Approximation of the two-dimensional Dirichlet problem by continuous and discrete problems on one-dimensional networks

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CONFLUENTES MATHEMATICI

Maryse BOURLARD-JOSPIN, Serge NICAISE, and Juliette VENEL

Tome 7, n^o1 (2015), p. 13-33.

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7, 1 (2015) 13-33

APPROXIMATION OF THE TWO-DIMENSIONAL DIRICHLET PROBLEM BY CONTINUOUS AND DISCRETE PROBLEMS ON

ONE-DIMENSIONAL NETWORKS

MARYSE BOURLARD-JOSPIN, SERGE NICAISE, AND JULIETTE VENEL

Abstract. We show that the solution of the two-dimensional Dirichlet problem set in a plane domain is the limit of the solutions of similar problems set on a sequence of one-dimensional networks as their size goes to zero. Roughly speaking this means that a membrane can be seen as the limit of rackets made of strings. For practical applications, we also show that the solutions of the discrete approximated problems (again on the one- dimensional networks) also converge to the solution of the two-dimensional Dirichlet problem.

1. Introduction

Approximation of multidimensional boundary value problems by discrete problems or by boundary value problems set on less dimensional ones is very important in practice. For discrete approximations, the most popular methods are the finite difference method or the finite element method, for which a lot of convergence results are proved [6, 23]. By the less dimensional approximation, we mean that a n-dimensional problem is approximated by a family of k-dimensional ones with k < n. For instance the approximation of boundary value problems set on objects of R³ with a small thickness by boundary value problems set on objects of dimension 1 or 2 was largely considered in the literature, see for instance [9, 7, 20].

In the same spirit, let us also mention homogenization techniques that analyze the limit process of problems set onn-dimensional domains of thickness to problems still set on domains of dimensionn[8].

The problems studied in this paper have some common properties with the above approaches since we will approach a two-dimensional problem by a family of continuous 1-dimensional problems but as each continuous 1-dimensional problem can be approximated by a discrete one, we also examine the limit of these discrete problems. The approximation of the low frequency spectrum of such problems was performed in [13, 12] (see also [19] for the plate problem), but to our best knowledge the approximation of the boundary value problem itself was not yet performed. Hence our goal is to fill this gap and to show that indeed the solutions of the continuous and discrete one-dimensional problems converge to the solution of the two-dimensional problem. More precisely, we first prove some error estimates between the solution of the two-dimensional problem in an arbitrary domain and the solutions of the continuous one-dimensional problems. Further we propose a numerical scheme based on the resolution of discrete one-dimensional problems and obtain error estimates similar to the standard two-dimensional finite element method. Our approach can be considered as an attractive alternative to the standard ones since its associated stiffness matrix is easier to compute and keeps the same properties (symmetry, positive definiteness and sparsity). Finally it can be used for domains with curved boundaries since no triangulation is needed.

The schedule of the paper is as follows: We recall in Section 2 the Dirichlet problem in the unit square as well as its continuous counterparts on networks that approach the square as the size goes to zero. An error estimate between the solutions of these continuous problems is proved in section 3 by using the second Strang lemma. Similarly section 4 is devoted to the error analysis between the exact solution in the unit square with the finite element approximations on the networks.

Math. classification:35R02, 35B40, 65N30.

13

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In section 5 we extend some of our previous results to the Dirichlet problem set on an arbitrary domain of the plane. Finally in section 6 some numerical tests are presented that confirm our theoretical results.

Let us finish this introduction with some notation used in the remainder of the paper: On D, the L²(D)-norm will be denoted by k · k_D. The usual norm and seminorm of H^s(D) (s >0) are denoted by k · ks,D and | · |s,D, respectively.

Finally, the notation a .b means the existence of a positive constantsC, which is independent of the size h of the edges of the network (see below) and of the considered quantitiesaandb such thata6Cb.

2. The continuous problems

2.1. The continuous two-dimensional problem. LetS denote the unit square ]0; 1[×]0; 1[ and∂Sits boundary. On this domain, we consider the Dirichlet problem

−∆u = f in S

u = 0 on∂S (2.1)

withf ∈ C(S).

According to Lax-Milgram lemma, there exists a unique weak solutionu∈H₀¹(S) of this problem, namelyu∈H₀¹(S) is the unique solution of

Z

S

∇u· ∇v dx= Z

S

f v dx, ∀v∈H₀¹(S).

According to Theorem 5.1.3.5 of [11], this solution belongs to W^2,p(S), for all p >2, and iff belongs toW^1,p(S), withp >2, is such thatf is zero at each corner ofS, then this solution belongs toW^3,p(S), hence in particular toH³(S).

