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On the convergence analysis of a time dependent elliptic equation with discontinuous coefficients

  • Mohamed Abdelwahed EMAIL logo and Nejmeddine Chorfi
Published/Copyright: October 16, 2019

Abstract

In this paper, we consider a heat equation with diffusion coefficient that varies depending on the heterogeneity of the domain. We propose a spectral elements discretization of this problem with the mortar domain decomposition method on the space variable and Euler’s implicit scheme with respect to the time. The convergence analysis and an optimal error estimates are proved.

MSC 2010: 35J57; 65M70

1 Introduction

This paper is devoted to the numerical analysis of the mortar spectral element discretization of the heat equation in an heterogenous medium with a variable diffusion coefficient λ formulated by the problem (1),

utdiv(λgradu)=f in Ω×]0,T[u=0 in Ω×]0,T[u(.,0)=u0on Ω. (1)

The connected two-dimentional domain Ω is open and bounded with a Lipschtiz-continuous boundary ∂Ω. Let T be a fixed positive real. We suppose that the function λ is positive and does not depend on time.

This problem was handled in some previous works on different cases [1]. The case where the function λ is not globally continuous is presented in [2, 3, 4]. The a priori and a posteriori analysis were proposed based on the finite element method and the spectral discretization. For the case where λ is piecewise constant and such that the ratio of its maximal value to its minimal value is large enough, the discretization of the stationary problem is studied in [5] by conforming finite elements and in [6] by the mortar spectral discretization. In the present work, we consider a non stationary problem with λ is piecewise constant. We proceed to the domain decomposition in two steps. Firstly, We associate a decomposition based on the value of λ (i.e. λ is constant on each sub-domain). Secondly, each obtained sub-domain is itself decomposed on rectangles using the mortar spectral method. The later is considered as the most suitable method for handling nonconforming decomposition (i.e the intersection of two sub-domains is not restricted to be a corner or a whole edge of both of them) [7]. The number of sub-domains can be highly reduced thanks to the non-conformity property. We refer to [8] for a first application of this method to discontinuous coefficient in the finite element method.

The discretization in time of our problem is based on the implicit Euler method. We prove that the semi-discrete problem on time is well posed and we give a time error estimate of order one. On each sub-domain, we consider a spectral discretization which approaches the solution by high degree polynomials. Since the basis of polynomials are tensorized, the sub-domains are rectangles. Different degrees of polynomials are chosen on each sub-domain according to the different values of λ. We prove that the discrete problem is well posed and we show an optimal error estimate for a good choice of domain decomposition. An outline of the paper is as follows:

  • In section 2 we present the continuous problem and some regularity results.

  • The section 3 is about the analysis and the error estimate of the semi-discrete problem on time.

  • The mortar spectral element discretization is developed in section 4.

  • In section 5, we perform the estimation of the error.

  • Section 6 is an annex of the proof of the error estimation since it is quite technical.

2 The continuous problem

We denote by x = (x, y) the generic point in ℝ2, and we suppose that there exists a finite number of sub-domain Ωi , 1 ≤ iI such that:

  1. Ω¯=i=1IΩ¯i,ΩiΩj=,1i<jI,
  2. the restriction of λ to each Ω¯i is continuous on Ωi , 1 ≤ iI,

  3. λ is bounded on each Ω¯i , let

    λimax=supxΩiλ(x)andλimin=minxΩiλ(x).

    We define

    λmax=max1iIλimax,andλmin=max1iIλimin. (2)

Let Hs(Ω), s > 0, the Sobolev spaces associated with the norm ∥ . ∥s,Ω and the semi-norm | . |s,Ω. The space H01 (Ω) stands for the closure in H1(Ω) of the space of infinitely differentiable functions with compact support in Ω and H–1(Ω) is its dual space. The scalar product and its associate norm on the space L2(Ω) are denoted by (., .) and ∥ . ∥0,Ω. H12 (∂Ω) is the space of trace of functions in H1(Ω). Let y∂Ω, H0012 (y) is the space of functions in H12 (y) such that their extension by zero to ∂Ω/y belongs to H12 (∂Ω).

We introduce some notions to clarify the spaces of functions that depend on time. The function u(x, t), defined on the domain Ω×]0, T[, can be written as:

u:]0,T[Xtu(t)=u(.,t)

where X is a separable Banach space. We define 𝓒j(0, T; X) the set of time 𝓒j classes functions with a value on X. 𝓒j(0, T; X) is a Banach space for the norm

uCj(0,T;X)=sup0tTl=0jtluX

where tlu is the partial derivative of order l in time of the function u. We define also the spaces

Lp(0,T;X)={vmesurableon]0,T[suchthat0Tv(t)Xpdt<}

and

Hs(0,T;X)={vL2(0,T;X);kvL2(0,T;X);ks}.

