Overlapping domain decomposition methods for elliptic ... - Springer Link

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 121, No. 4, November 2011, pp. 481–493. c Indian Academy of Sciences 

Overlapping domain decomposition methods for elliptic quasi-variational inequalities related to impulse control problem with mixed boundary conditions MOHAMED HAIOUR1 and SALAH BOULAARAS2 1 Department of Mathematics, Faculty of Science, University of Annaba, B.P. 12 Annaba 23000, Algeria 2 Hydrometeorological Institute of Formation and Research, B.P. 7019 Seddikia, Oran 31025, Algeria E-mail: [email protected]; [email protected]

MS received 23 April 2010; revised 6 July 2011 Abstract. In this paper we provide a maximum norm analysis of an overlapping Schwarz method on non-matching grids for quasi-variational inequalities related to impulse control problem with mixed boundary conditions. We provide that the discretization on every sub-domain converges in uniform norm. Furthermore, a result of approximation in uniform norm is given. Keywords. Domain decomposition; geometrical convergence; quasi-variational inequalities; impulse control; error analysis.

1. Introduction Schwarz method has been invented by Herman Amandus Schwarz in 1890. This method has been used to solve the stationary or evolutionary boundary value problems on domains which consists of two or more overlapping subdomains (see [1–3, 6, 8–12]). The solution is approximated by an infinite sequence of functions which results from solving a sequence of stationary or evolutionary boundary value problems in each of the subdomain. In this work we provide a maximum norm analysis of an overlapping Schwarz method on non-matching grids for quasi-variational inequalities related to impulse control problem with respect to the mixed boundary conditions. We can state our problem as follows: Find u ∈ H 1 () solution of ⎧ −u − f ≤ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u − Mu ≤ 0 Mu ≥ 0, ⎪ ⎨ ⎪ (u − f ) (u − Mu) = 0 in , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂u = ϕ in 0 and u = 0 in / 0 , ∂η 481

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Mohamed Haiour and Salah Boulaaras

where  is a smooth bounded domain of R2 with boundary  and f is a regular function and M is an operator given by Mu = k +infξ ≥0,x+ξ ∈¯ u (x + ξ ) where k > 0 and ξ ≥ 0 means that ξ = (ξ1 , ξ2 , . . . , ξn ) with ξi ≥ 0, and 0 is the part of the boundary defined by ¯ Finally, ∂u = ∇u · η, such that η is / }. 0 = {x ∈ ∂ =  such that ∀ξ > 0, x + ξ ∈ ∂η the normal vector. The symbol (., .) stands for the inner product in L 2 (), (., .)0 stands for the inner product in L 2 (0 ). On the analytical side, quasi-variational inequalities have been extensively studied in the last three decades (see [1, 2, 7–11]) and for the numerical approximations and computational aspects we have a few results (see [3–6, 12]). In the present paper, we provide a new approach for the finite element approximation of an overlapping Schwarz method on non-matching grids for the quasi-variational inequalities related to impulse control problem. We consider a domain which is the union of two overlapping subdomains, where each sub-domain has its own generated triangulation. The grid points on the subdomain boundaries need not match the grid points from the other sub-domains. Under a discrete maximum principle [10], we show that the discretization on each sub-domain converges quasi-optimally in the L ∞ -norm. The paper is organized as follows: In §2, we state the continuous alternating Schwarz sequence for quasi-variational inequalities and define their respective finite element counterparts in the context of overlapping grids. In §3, we provide the error analysis of the overlapping domain decomposition methods, where simple proofs to main fundamental theorems are proved. Then the geometrical convergence of the problem is established. Furthermore, an error estimate for each sub-domain is derived in a uniform norm.

2. The Schwarz method for the elliptic quasi-variational inequalities. In this section, we introduce some definitions and some classical results related to quasivariational inequalities. 2.1 Elliptic quasi-variational inequalities Let  be a convex domain in R2 with sufficiently smooth boundary ∂. We consider the following obstacle problem ⎧ −u − f ≤ 0, ∀u ∈ H 1 () , in , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u − Mu ≤ 0, Mu ≥ 0, ⎪ ⎪ ⎪ ⎨ Mu = k + inf u (x + ξ ) , ⎪ ⎪ ¯ ξ ≥0,x+ξ ∈ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂u ⎪ ⎪ ⎩ = ϕ in 0 and u = 0 in / 0 . ∂η We are also given the right-hand side f such that g ∈ L ∞ () .

(2.1)

Quasi-variational inequalities related to impulse control problem

483

M is an operator given by Mu = k +

inf

¯ ξ ≥0,x+ξ ∈

u (x + ξ ) .