2.2. The associated problem on networks. Now we intend to consider a similar problem set on a family of networks included inS. First we need to introduce some notation: For anyn∈N,n>2, leth= 1/nand introduce the networkRhdefined by

Rh = {]kh; (k+ 1)h[×{`h};∀k∈ {0, . . . , n−1},∀`∈ {1, . . . , n−1}}

∪ {{kh}×]`h; (`+ 1)h[;∀k∈ {1, . . . , n−1},∀`∈ {0, . . . , n−1}}.

The edges of R_h are the intervals ]kh; (k+ 1)h[×{`h} or {kh}×]`h; (`+ 1)h[ but will be quite simply denoted bye_i, in other words,

Rh={ei;i= 1, . . . , Nh}, withNh= 2n(n−1).

We directly check that the size (or length) of each edge of the network Rh is h.

We further write Nh for the set of nodes of Rh. Moreover we need to distinguish between nodes included intoS or into∂S, so we set

N_h^int = {(kh;`h);∀k, `∈ {1, . . . , n−1}},

N_hêxt = {(0;`h); (1;`h); (`h; 0); (`h; 1);∀`∈ {0, . . . , n}}, Nh = N_hînt∪ N_hêxt.

It remains a last notation to indicate the set of edges adjacent to a given node:

∀v∈ Nh, Iv={i∈ {1, . . . , Nh}such thatv∈ei}.

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Our aim is to approximate the solution uof the continuous problem (2.1) by the solutionu_h= (u_i)_i=1,...,N_h ∈

N_h

Y

i=1

H²(e_i) of the following problem:











−u⁰⁰_i = fei onei ∀i= 1· · ·Nh, ui(v) = 0 ∀v∈ N_h^ext,∀i∈ Iv, u_i(v) = u_j(v) ∀v∈ N_h^int,∀i, j∈ Iv,

X

i∈Iv

∂ui

∂ν_i(v) = 0 ∀v∈ N_h^int,

(2.2)

where

fei= 1

2γif. (2.3)

In the whole paper we use the abuse of notationu⁰⁰ for ∂²u

∂x² or ∂²u

∂y² according to the kind of the edge (horizontal or vertical). A similar abuse of notation will be used for the first order derivatives. Furthermore, _∂ν^∂

i andγ_i represent respectively the outer normal derivative operator and the trace operator on the edge ei. The last equation of problem (2.2) is nothing else but Kirchoff’s law. System (2.2) is a Dirichlet problem on the networkRhthat was largely studied in the literature, see [1, 2, 4, 5, 16, 15, 17, 18, 21] and the references there.

2.3. Variational formulation on the networks. The variational space associated with problem (2.2) is

Vh={uh= (ui)i=1,...,N_h ∈

N_h

Y

i=1

H¹(ei) s.t.

u_i(v) =u_j(v) ∀v∈ N_h^int,∀i, j∈ Iv, u_i(v) = 0 ∀v∈ N_h^ext,∀i∈ I_v},

(2.4)

equipped with the norm:

||u||_h=|u|_1,R_h=

"_N_h X

i=1

Z

e_i

(u⁰_i(x))²dx

#^1/2

. (2.5)

Due to the Dirichlet boundary conditions, the H¹-norm and its semi-norm are equivalent onV_h.

Lemma 2.1. — For everyw∈Vh, we have

||w||_R_h 6|w|_1,R_h, (2.6)

as well as

kwk∞,R_h:= sup

(x,y)∈R_h

|w(x, y)|6|w|1,R_h. (2.7) Proof. —Let us denoteL_`={(x, `h),0< x <1}andC_k ={(kh, y),0< y <1}.

Then

Rh=

n−1

[

`=1

L`

!

∪

n−1

[

k=1

Ck

!

. (2.8)

Asw(0, `h) = 0, we have for allx∈]0; 1[

|w(x, `h)|=

Z x 0

∂w

∂x(t, `h)dt 6

∂w

∂x _L

`

, (2.9)

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according to the Cauchy-Schwarz inequality. Then

||w||²_L_` = Z 1

0

|w(x, `h)|²dx6

∂w

∂x

2

L_`

6|w|²_1,L_`. (2.10) In the same way, we can check thatkwk²_C

k 6|w|²_1,C

k and by summing up these two inequalities we obtain the expected estimate (2.6).

The estimate (2.7) is a direct consequence of (2.9) and its counterpart inC_k. Now we define a bilinear formah onVh by

a_h:V_h×V_h→R: (u_h, w_h)→a_h(u_h, w_h) =

Nh

X

i=1

Z

e_i

u⁰_i(x)w⁰_i(x)dx, (2.11) that is clearly continuous and coercive onVh according to Lemma 2.1.