Lp(0, T; X) is a Banach space for the norm

vLp(0,T;X)=(0Tv(t)Xpdt)1p,for1p<+sup0tTv(t)X,forp=+,

and Hs(0, T; X) is an Hilbert space for the following scalar product:

(u,v)=((u,v)L2(0,T;X)+k=0s(ku,kv)L2(0,T;X))12.

Problem (1) admits the equivalent variational formulation:

For t ∈ ]0, T[ and fL2(0, T; H–1(Ω)), find u ∈ 𝓒0(0, T; H01 (Ω)) ∩ L2(0, T; H01 (Ω)) such that: for all v H01 (Ω)

Ωut(x,t)v(x)dx+i=1IΩiλ(x)u(x,t)v(x)dx=<f(.,t),v> (3)

where < ., . > is the duality product between H01 (Ω) and H–1(Ω).

We introduce the energy norm

uλ(T)=(u0,Ω2+i=1I0Tλ(x)12u(x,t)0,Ωi2dt)12. (4)

We recall the following proposition (see [9], chap 3 for its proof).

Proposition 1

For fL2(0, T; H–1(Ω)) and u0L2(Ω), the problem (3) has a unique solution uL2(0, T; H01 (Ω)) verifying the following estimation:

uλ(T)(u00,Ω2+(1λmin)fL2(0,T;H1(Ω))2)12. (5)

We have the regularity result proved in ([5], Prop 2.2) and [10].

Proposition 2

We suppose that the restriction of the function λ on each sub-domain Ωi , 1 ≤ iI is constant. There exists a real 0 < s0 < 12 , depending on the geometry of the domain and the quotient λmaxλmin , such that for fL2(0, T; Hs–1(Ω)), the solution u of problem (3) belongs to L2(0, T; Hs+1(Ω) ∩ H01 (Ω)) for any 0 ≤ ss0.

Remark 1

The maximum value of the real s0 is bounded in the following way (see [10])

0s0min(12,CLog(1λminλmax)),

where C is a constant that depends only on the domain Ω.

3 The time semi discrete problem

To find the discrete problem in time, we introduce a partition of the interval [0, T]. Let [tn–1, tn] the sub-interval of this partition, such that 0 = t0 < t1 < … < tn–1 < … < tM = T where M is a positive integer. We notice τ = tntn–1, 1 ≤ nM the step of the partition that we suppose constant. We denote by v(., tn) = vn, 0 ≤ nM. We define the function vτ which is affine on each interval [tn–1, tn] by

vτ(.,t)=vntntτ(vnvn1). (6)

Using Euler implicit method, the semi discrete problem is written as follows:

unun1τdiv(λun)=fn,inΩ,1nMun=0,onΩ,1nMu0=u0,inΩ. (7)

The problem (7) has the equivalent variational formulation: Find (un)0≤nML2(Ω) × H01 (Ω)M such that for all v H01 (Ω),

Ωun(x)v(x)dx+τi=1IΩiλ(x)un(x)v(x)dx=Ωun1(x)v(x)dx+τΩfn(x)v(x)dx. (8)

Let the bilinear form an(., .) and the linear form Ln(.) defined respectively by

an(un,v)=Ωun(x)v(x)dx+τi=1IΩiλ(x)un(x)v(x)dx

and

Ln(v)=Ωun1(x)v(x)dx+τΩfn(x)v(x)dx.

It is easy to prove that the bilinear form an(., .) is continuous on the space H01 (Ω) × H01 (Ω), coercive on the space H01 (Ω) and that the linear form Ln is continuous on the space H01 (Ω). So according to the Lax Milgram theorem, we deduce the following proposition.

Proposition 3

For any function f in 𝓒0(0, T; H–1(Ω)) and u0L2(Ω), problem (8) has a unique solution (un)0≤nML2(Ω) × ( H01 (Ω))M.

If we take v = un in problem (8), we deduce the following inequality:

un0,Ω2+τi=1Iλ12un0,Ωi2un10,Ω2+τλminfn1,Ω2,

and by making the sum on n we conclude:

un0,Ω2+τj=1ni=1Iλ12uj0,Ωi2u00,Ω2+τλminj=1nfj1,Ω2. (9)

Proposition 4

For f in 𝓒0(0, T; H–1(Ω)) and u0H1(Ω), the solution (un)0≤nM of problem (8) satisfies the following estimation

14(u00,Ω2+τλminj=1nfj1,Ω2)uτλ2u00,Ω2+τλminj=1nfj1,Ω2+12τλ12u00,Ω2. (10)

Proof

To prove the estimation (10), we have to compare the two terms Ak=tk1tkλ12uτ(.,t)0,Ω2dt andBk=τλ12uk0,Ω2.