(2.2)

Applying Green’s formula, (2.1) can be transformed to the following elliptic variational inequalities ⎧ a (u, v − u) − ( f, (v − u)) − (ϕ, (v − u))0 ≥ 0, on , v ∈ V ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u − Mu ≤ 0 Mu ≥ 0, ⎪ ⎪ ⎪ ⎨ . Mu = k + inf u (x + ξ ) , ⎪ ⎪ ¯ ξ ≥0,x+ξ ∈  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂u ⎪ ⎪ = ϕ in 0 and u = 0 in / 0 , ⎩ ∂η (2.3) where V is the non empty convex set defined by   ∂v V = v ∈ H 1 () : = ϕ in 0 , v ≤ Mu on  and v = 0 in / 0 , ∂η (2.4) where ϕ is a regular function defined on 0 . Thus, it can be easily deduce that  a (u, v) = ∇u · ∇vdx = (∇u, ∇v) , 

 ( f, v) =

f · vdx 

and

 (ϕ, v)0 =

ϕ · vdσ. 0

Theorem 1 [7]. Under the previous assumptions, the problem (2.1) has an unique solution. Moreover it satisfies: u ∈ W 2, p ,

2 ≤ p ≤ ∞.

2,∞ Theorem 2 [7]. If ϕ ∈ W 2,∞ (∂) and Mϕ ∈ Wloc () (resp. ϕ is locally 1,∞ 2,∞ semi-concave). Then u ∈ Wloc () (resp. u ∈ Wloc ()).

Let V h be the space of finite elements consisting of continuous piecewise linear functions and K ϕh the non-empty discrete convex set associated to K ϕ defined as K ϕh = {u h ∈ V h : u h = πh ϕ on 0 , u h ≤ rh Mu h on  and u = 0 in / 0 }, (2.5)

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Mohamed Haiour and Salah Boulaaras

where πh is an interpolation operator on 0 , and rh is the usual finite element restriction operator on . The discrete counterpart of (2.3) consists of finding u h ∈ K ϕh such that ⎧ ⎪ ⎪ a (u h , vh − u h ) − ( f, (vh − u h )) − (ϕ, (vh − u h ))0 ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ inf u h (x + ξ ) , Mu h = k + ⎪ ⎪ ⎪ ¯ ξ ≥0,x+ξ ∈ ⎨ (2.6) ⎪ ⎪ u h − rh Mu h ≤ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂u h ⎪ ⎪ = ϕ in 0 and u h = 0 in / 0 . ⎩ ∂η The lemma below establishes a monotonicity property of the solution of (2.3) with respect to the obstacle defined as an impulse control problem. Lemma 1 [7]. If we have u h ≤ u˜ h in the K ϕh , then Mu h ≤ M u˜ h and we have ∀u ∈ K ϕ , λ ∈ R, M (u + λ) = Mu + λ.

(2.7)

Remark 1. Under the previous hypotheses we have the following inequality

Mu − M u

˜ L ∞ () ≤ u − u

˜ L ∞ () ,

∀u, u˜ ∈ K ϕ .

(2.8)

Proof. We have u ≤ u˜ + u − u

˜ L ∞ () . Using the Lemma 1, we get

Mu ≤ M u˜ + u − u

˜ L ∞ () = M u˜ + u − u

˜ L ∞ () , thus Mu − M u˜ ≤ u − u

˜ L ∞ () . Similarly, interchanging the roles of Mu and M u, ˜ we get M u˜ − Mu ≤ u − u

˜ L ∞ () . Hence

Mu − M u

˜ L ∞ () ≤ u − u

˜ L ∞ () .  Notation 1. Let (Mξ, ϕ) , (M ξ˜ , ϕ) ˜ be a pair of data, and ξ = σ (Mξ, ϕ) , ξ˜ = ˜ σ (M ξ , ϕ) ˜ be the corresponding solutions to the following quasi-variational inequalities (QVI): a (ξ, v − ξ ) ≥ ( f, v − ξ ) + (ϕ, (v − ξ ))0 ,

∀v ∈ K ϕ

Quasi-variational inequalities related to impulse control problem

485

and ˜ (v − ξ ))0 , a(ξ˜ , v − ξ˜ ) ≥ ( f, v − ξ˜ ) + (ϕ,

∀v ∈ K ϕ˜ .

Then, the following comparison result holds. Lemma 2. If ϕ ≥ ϕ, ˜ then σ (Mξ, ϕ) ≥ σ (M ξ˜ , ϕ). ˜ Proof. Let v = min(0, ξ − ξ˜ ). In the region where v is negative (v < 0), we have ξ ≤ ξ˜ ≤ M ξ˜ ≤ Mξ, which means that the obstacle is not active for u. So, for v, we have a (ξ, v) = ( f, v) + (ϕ, v)0 ,

(2.9)

ξ˜ + v ≤ M ξ˜ .