Proposition 2.2. — The variational formulation of problem (2.2) is to find u_h∈V_h solution of

∀wh∈V_h, a_h(u_h, w_h) =F(w_h), (2.12) with

F(wh) =

N_h

X

i=1

Z

ei

fei(x)wi(x)dx. (2.13) Proof. —The proof is quite standard (cf. Lemma 2.2.12 in [2] for instance), we give it for the sake of completeness. Let us assume that there exists a solution u_h = (u_i)_i=1,...,N_h ∈

N_h

Y

i=1

H²(e_i) of problem (2.2). Obviously u_h belongs to V_h. Moreover uh is solution of (2.12). Indeed, let wh = (wi)i=1,...,N_h ∈ Vh, then we have for alli∈ {1, ..., Nh},

− Z

ei

u⁰⁰_i(x)wi(x)dx= Z

ei

fei(x)wi(x)dx.

Integrating by parts, we obtain Z

e_i

u⁰_i(x)w⁰_i(x)dx−[u⁰_i(v)w_i(v)]^v=v_v=vⁱ²

i1= Z

e_i

fe_i(x)w_i(x)dx, (2.14) wherevi1andvi2∈ Nh are such thati∈ Iv_i1∩ Iv_i2.

We claim that

N_h

X

i=1

[u⁰_i(v)wi(v)]^v=v_v=vⁱ²

i1 = 0. (2.15)

In fact, we have

i1= ∂u_i

∂νi

(vi1)wi(vi1) +∂u_i

∂νi

(vi2)wi(vi2), (2.16) and consequently,

Nh

X

i=1

i1 = X

v∈N_h^ext

X

i∈Iv

∂ui

∂νi

(v)wi(v) + X

v∈N_h^int

X

i∈Iv

∂ui

∂νi

(v)wi(v). (2.17) Ifv ∈ N_h^ext, thenw_i(v) = 0, for all i∈ I_v and therefore the second term of (2.17) is zero. Ifv∈ N_h^int, then

X

i∈Iv

∂ui

∂ν_i(v)wi(v) =wj(v)X

i∈Iv

∂ui

∂ν_i(v) (2.18)

for anyj∈ Iv, sincewhis continuous at the nodes. Then, using Kirchoff’s law, the right-hand side of the identity (2.18) is equal to zero and the first term of (2.17) is zero. Hence (2.15) is established and we conclude with (2.14) and (2.15).

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3. An approximation result between the continuous problems In this section, we analyze the error between the solutionuof problem (2.1) and the solutions u_h of (2.12). For that purpose, we make use of the second Strang lemma (see below). Hence we first estimate the consistency error:

Theorem 3.1. — Let u denote the solution of (2.1), and u_h the solution of (2.12). Ifu∈H³(S), then

sup

w∈Vh

|ah(u, w)−F(w)|

||w||h .

√

h||u||3,S. (3.1) Proof. —Since u∈ H³(S), for all i = 1, ..., N_h, u_i = γ_iuhas a meaning and sinceuis also continuous, its restriction to R_h, still denoted byu, belongs toV_h. Fix w= (wi)i=1,...,N_h ∈Vh. It can be shown, as in the proof of Proposition 2.2, that

ah(u, w) =−

Nh

X

i=1

Z

e_i

u⁰⁰_i(x)wi(x)dx.

Then, thanks to (2.13),

ah(u, w)−F(w) =−

N_h

X

i=1

Z

e_i

(u⁰⁰_i(x) +fei(x))wi(x)dx. (3.2) For everyv∈ N_h, if (ξ, ϕ) are the coordinates ofv, we define the rectangle

C_v^h=

]ξ−h 2, ξ+h

2[×]ϕ−h 2, ϕ+h

2[

∩ S and its intersection withRh

R_v^h=C_v^h∩ R_h. (3.3)

Ifv ∈ N_h^int, R^h_v is a cross, while if v∈ N_h^ext,R_v^h is a half edge. The identity (3.2) can be rewritten as

a_h(u, w)−F(w) =− X

v∈N_h^int

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx− X

v∈N_h^ext

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx.

(3.4) Step 1 : Case of the interior nodes

Fixv∈ N_h^int. We define the reference squareCb=]−¹₂;¹₂[×]−¹₂;¹₂[ and the reference crossRb= ({0}×]−¹₂;¹₂[)∪(]−¹₂;¹₂[×{0}). We consider the change of variables

φ:Cb→C_v^h: ˆx→x=φ(ˆx) =v+hˆx.

Note thatφ(R) =b R^h_v and Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx=h Z

Rb

(u⁰⁰(φ(ˆx)) + ˜f(φ(ˆx)))w(φ(ˆx))dˆx.