According to the definition of the function uτ defined in (6), we have ∀xΩ

tk1tkuτ(x,t)2dt=τ3(uk(x)2+uk1(x)2+uk(x).uk1(x)),

where ’.’ is the scalar product in ℝ2 and | . | its associate norm.

Then

Ak=τ3(λ12uk(x)0,Ω2+λ12uk1(x)0,Ω2+(λ12uk(x),λ12uk1(x))).

Given that xy14x2y2, thus

Akτ4λ12uk0,Ω2=14Bk.

We deduce the first inequality of (10) by doing the sum on k.

Now using the fact that xy12x2+12y2, we conclude that

Akτ2(λ12uk0,Ω2+λ12uk10,Ω2).

By doing the sum on k and using the estimation (9) we prove the second inequality of (10).□

We define the norm ∥ . ∥n by:

unn=(un0,Ω2+τj=1ni=1Iλ12uj0,Ωi2)12. (11)

The a priori error estimate is the object of the following theorem.

Theorem 3.1

If the solution u of problem (3) verifies that t2 u(., t) ∈ L2(0, T, H–1(Ω)), then

uuτncτuH2(0,T,H1(Ω)) (12)

where c is a positive constant.

Proof

Let ej = u(tj) – uj, 1 ≤ jM and e0 = 0. Taking t = tj in problem (3), we obtain

Ωut(x,tj)v(x)dx+i=1IΩiλ(x)u(x,tj)v(x)dx=Ωf(x,tj)v(x)dx.

Since

Ω(tj1tjutdt)v(x)dx=Ω(u(x,tj)u(x,tj1))dx,

then using the variational formulation (8), we deduce that for any v H01 (Ω), the sequence (ej), 1 ≤ jM is a solution of problem

Ωej(x)v(x)dx+τi=1IΩiλ(x)ej(x)v(x)dx=Ωej1(x)v(x)dx+τ(Ω(1τtj1tjut(x,t)dtut(x,tj))v(x)dx).

We remark that the error ej is the solution of problem (8) for a data function

fj=1τtj1tjut(x,t)dtut(x,tj).

By applying the mean value theorem and (9), we obtain, for t2 u(., t) ∈ L2(0, T, H–1(Ω)), the following estimation

en0,Ω2+τj=1ni=1Iλ12ej0,Ωicτt2uL2(0,T,H1(Ω))2,

where c is a positive constant.

We conclude by using the proposition 4.□

4 The mortar spectral element discretization

In this section we consider the function λ piecewise constant. The spectral discretization requires that the elements be rectangles, which leads us to make another partition without overlapping of the domain Ω

Ω¯=i=1i=IΩ¯i,ΩiΩj=,ij. (13)

We suppose the function λ is constant on each Ωi, 1 ≤ iI. We remark that for any 1 ≤ iI, there exits 1 ≤ jI, such that Ωi Ωj and I > I.

To explain this problem, we take the case where I = 2. This means that Ω is composed of two heterogeneous regions (see figure 1). To handle this domain by spectral discretization, we need 5 rectangles (I = 5).

Fig. 1 
The domain Ω
Fig. 1

The domain Ω

However, 9 rectangles are required for a conforming decomposition. We mean by conforming that if the intersection of two rectangles Ωi and Ωj, ij is not empty, it is necessarily equal to a corner or to a hole edge of Ωi and Ωj.

We suppose that the intersection of each boundary ∂Ωi of the sub-domain Ωi with the boundary ∂Ω of the domain Ω is a corner or a hole edge of Ωi. The skeleton of the decomposition

S=i=1IΩiΩ

is equal to

S¯=m=1Mym¯,ymym=,1mmM, (14)

where ym is called mortar, which is equal to a hole edge of one sub-domain Ωi that we note Ωi(m). Let ℙNi(Ωi), Ni ≥ 2, 1 ≤ iI, the space of the polynomial functions defined on Ωi, with degree ≤ Ni, for the two variables x and y.

