𝑢¹_ℎ11<𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11 (217) or (b)𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11<𝑢₁−𝑢¹_ℎ11. (218) That is 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²logℎ (219) or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11. (220) It is clear that only case (b) is true because it coincides with (208) and by adding𝐶ℎ²|logℎ|in (220) we get 11−𝜌𝐶ℎ²logℎ=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11 (221) comparing this last inequalty with (215), we distingush the two following possibilities 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ (222) or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22 (223) which imply 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11 (224) or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11. (225) 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 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (226) Case  3 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (227) By multiplying (227) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (228) We remark that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_2<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_2<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗then (c)max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (229) or (d)𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (230) That is max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (231) or 𝜌1−𝜌𝐶ℎ²logℎ<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (232) Which implies max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ (233) or 𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (234) It is clear that only case (d) is possible. By adding𝐶ℎ²|logℎ|in (232) we get 𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (235) so 11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (236) comparing the last inequality with (227), we consider the two following possibilties 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ (237) or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22 (238)
that is 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (239)
or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (240)
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 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (241)
So case 1 is impossible and in both cases 2 and 3 we obtain 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (242)
Equations (208) and (242) imply max_1≤𝑖<3𝑢_𝑖−𝑢¹_ℎ𝑖𝑖=11−𝜌𝐶ℎ²logℎ. (243)
We adopt the same approach for the other subdomainsΩ_𝑖,𝑖=3,…,𝑚in order, we get 𝑢_𝑖−𝑢¹_ℎ𝑖𝑖=11−𝜌𝐶ℎ²logℎ,𝑖=1,…,𝑚. (244)
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 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (245)
We have to distinguish the two following cases (1)max𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₁−𝑢¹_ℎ11 (246)
or (2)max𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (247)
Case  1 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11. (248)
So 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ,𝐶ℎ²logℎ<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11≤𝜌1−𝜌𝐶ℎ²logℎ. (249)
We remark that (249) coincides with (70). Case  2 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (250)
By multiplying (250) by𝜌we get 𝜌𝑢₁−𝑢¹_ℎ11≤𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (251)
It is clear that𝜌‖𝑢₁−𝑢¹_ℎ1‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_1<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_1<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗, so (a)max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (252)
or (b)𝜌𝐶ℎ²logℎ+𝜌max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (253)
Which implies max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (254)
or 𝜌1−𝜌𝐶ℎ²logℎ<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (255)
Then 𝐶ℎ²logℎ<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (256)
or 𝐶ℎ²≤𝜌logℎ1−𝜌𝐶ℎ²logℎ<max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (257)
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 max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (258)
Then case 2 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_1<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²=1logℎ1−𝜌𝐶ℎ²logℎ. (259)
So in both cases 1 and 2 we get 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ. (260)
Similarly applying (73) to subdomain 2 we get in this situation (B) 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²𝜌𝑢logℎ+max₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (261)
So we have to distinguish between the three following cases (1)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₂−𝑢¹_ℎ22 (262)
or (2)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝑢₁−𝑢¹_ℎ11 (263)
or (3)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (264)
Case  1 implies 𝑢₁−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22. (265)
Then 𝑢₁−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²,𝑢logℎ₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²≤1logℎ1−𝜌𝐶ℎ²logℎ,𝐶ℎ²logℎ<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²logℎ. (266)
Case  2 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (267)
By multiplying (267) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (268)
It is clear that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌‖𝑢₁−𝑢¹_ℎ1‖₁and‖𝑢₁−𝑢¹_ℎ1‖₁so (a)𝑢₁−𝑢¹_ℎ11<𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11 (269)
or (b)𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11<𝑢₁−𝑢¹_ℎ11. (270)
That is 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²logℎ (271)
or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11. (272)
Thus 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²<1logℎ1−𝜌𝐶ℎ²logℎ (273)
or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ. (274)
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 𝑢₁−𝑢¹_ℎ11=𝜌1−𝜌𝐶ℎ²logℎ. (275)
With 𝐶ℎ²logℎ<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11=𝜌1−𝜌𝐶ℎ²logℎ, (276)
So case  2 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ (277)
Then 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²logℎ. (278)
Case  3 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (279)
By multiplying (279) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (280)
We remark that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_2<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_2<𝑗‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗then (c)max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (281)
or (d)𝜌𝐶ℎ²logℎ+𝜌max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (282)
That is max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (283)
or 𝜌1−𝜌𝐶ℎ²logℎ<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (284)
Which implies 𝐶ℎ²logℎ≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (285)
or 𝐶ℎ²≤𝜌logℎ1−𝜌𝐶ℎ²logℎ<max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (286)
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. max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ (287)