Let us set ˆu=u◦φ, ˆw=w◦φ, then ˆu⁰ =h(u⁰◦φ). In the same way, u⁰⁰◦φ= 1

h²uˆ⁰⁰and (∆u)◦φ= 1

h²∆ˆu. (3.5)

Owing to the definition (2.3) of ˜f, ˜f◦φ=−¹₂_h¹2∆ˆuand finally Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx= 1 h

Z

Rb (ˆu⁰⁰−1

2∆ˆu)(ˆx) ˆw(ˆx)dˆx=h⁻¹(I₁+I₂), (3.6) where

I1= Z

Rb (ˆu⁰⁰−1

2∆ˆu)(ˆx)( ˆw(ˆx)− Mw)dˆˆ x (3.7) and

I₂= Z

Rb (ˆu⁰⁰−1

2∆ˆu)(ˆx)(Mw)dˆˆ x, (3.8)

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with the constant

Mwˆ = Z

Rb ˆ

w(ˆx)dˆx. (3.9)

Let us begin with the estimate ofI1. With (3.7) and the Cauchy-Schwarz inequality,

|I₁|6||ˆu⁰⁰−1 2∆ˆu||

Rb||wˆ− Mw||ˆ

Rb. (3.10)

Moreover we have

||ˆu⁰⁰−1 2∆ˆu||

Rb. X

|α|=2

||D^αu||ˆ

Rb. X

|α|=2

||D^αˆu||_1,

Cb. (3.11) by using a trace theorem [11, Thm 1.5.1.2]. We recall that due to the Poincaré- Friedrichs inequality,

||wˆ− Mw||ˆ

Rb.|w|ˆ

1,^Rb

. (3.12)

Thanks to (3.10), (3.11) and (3.12), we have shown

|I1|.|w|ˆ_1,

Rb X

|α|=2

||D^αu||ˆ _1,

Cb. (3.13)

Now in order to estimateI2, we need the following lemma that can be proved by easy computations.

Lemma 3.2. —

∀ˆp∈P2(R),b Z

Rb (ˆp⁰⁰−1

2∆ˆp)(ˆx)dˆx= 0, (3.14) whereP2(R)b represents the set of polynomials of degree at most 2 onR.b

Owing to (3.8) and (3.14), for all ˆp∈P2(R),b I2=

Z

Rb

[(ˆu−p)ˆ⁰⁰−1

2∆(ˆu−p)](ˆˆ x)(Mw)dˆˆ x.

According to the Cauchy-Schwarz inequality, and sinceMwˆ is a constant,

|I₂| . |Mw|k(ˆˆ u−p)ˆ⁰⁰−1

2∆(ˆu−p)kˆ

Rb.|Mw|ˆ X

|α|=2

||D^α(ˆu−p)||ˆ

Rb . |Mw|ˆ X

|α|=2

||D^α(ˆu−p)||ˆ _1,

Cb.|Mw| ||ˆˆ u−p||ˆ _3,

Cb

by using the same trace theorem as previously. Let ˆpbe the orthogonal projection of ˆuonP2(R) for theb H³(C)-norm, thenb

||uˆ−p||ˆ _3,

Cb.|ˆu|_3,

Cb. (3.15)

Moreover, due to (3.9),|Mw|ˆ .||w||ˆ

Rb, so the three last inequalities imply that

|I₂|.||w||ˆ

Rb|ˆu|

3,^Cb

. (3.16)

Now we recall the next lemma that specifies the change ofH^m-semi-norms from a domain to a reference domain [6].

Lemma 3.3. — Considerm∈Nand let us denotefˆ=f◦φ. Then

∀f ∈H^m(R^h_v),|f|_m,Rh

v = 1

h^m−1/2|fˆ|

m,^Rb and

∀f ∈H^m(C_v^h),|f|_m,Ch

v = 1

h^m−1|fˆ|_m,

Cb.

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By (3.13),

|I1|.|w|ˆ _1,

Rb



 X

|α|=2

||D^αu||ˆ

Cb+ X

|α|=2

|D^αu|ˆ_1,

Cb



. (3.17)

It follows from Lemma 3.3 and (3.5) that

|I1|.h²√

h|w|_1,Rh v



 X

|α|=2

h⁻¹||D^αu||_Ch

v + X

|α|=2

|D^αu|_1,Ch v



, and finally,

|I1|.h^3/2|w|_1,Rh

v||u||_3,Ch

v. (3.18)

According to Lemma 3.3, (3.16) leads to

|I2|.h^3/2||w||_Rh v|u|_3,Ch

v. (3.19)

Gathering the results (3.6), (3.18) and (3.19), we have proved that

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx

.√

h |w|_1,Rh

v||u||_3,Ch

v +||w||_Rh v|u|_3,Ch

v

. (3.20) Step 2 : Case of the exterior nodes

Fixv∈ N_h^ext and let us denoteRb_1/2=φ⁻¹(R^h_v) andCb_1/2=φ⁻¹(C_v^h).