We define the mortar discrete space 𝕏δ, (δ = (N1, …, NI) is the discretization parameter) as the space of functions uδ such that (see [7]):

  • uδ/Ωi, 1 ≤ iI, belongs to the polynomial space ℙNi(Ωi),

  • uδ vanishes on the boundary ∂Ω,

  • let ϕ the mortar function such that ϕ/ym = uδ/Ωi(m)/ym, for each Ωi, 1 ≤ iI and an edge Γ of Ωi, which is not included on the boundary ∂Ω, we have the following matching condition:

    χδPNi2(Γ),Γ(uδ/Ωiϕ)(ξ)χδ(ξ)dξ=0, (15)

where ℙNi–2(Γ) is the space of polynomials with degree ≤ (Ni – 2), defined on Γ. Since Γ does not always coincide with a mortar ym, 1 ≤ m ≤ 𝓜, this allows us to say that the discretization is not conforming (𝕏δ is not a subspace of H1(Ω)).

We remind the Gauss-Lobatto quadrature formula on the interval Λ = ]–1, 1[:

If N ≥ 2 is an integer, let ϵ0 = –1 and ϵN = 1, there exists a unique set of nodes ϵk, 1 ≤ k ≤ (N – 1) and weights ϱk, 0 ≤ kN, such that:

ψP2N1(Λ),11ψ(ξ)dξ=k=1Nψ(ϵk)ϱk. (16)

Hereinafter, We recall the following property (see [11]):

κNPN(Λ),κN0,Λ2k=0NκN2(ϵk)ϱk3κN0,Λ2. (17)

We find the value of the nodes and weights ϵikx and ϱikx (respectively ϵiky and ϱiky ) in the direction x (respectively in the direction y) by homothety and translation of the domain Ωi to the reference domain Λ2. So, we have the discrete scalar product defined as:

For φ and ψ continuous on each Ωi, 1 ≤ iI

(φ,ψ)δ=i=1I(φ,ψ)Ni, (18)

where

(φ,ψ)Ni=k=0Nil=0Niφ(ϵikx,ϵily)ψ(ϵikx,ϵily)ϱikxϱily.

Let φd,Ω=(φ,φ)δ12 the associate discrete norm.

We introduce the auxiliary space

Yδ={κδL2(Ω);κδ/ΩiPNi(Ωi);1iI}

and ℑδ the Lagrange interpolation operator defined as:

For all φ ∈ 𝕏δ such as φ/Ωi, 1 ≤ iI is continuous on Ωi, ℑδ(φ) ∈ 𝕐δ, with Iδ(φ)(ϵikx,ϵily)=φ(ϵikx,ϵily).

We suppose for any 0 ≤ nM, fn is continuous on each sub-domain Ωi, 1 ≤ iI. Then we define the discrete problem:

Find uδn ∈ 𝕏δ for each 1 ≤ nM, such that

uδ0=Iδ(u0),

and

vδXδ,aδn(uδn,vδ)=Lδn(vδ). (19)

The bilinear form aδn (., .), and the linear form Lδn (.), for 1 ≤ nM are defined as:

aδn(uδn,vδ)=(uδn,vδ)δ+τi=1Iλi(uδn,vδ)Ni (20)

and

Lδn(vδ)=(uδn1,vδ)δ+τ(fn,vδ)δ. (21)

Since the discretization is not conforming, we define the broken energy norm on 𝕏δ

vδXδ=(vδ0,Ω2+τi=1Iλivδ1,Ωi2)12. (22)

Lemma 1

There exist two constants c1 and c2 independent of δ such that for all vδ in 𝕏δ, we have the following equivalence:

c1min(1,λmin)i=1Ivδ1,Ωi2vδXδc2max(1,λmax)i=1Ivδ1,Ωi2. (23)

Proof

From (22) we deduce that

(vδ0,Ω2+τλmini=1Ivδ1,Ωi2)12vδXδ(vδ0,Ω2+τλmaxi=1Ivδ1,Ωi2)12.

We conclude (23) since

(i=1Ivδ1,Ωi2)12and(i=1Ivδ1,Ωi2)12

are equivalent with constants c1 and c2 independent of δ (see [12]).□

We prove using (17), Cauchy-Schwarz inequality and lemma 1 that the bilinear form aδn (., .) is continuous on 𝕏δ × 𝕏δ, coercive on 𝕏δ and that the linear form Lδn (.) is continuous on 𝕏δ. Using the Lax Milgram theorem, we obtain the following result.

Theorem 1

For f continuous on Ω×[0, T] and u0 continuous on Ω, problem (19) has a unique solution (uδn)0nM in 𝕐δ × (𝕏δ)M such that

uδn0,Ω2+τj=1ni=1Iλiuδj0,Ωi9INu00,Ω2+81C2λmaxλminj=1nINfj0,Ω

where C is the Poincaré-Friedrichs constant which only depends on the domain Ω.