Then in case 3 we obtain 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ (288)
Thus 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²logℎ. (289)
With 𝑢₁−𝑢¹_ℎ11≤max_2<𝑗𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌1−𝜌𝐶ℎ²logℎ. (290)
We adopt the same approach for the other subdomainsΩ_𝑖,𝑖=2,…,𝑚in order, we get 𝑢_𝑖−𝑢¹_ℎ𝑖𝑖≤11−𝜌𝐶ℎ²logℎ,𝑖=1,…,𝑚. (291)
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 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²𝜌𝑢logℎ+max₁−𝑢_ℎ^𝑛+1₁₁,max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢^𝑛_ℎ𝑗𝑗 (292)
We have to distinguish the two following cases (1)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₁−𝑢¹_ℎ11 (293)
or (2)max𝜌𝑢₁−𝑢¹_ℎ11,max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (294)
Case  1 implies 𝑢₁−𝑢¹_ℎ11≤11−𝜌𝐶ℎ²logℎ,max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₁−𝑢¹_ℎ11≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (295)
We remark that (295) contradicts (71) so case 1 is impossible. Case  2 implies 𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (296)
By multiplying (296) by𝜌we get 𝜌𝑢₁−𝑢¹_ℎ11≤𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₁−𝑢¹_ℎ11≤max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (297)
It is clear that𝜌‖𝑢₁−𝑢¹_ℎ1‖₁is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{1<𝑗,𝑗∈𝐽}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_{1<𝑗,𝑗∈𝐽}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗, so (a)max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (298)
or (b)𝜌𝐶ℎ²logℎ+𝜌max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (299)
Which implies max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (300)
or 𝜌1−𝜌𝐶ℎ²logℎ<max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (301)
Then 𝐶ℎ²logℎ≤max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (302)
or 𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ≤max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (303)
We can see that only case (b) is possible and by adding𝐶ℎ²|logℎ|in (301) we get 𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (304)
thus 11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (305)
comparing this last inequalty (305) with (296), we deduce the two following possibilities 𝑢₁−𝑢¹_ℎ11<11−𝜌𝐶ℎ²logℎ (306)
or 11−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11, (307)
which implies 𝑢₁−𝑢¹_ℎ11<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (308)
or 11−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11≤𝐶ℎ²logℎ+max_{1<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (309)
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 𝑢₁−𝑢¹_ℎ11=11−𝜌𝐶ℎ²logℎ. (310)
Similarly applying (73) to subdomain 2 in conjunction with (71) we get 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²𝜌𝑢logℎ+max₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (311)
So we have to distinguish between the three following cases (1)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝜌𝑢₂−𝑢¹_ℎ22 (312)
or (2)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=𝑢₁−𝑢¹_ℎ11 (313)
or (3)max𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗=max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (314)
Case  1 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝜌𝑢₂−𝑢¹_ℎ22,𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22. (315)
Then 𝑢₂−𝑢¹_ℎ22≤11−𝜌𝐶ℎ²,1logℎ1−𝜌𝐶ℎ²logℎ=𝑢₁−𝑢¹_ℎ11≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²logℎ,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝜌𝑢₂−𝑢¹_ℎ22≤𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ. (316)
Equation (316) contradicts (71) so case 1 is impossible. Case  2 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (317)
By multiplying (317) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11,𝜌𝑢₂−𝑢¹_ℎ22≤𝑢₁−𝑢¹_ℎ11,max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗≤𝑢₁−𝑢¹_ℎ11. (318)
It is clear that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌‖𝑢₁−𝑢¹_ℎ1‖₁and‖𝑢₁−𝑢¹_ℎ1‖₁so (a)𝑢₁−𝑢¹_ℎ11<𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11 (319)
or (b)𝜌𝐶ℎ²logℎ+𝜌𝑢₁−𝑢¹_ℎ11<𝑢₁−𝑢¹_ℎ11. (320)
That is 𝑢₁−𝑢¹_ℎ11<𝜌1−𝜌𝐶ℎ²logℎ (321)
or 𝜌1−𝜌𝐶ℎ²logℎ<𝑢₁−𝑢¹_ℎ11. (322)
It is clear that only case (b) is true because it coincides with (310) and by adding𝐶ℎ²|logℎ|in (322) we get 11−𝜌𝐶ℎ²logℎ=𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11 (323)
comparing this last inequalty with (317), we distingush the two following possibilities 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ (324)
or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22 (325)
which imply 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11 (326)
or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+𝑢₁−𝑢¹_ℎ11. (327)
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 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (328)
Case 3 implies 𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (329)
By multiplying (329) by𝜌we get 𝜌𝑢₂−𝑢¹_ℎ22≤𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝜌𝑢₂−𝑢¹_ℎ22≤max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗,𝑢₁−𝑢¹_ℎ11≤max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (330)
We remark that𝜌‖𝑢₂−𝑢¹_ℎ2‖₂is bounded by both𝜌𝐶ℎ²|logℎ|+𝜌max_{2<𝑗,𝑗∈𝐽}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗andmax_{2<𝑗,𝑗∈𝐽}‖𝑢_𝑗−𝑢⁰_ℎ𝑗‖_𝑗then (c)max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (331)
or (d)𝜌𝐶ℎ²logℎ+𝜌max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (332)
That is max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ (333)
or 𝜌1−𝜌𝐶ℎ²logℎ<max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (334)
Which implies max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗<𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ (335)
or 𝜌1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ≤max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (336)
It is clear that only case (d) is possible. By adding𝐶ℎ²|logℎ|in (334) we get 𝐶ℎ²+𝜌logℎ1−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (337)
so 11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (338)
comparing the last inequality with (329), we consider the two following possibilties 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ (339)
or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22 (340)
that is 𝑢₂−𝑢¹_ℎ22<11−𝜌𝐶ℎ²logℎ<𝐶ℎ²logℎ+max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗 (341)
or 11−𝜌𝐶ℎ²logℎ<𝑢₂−𝑢¹_ℎ22≤𝐶ℎ²logℎ+max_{2<𝑗,𝑗∈𝐽}𝑢_𝑗−𝑢⁰_ℎ𝑗𝑗. (342)
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 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (343)
So case 1 is impossible and in both cases 2 and 3 we obtain 𝑢₂−𝑢¹_ℎ22=11−𝜌𝐶ℎ²logℎ. (344)
Equations (310) and (344) imply max_1≤𝑖<3𝑢_𝑖−𝑢¹_ℎ𝑖𝑖=11−𝜌𝐶ℎ²logℎ. (345)
We adopt the same approach for the other subdomainsΩ_𝑖,𝑖=3,…,𝑚in order, we get 𝑢_𝑖−𝑢¹_ℎ𝑖𝑖=11−𝜌𝐶ℎ²logℎ,𝑖=1,…,𝑚. (346)
The remainder of the proof is by induction and is by adopting the same approach used in iteration one.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.