We show as (3.6) that Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx= 1 h

Z

Rb_1/2

(ˆu⁰⁰−1

2∆ˆu)(ˆx) ˆw(ˆx)dˆx. (3.21) Using the Cauchy-Schwarz inequality, this implies

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx

6 1

h||ˆu⁰⁰−1 2∆ˆu||

Rb_1/2||w||ˆ

Rb_1/2. (3.22) Arguing as for (3.11), sinceu∈H³(S), we get

||ˆu⁰⁰−1 2∆ˆu||

Rb_1/2 . X

|α|=2

||D^αˆu||

1,^Cb1/2

. (3.23)

On the other hand,w∈Vh implies that ˆw(0) = 0, so it can be proved as in Lemma 2.1 that

||w||ˆ

Rb_1/2 .|w|ˆ _1,

Rb_1/2. (3.24)

Thanks to (3.22), (3.23) and (3.24), we have

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx

. 1 h|w|ˆ_1,

Rb_1/2



 X

|α|=2

||D^αu||ˆ

Cb_1/2+ X

|α|=2

|D^αu|ˆ_1,

Cb_1/2



. (3.25) Using Lemma 3.3 and the identity (3.5), it comes

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx

. 1 h

√

h|w|_1,Rh vh²



 X

|α|=2

h⁻¹||D^αu||_Ch

v + X

|α|=2

|D^αu|_1,Ch v



 .√

h|w|_1,Rh

v||u||_3,Ch

v. (3.26)

Step 3 : Conclusion The identity (3.4) leads to

|ah(u, w)−F(w)|6 X

v∈N_h^int

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx

+X

v∈N_h^ext

Z

R^h_v

(u⁰⁰+ ˜f)(x)w(x)dx .

(3.27)

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Summing (3.20) for allv ∈ N_h^int and (3.26) for allv ∈ N_h^ext, we deduce from the previous inequality

|ah(u, w)−F(w)| .

√

h X

v∈N_h^int

|w|_1,Rh

v||u||_3,Ch

v +||w||_Rh v|u|_3,Ch

v

+ √

h X

v∈N_h^ext

|w|_1,Rh

v||u||_3,Ch v

. √

h X

v∈N_h^int

||w||_Rh v|u|_3,Ch

v +√

h X

v∈Nh

|w|_1,Rh

v||u||_3,Ch v. By the discrete Cauchy-Schwarz inequality, we obtain

|a_h(u, w)−F(w)|.√

h(|w|_1,R_h||u||_3,S+||w||_R_h|u|_3,S). (3.28) We conclude the proof thanks to Lemma 2.1 and inequality (3.28).

Now we recall the following result which is a consequence of the second Strang Lemma and can be found for example in [6, Thm 4.2.2].

Lemma 3.4. — Letudenote the solution of (2.1) supposed to belong toV_h, and letuh be the solution of (2.12). Then

||u−uh||h. sup

wh∈Vh

|ah(u, wh)−F(wh)|

||wh||h

. (3.29)

Remark 3.5. — Note that the upper bound in the second Strang Lemma con- tains another term, namely infv_h∈V_h||u−vh||h, called the “interpolation error".

Here only the "consistency error" term appears as we have assumed that u∈Vh, the interpolation error being obviously equal to zero.

Corollary 3.6. — Letudenote the solution of (2.1), and letuhbe the solution of (2.12). Ifu∈H³(S), then

||u−uh||_1,R_h.√

h||u||_3,S, (3.30)

and

||u−uh||_∞,R_h.√

h||u||_3,S. (3.31)

Proof. —We deduce from Theorem 3.1 and Lemma 3.4 that

|u−uh|1,R_h .

√

h||u||3,S. (3.32)

Now the estimates (3.30) and (3.31) are a direct consequence of Lemma 2.1 since

u−uh∈Vh.

4. The finite element method on the networks

In the previous section, we have checked thatuh is a good approximation ofu.

However, problem (2.2) is still set in an infinite dimensional space and except for some specific right-hand sides ˜f, its solutionuh cannot be computed analytically.

Hence in practice problem (2.2) has to be discretized. Here we choose the finite element method and propose to deal with two different cases according to the regularityH³(S) or C³( ¯S) of the solutionu.

4.1. A less regular solution. Here we assume that the solution of the continuous problem (2.1)ubelongs toH³(S) andf is a continuous function inS. Let P1(ei) denote the set of polynomials of degree at most 1 onei, for alli∈ {1, . . . , Nh}. We define the discrete variational space

Wh={wh= (wi)i=1,...,N_h∈Vh s.t. wi ∈P1(ei),∀i= 1, . . . , Nh}. (4.1) LetUh∈Wh be the solution of the finite element problem

ah(Uh, wh) =F(wh),∀wh∈Wh. (4.2)

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In order to compareuandU_hinS, we will use an interpolantI_huofuand a lifting R_hU_h ofU_h defined as follows: Let us denoteK_k,`^h =]kh,(k+ 1)h[×]`h,(`+ 1)h[, for eachk, `∈ {0, . . . , n−1}. Observe that

S=

n−1

[

k=0 n−1

[

`=0

K_k,`^h

!