Proof 1

Let vδ = uδn in problem (19). Using the Cauchy-Schwarz inequality, we have

|uδn|d,Ω2+τi=1Iλi|uδn|d,Ωi2|uδn1|d,Ω|uδn|d,Ω+|INfn|d,Ω|uδn|d,Ω.

Using (17), Poincaré-Friedrichs inequality and the fact that aba22μ+μb22,μ>0

12|uδn|d,Ω2+τi=1Iλiuδn0,Ωi2|uδn1|d,Ω22+9INfn0,Ω22μ+λmax9C2μ2λmini=1Iλiuδn0,Ωi2,

where C is the Poincaré-Friedrichs constant.

Doing the sum on n and using (17)

12uδn0,Ω2+τj=1ni=1Iλiuδj0,Ωi29INu00,Ω22+92μj=1nINfj0,Ω2+λmax9C2μ2λminj=1ni=1Iλiuδn0,Ωi2.

Then we conclude by choosing μ=λmin9C2λmax.

5 Error estimate

For 1 ≤ nM, we recall that un is the solution of problem (7). In the case where λ is piecewise constant, problem (7) is written:

un+τλΔun=un1+τfn inΩun=0 onΩu0=u0 inΩ. (24)

Multiplying the first equation in (24) by vδ ∈ 𝕏δ and integrating by parts gives

an(un,vδ)=Ln(vδ)+i=1IΩiλi(niun)vδdξ,

where ni is the unit normal vector to ∂Ωi.

If we define [vδ] the jump of vδ through the skeleton S, we obtain

an(un,vδ)=Ln(vδ)+12Sλ(nun)[vδ]dξ. (25)

Proposition 5

If f and u0 are respectively continuous on Ω × [0, T] and Ω, then the error estimate between (un)0≤nML2(Ω) × ( H01 (Ω))M solution of (8) and (uδn)0nMYδ×(Xδ)M solution of (19) is

unuδnnc(infvδnXδunvδnn+[infvδ0Yδu0vδ00,Ω+λmaxλminj=1n(Ea1,j+Ea2,j+Efj+Ecj)]) (26)

where

Ea1,j=1τsupwδXδΩ(ujuj1)(x)wδ(x)dx(vδjvδj1,wδ)δwδxδ,Ea2,j=supwδXδi=1Iλi(Ωiujwδdx(vδj,wδ)Ni)wδXδ,Efj=supwδXδΩfj(x)wδ(x)dx(f,wδ)δwδXδ,Ecj=supwδXδSλnuj[wδ]dξwδXδ

and c is a positive constant independent of δ.

Proof 2

Let (vδn)0nMXδ. By triangular inequality

unuδnnunvδnn+vδnuδnn.

To estimate the term uδnvδnn, we consider the two problems (25) and (19) for wδ ∈ 𝕏δ:

Ωun(x)wδ(x)dx+τi=1IλiΩiun(x)wδ(x)dx=Ωun1(x)wδ(x)dx+τΩfn(x)wδ(x)dx+12Sλnun[wδ]dξ

and

(uδn,wδ)δ+τi=1Iλi(uδn,wδ)Ni=(uδn1,wδ)δ+τ(fn,wδ)δ.

By doing the difference term by term, we obtain

(uδnvδn,wδ)+τi=1Iλi((uδnvδn),wδ)Ni=(uδn1vδn1,wδ)δ+τLn(wδ)

where

Ln(wδ)=1τ[Ω(unun1)wδdx(vδnvδn1,wδ)δ]+i=1Iλi(Ωiunwδdx(vδn,wδ)Ni)+Ωfnwδdx(fn,wδ)δ+12Sλnun[wδ]dξ.

We remark that 𝓛n is linear and continuous on 𝕏δ which is an Hilbert space for the scalar product (., .)δ. Then, by Riesz theorem, there exists a unique element gδn ∈ 𝕏δ such that

Ln(wδ)=(gδn,wδ)δ.

Therefore, we conclude that zδn=uδnvδn is solution to problem (8) with a data function equal to gδn and zδ0 = ℑδu0 vδ0 . Consequently using theorem 1, we have

uδnvδnncIδu0vδ00,Ω2+λminλmaxj=1ngδj0,Ω21/2.

We remark that

gδj0,ΩcsupwδXδ(gδj,wδ)δwδXδ,

which permits to conclude (26).

Let for each mortar ymS, 1 ≤ m ≤ 𝓜, ζ(m) is the set of subscripts i, 1 ≤ iI, such that ∂Ωiym has a positive measure. By estimating each term in (26), we obtain the following result proved in section 6.

Theorem 2

For λ constant on each Ωi, 1 ≤ iI. Let f such that f/Ωi ∈ 𝓒0(0, T; Hσi(Ωi));

σi > 1, u0 is such that u0/ΩiHμi(Ωi);μi > 1 and the solution (un)0≤nM of problem (8) is such that u/Ωin Hsi+1(Ωi); si ≥ 0. Then the error between (un)0≤nM and (uδn)0nM solution of problem (19) is

unuδnncτ[(1+β+βδ)λmaxλmini=1IλiNi2silog(Ni)unsi+1,Ωi21/2+1min(1,λmin)1/2i=1INi2σifC0(0,T;Hσi(Ωi))21/2+i=1INi2μiIδu0μi,Ωi21/2], (27)

where c is a positive constant independent of δ,

β=max1mMmaxkζ(m)λkλi(m)1/2

and

βδ=max1mMmaxkζ(m)λkNi(m)λi(m)Nk1/2.

Remark 2

For a conforming decomposition, the term βδ vanishes. Since there is no restriction on choosing of mortars, we opt, if it is possible, for those leading to

kζ(m),λkλi(m). (28)

Thus β ≤ 1. If it is not possible, we choose the mortars and the degrees of approximation polynomials such that

kζ(m),λkNk1λi(m)Ni(m)1, (29)

which may force us to make a small change in the decomposition. So we can optimize the estimation (27) without forcing the conformity of the decomposition.

6 Annex

This section is devoted to the proof of theorem 2. We need to estimate each term in (26).

6.1 Estimation of Ea1,j

We pose κj = ujuj–1. By remarking that ΠNi11κj=vδjvδj1, we deduce that

Ea1,jκjΠNi11κj0,Ω

where ΠNi11 is the orthogonal projection from H1(Ω) to PNi–1(Ω).

The approximation properties of the operator ΠNi11 are well known (see [11], Theorem 7.3 or [13], Proposition 2.6). This permit to obtain

Ea1,jcNisiκjsi,Ωi,

where, κjHsi(Ωi); for si ≥ 1.

6.2 Estimation of Ea2,j

Using the exactness of the quadrature formula for a polynomial of degree ≤ 2N – 1, we write:

i=1Iλi[Ωiujwδdx(vδj,wδ)Ni]=i=1Iλi[Ωi(ujΠNi11uj)wδdx((vδjΠNi11uj),wδ)Ni].

By the triangular and Cauchy-Schwarz inequalities, we obtain:

supwδXδi=1Iλi[Ωiujwδdx(vδj,wδ)Ni]wδXδcλmax[i=1I|ujΠNi11uj|1,Ωi+|vδjΠNi11uj|1,Ωi],

then we conclude by the properties of operator ΠNi11 .

6.3 Estimation of Efj

Let ΠNi–1 the orthogonal projection from L2(Ωi) to ℙNi–1(Ωi). We have by the exactness of the quadrature formula, for a polynomial of degree ≤ 2Ni – 1,

Ωfj(x)wδ(x)dx(fj,wδ)δ=i=1I[Ωi(fjΠNi1fj)(x)wδ(x)dx(IδfjΠNi1fj,wδ)Ni]

for all wδ ∈ 𝕏δ.

Using (17) in each direction, we obtain

Ωfj(x)wδ(x)dx(fj,wδ)δ[10(i=1IfjΠNi1fj0,Ω2)1/2+9fjIδfj0,Ω2]wδ0,Ω.

Using lemma 1 to bound ∥wδ0,Ω by ∥wδ𝕏δ and the approximation properties of operator ΠNi–1 (see [11], Theorem 7.1) and ℑδ (see [11], Theorem 14.2) for fjHσi(Ωi); σi > 1, we obtain

supwδXδΩfj(x)wδ(x)dx(fj,wδ)δwδXδc(1min(1,λmin))1/2(i=1INi2σifjσi,Ωi2)1/2.

6.4 Estimation of Ecj

Let the operator

λnu/Ωi=λiniu,1iI.

This operator is discontinuous through the skeleton S. It means that if Γ = ∂Ωi∂Ωi, 1 ≤ i < i′ ≤ I, we have

λnu/Ωi+λnu/Ωi=0 on Γ.