∪ Rh, (4.3)

and therefore the set of ¯K_k,`^h is a triangulation of S. Hence let Ihu denote the Lagrange interpolation ofurelated to this triangulation, namelyIhuis the function such that its restriction to K_k,`^h belongs to Q1(K_k,`^h ) (where Q1 is the space of polynomials in (x, y) of degree at most 1 in each variablexandy) and that coincides withuat each nodev∈ Nh. As a consequenceIhuis continuous onS. Finally we defineRhUh=IhUhin the sense that its restriction toK_k,`^h fulfilsRhUh∈Q1(K_k,`^h ) and

RhUh(v) =Uh(v),∀v∈K_k,`^h ∩ Nh. (4.4) ThusRhUh coincides withUh onRh and is continuous onS.

Now we aim at approximatingubyUh. The estimate of the error is made with the help of the following three lemmas.

Lemma 4.1. —

|Ihu−RhUh|_1,S .√

h|Ihu−Uh|_1,R_h. (4.5) Proof. —According to Lemma 3.3, with ˜Φ : ˆC→K_k,`^h : ˆx7→(kh, `h) +h(ˆx+¹₂), we have

|Ihu−R_hU_h|_1,Kh

k,` .|dI_hu−R\_hU_h|_1,

Cb. (4.6)

Let us denoteQ⁰₁(C) =b {q∈Q1(C),b R

∂^Cb q= 0}.

Takeq∈Q₁(C), thenb

|q|_1,

Cb=|πq|_1,

Cb6kπqk_1,

Cb, where πq = q−¹₄R

∂^Cb

q ∈ Q⁰₁(C). Note thatb Q⁰₁(C) is a finite dimensional spaceb and| · |_1,∂

Cb is a norm on this space. So kπqk1,^Cb.|πq|

1,∂^Cb

=|q|

1,∂^Cb . We have thus proved that

∀q∈Q1(C),b |q|_1,

Cb.|q|_1,∂

Cb. (4.7)

AsIdhu−R\hUh∈Q1(C), thanks to (4.6) and (4.7), we haveb

|Ihu−RhUh|_1,Kh

k,`.|dIhu−R\hUh|

1,∂^Cb , and owing to Lemma 3.3 again,

|I_hu−R_hU_h|_1,Kh k,` .√

h|I_hu−U_h|_1,∂Kh

k,`. (4.8)

Collecting the pieces, we obtain

|I_hu−R_hU_h|²_1,S = X

k,`

|I_hu−R_hU_h|²_1,Kh k,`

. hX

k,`

|Ihu−U_h|²_1,∂Kh k,`

. h|Ihu−Uh|²_1,R_h,

the last inequality following from the fact that each edge ofRh is in the boundary

of two domainsK_k,`^h .

Lemma 4.2. — Ifu∈H²(S), then

|Ihu−u|_1,R_h .√

h|u|_2,S. (4.9)

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Proof. —Using Lemma 3.3,

|Ihu−u|_1,∂Kh

k,` =h^−1/2|dI_hu−u|b_1,∂

Cb6h^−1/2||dI_hu−bu||_1,∂

Cb.

Thanks to a trace theorem (see Theorem 1.5.2.1 in [11] for instance), this leads to

|Ihu−u|_1,∂Kh

k,` .h^−1/2||dI_hu−bu||_2,

Cb. (4.10)

By the classical interpolation error estimate (see for instance Theorem 3.1.6 in [6]), we have

||dIhu−u||b _2,

Cb.|ˆu|_2,

Cb. Owing to Lemma 3.3 again,

||dI_hu−u||b _2,

Cb.h|u|_2,Kh

k,`. (4.11)

Then (4.10) and (4.11) imply

|I_hu−u|_1,∂Kh k,` .√

h|u|_2,Kh k,`.

We conclude the proof by squaring this inequality and summing up for k, ` ∈

{0, . . . , n−1}.

Lemma 4.3. — Let us assume thatf ∈ C(S). Then

||f||Rh6√

2h^−1/2||f||∞,S. (4.12) Proof. —Let us use the notation of the proof of Lemma 2.1. Then

||f||²_L

` = Z 1

0

|f(x, `h)|²dx6||f||²_∞,S. (4.13) Obviously we have the same estimate for||f||²_C

k. This leads to

||f||²_R_h =

n−1

X

`=1

||f||²_L_`+

n−1

X

k=1

||f||²_C_k 62n||f||²_∞,S.

Sinceh= 1/n, we obtain the expected result.