Lemma 2

For uj in 𝕏δ, 1 ≤ jn, such that u/ΩijHsi+1(Ωi);si>0, we have:

supwδXδSλnuj[wδ]dξwδXδc(1+β)i=1IλiNi2si(log(Ni))ujsi+1,Ωi21/2

where β=max1mMmaxkζ(m)λkλi(m)1/2 and c is a positive constant independent of δ.

Proof 3

Let ψ the mortar function associated to wδ ∈ 𝕏δ. By the matching condition (15), for each edge Γ of ∂Ωi, 1 ≤ iI which is not mortar, we have:

Γλnuj[wδ]dξ=λiΓ(niujπNi2Γniuj)(wδ/Ωiψ)dξ

where πNi2Γ is the projection operator onNi–2(Γ).

We suppose now that the decomposition is conforming, then Γ = ΩiΩi, 1 ≤ ii′ ≤ I. We know that u/ΩijHsi+1(Ωi), we study the two following cases.

  1. If 0 < si < 12 , we define the trace of λi ni uj by duality:

    For vH1/2–si(Γ)

    Γλiniujvdξ=λiΩiΔujv~(x)dx+λiΩiujv~dx

    where is the lifting function in H1–si(Ωi) of v.

    Therefore, since πNi2Γ is the orthogonal operator from Hsi–1/2(Ωi) intoNi–2(Γ), we conclude then for any wNi–2 and ψNi–2 inNi–2(Γ),

    Γλnuj[wδ]dξλi1/2niujsi1/2,Γλi1/2[wδwNi21/2si,Γ+ψψNi21/2si,Γ].

    Thus, the approximation properties (see [11]) allows us to deduce

    Γλnuj[wδ]dξcNisiλi1/2niujsi1/2,Γλi1/2[wδ1/2,Γ+ψ1/2,Γ].

    We remark thatwδ1/2,Γ is bounded bywδ1,Ωi and since the decomposition is conforming, ∥ψ1/2,Γ is bounded by ∥wδ1,Ωi which permit to conclude.

  2. In the case where si > 12 , we define πNi2Γ the orthogonal projection operator from L2(Γ) toNi–2(Γ), then

    Γλnuj[wδ]dτλi1/2(niujπNi2Γ(niuj))0,Γλi1/2[wδ/ΩiwNi20,Γ+ψψNi20,Γ].

    We conclude thanks to the approximation properties of πNi2Γ (see [11], theorem 6.1).

The estimation in the case si = 1/2 is given by interpolation argument. In the case where Γ = ∂Ωi∂Ω, the mortar function ψH1/2(Γ) and a modification make appears (log(Ni))1/2 ([13], proposition 21).

Remark 3

The term (log(Ni))1/2 is negligible compared to Nisi when Ni is big enough. This term disappears when the edges of any sub-domain Ωi, 1 ≤ iI, which are not a mortar, are in the boundary ∂Ω.

6.5 Estimation of infvδnXδunvδnn

Lemma 3

Let un ∈ 𝕏δ; 1 ≤ nM such that u/Ωin Hsi+1(Ωi), si ≥ 2, then we have

infvδnXδunvδnnc(1+β+βδ)i=1IλiN2siunsi+1,Ωi21/2

where c is a positive constant independent on δ and

βδ=max1mMmaxkξ(m)λkNi(m)λi(m)Nk1/2

depends on δ.

Proof 4

The proof will be done in three steps to construct vδn ∈ 𝕏δ which represents the best approximation of un in the space 𝕏δ.

  1. Let πN0 the operator from H3/2(Λ) inN(Λ) which preserves ±1 (see [11], theorem 6.4), thenφHs(Λ), s ≥ 3/2 and 0 ≤ t ≤ 3/2,

    φπN0φt,ΛcNtsφs,Λ.

    Let πNi0 the operator defined from πN0 by translation and homothety. We choose vδn1 such that

    vδn1/Ωi=(πNi0,xπNi0,y)(u/Ωin),

    where πNi0,x and πNi0,y are the operators in the x- and y-directions on Ωi. We conclude then, if si ≥ 2 (see [11]):

    λi1/2unvδn1t,Ωicλi1/2Nitsi1unsi+1,Ωi, (30)

    where the function vδn1 vanishes on ∂Ω and coincides with un at the corners of the sub-domain Ωi, 1 ≤ iI.

  2. Let ϖ the set of vertices of Ωi which are not a boundary of an edge of an other sub-domain Ωi. For cϖ, ζ(c) is the set of subscripts i, 1 ≤ iI such that c∂Ωi. We denote by i(c) the index of ζ(c) such that

    iζ(c);λiλi(c)

    and for iζ(c)/{i(c)}, we define ψic the polynomial of minimal degree such that ψic (c) = 1 and vanishes on the edges of Ωi which do not contain c and on the other points of ϖ∂Ωi, then:

    vδn2=cϖiζ(c)/{i(c)}vδ1n/Ωi(c)vδ1n/Ωi(c)ψic.

    So, we have for 0 ≤ t ≤ 3/2

    λi1/2vδn2t,Ωicλi1/2cϖΩivδ1n/Ωi(c)vδ1n/Ωi(c)cλi1/2cϖΩiunvδn1L(Ωi)+unvδn1L(Ωi(c)).

    Thanks to the Gagliardo-Nirenberg inequality, for 0 < αi < 1/2

    λi1/2vδn2t,Ωicλi1/2(cϖΩiunvδn11αi,Ωi1/2unvδn11+αi,Ωi1/2+unvδn11αi,Ωi(c)1/2unvδn11+αi,Ωi(c)1/2

    Since λiλi(c) and using (30) (for t = 1 – αi and t = 1 + αi), we conclude that:

    λi1/2vδn2t,Ωicλi1/2Nisiunsi+1,Ωi+cϖΩiλi(c)1/2Nisi(c)unsi(c)+1,Ωi(c). (31)

    Therefore, we end by doing the sum on i, 1 ≤ iI.

  3. Let 𝓗i, 1 ≤ iI the set of edges of Ωi which are not include in ∂Ω and are not mortars.

    If φ is the mortar function associated to vδn1+vδn2, then

    vδn3=i=1IΓHiRiLπNiΓφ(vδn1+vδn2)/Ωi

where

  • RiL is the lifting operator fromNi(Γ) ∩ H001/2 (Γ) intoNi(Ωi), such thatφδ ∈ ℙNi(Γ) ∩ H001/2 (Γ) (see [14])

    RiLφδ1,ΩicφδH001/2(Γ), (32)
  • πNiΓ is the operator of projection from H001/2 (Γ) toNi(Γ) ∩ H001/2 (Γ), such thatφ H001/2 (Γ)

    Γ(φπNiΓφ)ψδ(τ)dξ=0ψδPNi2(Γ). (33)

Then following (31)

λi1/2vδn31,Ωicλi1/2ΓHiπNiΓφ(vδn1+vδn2)/ΩiH001/2(Γ).

Since πNiΓφ(vδn1+vδn2)/Ωi coincides with (vδn1+vδn2)/Ωi on Γ, then:

λi1/2vδn31,Ωicλi1/2ΓHi(un(vδn1+vδn2)H001/2(Γ)+(idπNiΓ)un(vδn1+vδn2)H001/2(Γ)) (34)

where [.] is the jump through Γ.

Using the trace theorem, (30) and (31), we conclude for t = 1 that

λi1/2un(vδn1+vδn2)H001/2(Γ)λi1/2(Nisiunsi+1,Ωi+mes(Ωiym)>0Ni(m)si(m)unsi(m)+1,Ωi(m)). (35)

To evaluate the second term in (34), we remark that the operator πNiΓ is equal to the operator of projection from H01 (Γ) intoNi(Γ) ∩ H01 (Γ). Then, ∀ψ H01 (Γ)

ψπNiΓψH001/2(Γ)cNi1/2un(vδn1+vδn2)1,Γ.

Therefore, we proceed as before with t = 3/2, we obtain

λi1/2(idπNiΓ)un(vδn1+vδn2)H001/2(Γ)cλi1/2Ni1/2(Ni1/2siunsi+1,Ωi+mes(Ωiym)>0Ni(m)1/2si(m)unsi(m)+1,Ωi(m)). (36)

This is where we introduce βδ.

To conclude the proof of Theorem 2, we choose vδn=vδn1+vδn2+vδn3 in 𝕏δ and combine (30), (31), (35) and (36).

Conclusion

This work concerns the numerical analysis of the mortar spectral elements method discretization of the heat equation with a diffusion coefficient λ, depending on the heterogeneity of the domain. To solve the problem of the solution singularity due to the discontinuity of λ, we use a non conform geometric decomposition of the domain. We prove an optimal error estimate that depends only on the local regularity of the solution. The numerical validation of this result will be the subject of a forthcoming work.

Acknowledgments

The authors acknowledge funding from the Research and Development (R&D) Program (Research Pooling Initiative), Ministry of Education, Riyadh, Saudi Arabia, (RPI-KSU).

References

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Received: 2019-05-06
Accepted: 2019-06-06
Published Online: 2019-10-16

© 2019 Mohamed Abdelwahed and Nejmeddine Chorfi, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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