Proposition 4.4. — Letuh ∈Vh denote the solution of (2.12) andUh ∈Wh

the solution of (4.2). Let us assume that the datumf belongs toC(S). Then

||uh−U_h||1,Rh .√

h||f||∞,S. (4.14)

Proof. —It can be proven (see for example Theorem 3.1.6 in [6]) that

||u_h−U_h||_1,R_h .h|u_h|_2,R_h. (4.15) Butuh is a solution of (2.2), so

|uh|_2,R_h =|f˜|_R_h= 1

2||f||_R_h. (4.16)

Due to Lemma 4.3,

|u_h|_2,R_h .h^−1/2||f||_∞,S. (4.17)

The aim then follows from (4.15) and (4.17).

Theorem 4.5. — Let Uh ∈ Wh denote the solution of (4.2), and RhUh be defined by(4.4). Let us assume that the solutionuof the continuous problem (2.1) belongs toH³(S), and the datumf belongs toC(S). Then

||u−RhUh||_1,S .h(||u||_3,S +||f||_∞,S). (4.18)

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Proof. —As the trace of I_hu−R_hU_h is equal to 0 on∂S, we have

||Ihu−RhUh||_1,S .|Ihu−RhUh|_1,S. (4.19) Lemma 4.1 leads to

||Ihu−RhUh||_1,S .

√

h|Ihu−Uh|_1,R_h .

√

h(|Ihu−u|1,R_h+|u−uh|1,R_h+|uh−Uh|1,R_h)(4.20). We deduce from Lemma 4.2, Corollary 3.6 and Proposition 4.4 that

||Ihu−RhUh||1,S .h(||u||3,S+||f||∞,S). (4.21) On the other hand, thanks to Theorem 3.1.6 of [6],

||u−I_hu||1,S .h|u|2,S. (4.22)

We conclude with (4.21) and (4.22).

4.2. A more regular solution. For more regular solutions, we will exploit the analogy with a finite difference scheme to get a pointwise convergence result.

For everyv∈ N_hînt, we define λv ∈Wh such thatλv(v) = 1 andλv(v⁰) = 0, for all v⁰∈ Nh such thatv⁰6=v. Remark that the support ofλv is included in{¯ei;i∈ Iv} and that the set{λv, v∈ N_hînt}forms a basis of the spaceWh. The stiffness matrix Mh of problem (4.2) is easily computed. More precisely, we enumerate the interior nodes v ∈ N_hînt line by line, namely let us denote v1 = (h, h), v2 = (2h, h), . . ., vn−1 = ((n−1)h, h), vn = (h,2h), vn+1 = (2h,2h), . . . , v2n−2 = ((n−1)h,2h), . . .v_(n−1)2 = ((n−1)h,(n−1)h). LetMh denote the stiffness matrix such that

(Mh)i,j=ah(λv_i, λv_j),∀i, j∈ {1, . . . ,(n−1)²}.

ThenM_his a symmetric matrix that can be written Mh= 1

h

A˜h (4.23)

where

A˜_h=







A1,1 A1,2 0 ... 0

A_2,1 A_2,2 A_2,3 . .. 0

0 A_3,2 A_3,3 . .. 0

... . .. . .. . .. ... 0 . . . 0 A_n−1,n−2 A_n−1,n−1







. (4.24)

The blocks Ak,l are symmetric matrices of dimension (n−1) and satisfy for all k ∈ {1, . . . , n−1}, A_k,k−1 = A_k−1,k = −In−1 (I_n−1 is the identity matrix of dimension (n−1)), and

Ak,k=







4 −1 ... 0

−1 4 . .. ... ... . .. . .. −1 0 . . . −1 4







. (4.25)

Setm= (n−1)²(for shortness we skip the dependence ofmonh), asUhbelongs toWh, it can be expressed in the basis (λv_k)k=1,...,m as follows:

Uh=

m

X

k=1

Uh(vk)λv_k. As usual,U_his the solution of problem (4.2) if and only if

M_hUf_h=Ff_h (4.26)

whereUfh= (Uh(v1), ..., Uh(vm))^> andFfh= (F(λv₁), ..., F(λv_m))^>.

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Now we want to check that the values ofU_hat the nodes are a good approximation of the values of u. To this end, we observe that M_h is closely related to the matrix obtained by using the finite difference method to approximate the continuous problem (2.1). Indeed ifD_hdenote the approximation of the solutionuof (2.1) with the finite difference method, thenD_h is solution of the linear system [14]

AhDh=Fh, (4.27)

where

Ah= 1

hMh= 1 h²

A˜h (4.28)

with ˜A_h defined by (4.24) andF_h= (f(v₁), ..., f(v_m))^>.

For further purposes, we state the following two results (see Lemma 6.2 of [3]

for the proof of the first result, the second one being proved in a fully similar way, see also Property 1.20 of [22]).

Proposition 4.6. — LetA∈R^m×m satisfying the following conditions (1)∀i6=j, a_ij 60, and

(2)∀i= 1, . . . , m,Pm

j=1a_ij >0,

thenA is a monotone matrix, i. e., ifX = (x_i)_i=1,...,m ∈R^m is such thatAX>0 (in the sense that(Ax)_i>0, for alli= 1, . . . , m), thenX >0.

Remark 4.7. — The result of Proposition 4.6 still holds if the assumption (2) is replaced by

(2’)Ais a regular matrix and for alli= 1, . . . , m,Pm

j=1a_ij>0.

Corollary 4.8. — A˜h andAh given by (4.28) are monotone matrices.

Proof. —Since ˜Ah is symmetric and positive definite, ˜Ah is a regular matrix.

Moreover, ˜Ah fulfils condition (1) of Proposition 4.6 and condition (2’) of Remark

4.7.

Proposition 4.9. — Consider u the solution of Problem (2.1) and suppose thatu∈ C³(S). SetU = (u(v₁), ..., u(v_m))^> and letD_h be the solution of equation (4.27). Then

A_h(U−D_h) =η(u), withη(u) = (η(u)(v1), ..., η(u)(vm))^>, where

η(u)(x_i, y_i) =−h 6

∂³u

∂x³(x_i+θ_i,1h, y_i)−∂³u

∂x³(x_i−θ_i,2h, y_i) +∂³u

∂y³(xi, yi+θi,3h)−∂³u

∂y³(xi, yi−θi,4h)

with someθ_i,j ∈]0; 1[and(x_i, y_i)being the coordinates ofv_i. Moreover, one has

||η(u)||_∞= max

i=1,...,m|η(u)(xi, yi)|.hM3, whereM₃=||D³u||_∞= max

(x, y)∈ S

|α|= 3

|D^αu(x, y)|.

Proof. —This result is just a consequence of Taylor’s formula. We refer the

reader to [14] for the details.

Lemma 4.10. — Let W = (w1, . . . , wm)^>, G = (g1, . . . , gm)^> ∈ R^m be such thatAhW =G. Then, for alli= 1, . . . , m,

|w_i|6 1

4[x_i(1−x_i) +y_i(1−y_i)]||G||_∞ (4.29) where(xi, yi)are the coordinates ofvi and||G||_∞= max

i=1,...,m|gi|.

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Proof. —Let us consider ˜wdefined by ˜w(x, y) = ¹₄(x(1−x) +y(1−y))˜h,with

˜h=||G||_∞. We notice that ^∂_∂x²^w₂^˜ = ^∂_∂y²^w₂^˜ =−¹₂˜h, and thus ˜w∈ C⁴(S) is solution of −∆ ˜w= ˜h in S

˜

w= 0 on∂S.

We writeD_h^wfor the solution of the following finite difference problem:

AhD^w_h = ˜H

where ˜H = ˜h(1,· · ·,1)^>. Owing to Proposition 4.9 and noticing thatη( ˜w) = 0, we get

Ah( ˜W −D_h^w) = 0

where ˜W = (wf1, ...,gwm)^> withwfi = ˜w(vi), for every i∈ {1, . . . , m}. Comparing the two last identities, we obtain

A_hW˜ = ˜H. (4.30)

Since for alli∈ {1, . . . , m},eh=||G||_∞>|g_i|, we deduce from (4.30) that (A_hW˜)_i>

|(AhW)i|. This implies that Ah( ˜W −W)>0 and Ah( ˜W +W)>0. As Ah is a monotone matrix, this leads to ˜W −W > 0 and ˜W +W > 0. In other words, for every i = 1, ..., m, fwi >|wi| and thanks to the definition of ˜w, we finally get

(4.29).

Proposition 4.11. — The finite difference problem (4.27) admits a unique solutionDh.

Assume that u∈ C³(S) and set Dh = (Dh(v1), ..., Dh(vm))^>, then for every i = 1, . . . , m, we have

|u(vi)−D_h(v_i)|.M₃h[x_i(1−x_i) +y_i(1−y_i)], (4.31) with M₃ = ||D³u||_∞, (xi, yi) denotes the coordinates of vi, and the numerical constant appearing here (and below) is independent ofu,handi.

Proof. —Due to Corollary 4.8, Ah is a monotone matrix and consequentlyAh

is regular. This implies that there exists a unique solutionDh of (4.27).

Owing to Proposition 4.9,

A_h(U−D_h) =η(u).

Let us apply Lemma 4.10 withW =U−DhandG=η(u). Then

|(U−D_h)(v_i)|6 1

4[x_i(1−x_i) +y_i(1−y_i)]||η(u)||∞.

The conclusion follows directly from the estimates of||η(u)||_∞given in Proposition

4.9.

Proposition 4.12. — Iff ∈ C¹(S), then

F(λv) h −f(v)

.hk∇fk_∞,S,∀v∈ N_h^int. Proof. —Letv∈ N_h^int, it is easy to prove that for alli∈ Iv,

Z

e_i

λv(x)dx= h

2. (4.32)