(2.10)

a(ξ˜ , v) ≥ ( f, v) + (ϕ, v)0 .

(2.11)

So

Subtracting (2.9) and (2.11) from each other, we obtain a(ξ − ξ˜ , v) ≥ 0.

(2.12)

a (v, v) = a(ξ − ξ˜ , v) = −a(ξ˜ − ξ, v) ≤ 0,

(2.13)

But

so v=0 and consequently, ξ ≥ ξ˜ 

which completes the proof. The proof for the discrete case is similar. PROPOSITION 1 Under the previous hypotheses, we have the following inequality

u − u

˜ L ∞ (i ) + ϕ − ϕ

˜ L ∞ (∂i ∩ j ) , ˜ L ∞ (i ) ≤ Mu − M u

such that i = j, i, j = 1, 2.

(2.14)

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Mohamed Haiour and Salah Boulaaras

Proof. Setting β = Mu − M u

˜ L ∞ (i ) + ϕ − ϕ

˜ L ∞ (∂i ∩ j ) , such that i = j, i, j = 1, 2,

(2.15)

we have Mu ≤ M u˜ + Mu − M u˜ ≤ M u˜ + |Mu − M u| ˜ ≤ M u˜ + Mu − M u

˜ L ∞ (i ) ≤ M u˜ + Mu − M u

˜ L ∞ (i ) + ϕ − ϕ

˜ L ∞ (∂i ∩ j ) . Hence Mu ≤ M u˜ + β. On the other hand, we have ϕ ≤ ϕ˜ + ϕ − ϕ˜ ≤ ϕ˜ + |ϕ − ϕ| ˜ ≤ ϕ˜ + ϕ − ϕ

˜ L ∞ (∂i ∩ j ) ≤ ϕ˜ + ϕ − ϕ

˜ L ∞ (∂i ∩ j ) + Mu − M u

˜ L ∞ (i ) , that is to say, ϕ ≤ ϕ˜ + β.

(2.16)

Since σ is increasing in L ∞ (), we have σ (Mu, ϕ) ≤ σ (M u˜ + β, ϕ˜ + β) ≤ σ (M u, ˜ ϕ) ˜ + β. Hence σ (Mu, ϕ) − σ (M u, ˜ ϕ) ˜ ≤ β. Similarly, interchanging the roles of the couples (Mu, ϕ) and (M u, ˜ ϕ), ˜ we get σ (M u, ˜ ϕ) ˜ − σ (Mu, ϕ) ≤ β. The proof for the discrete case is similar.



Theorem 3 [3, 4]. Under the previous hypotheses and the maximum principle [5], there exists a constant C independent of h such that

u − u h L ∞ () ≤ Ch 2 |log h|2 . 2.2 The continuous Schwarz sequences Let  be a bounded open domain in R2 and we assume that  is smooth and connected. Then we decompose  into two sub-domains 1 , 2 such that  = 1 ∪ 2

(2.17)

Quasi-variational inequalities related to impulse control problem

487

and u satisfies the local regularity condition u |i ∈ W 2, p (i )

(2.18)

and we denote by  = ∂, 1 = ∂1 , 2 = ∂2 , γ1 = ∂1 ∩ 2 , γ2 = ∂2 ∩ 1 , 1,2 = 1 ∩ 2 . We consider the model obstacle problem. Find u ∈ K ϕ such that ⎧ ⎪ ⎪ a (u, v − u) − ( f, (v − u)) − (ϕ, (v − u))0 ≥ 0, ∀v ∈ K ϕ , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u − Mu ≤ 0 Mu ≥ 0, ⎪ ⎪ ⎪ ⎨ Mu = k + inf u (x + ξ ) , ⎪ ⎪ ¯ ξ ≥0,x+ξ ∈ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂u ⎪ ⎪ = ϕ in 0 and u = 0 in / 0 , ⎩ ∂η

(2.19)

where f is a given function in L ∞ (), and we will assume in this section that f ≥ 0, ϕ  0. Indeed this enables us to make such an assumption by adding constants to u and ϕ and a positive function to f. We define the following process:

Let u 0 ∈ K ϕ be given, and we define the alternating Schwarz sequences u 2n+1 on 1 such that u 2n+1 ∈ K ϕ solves ⎧ a(u 2n+1 , v − u 2n+1 ) − ( f, (v − u 2n+1 ))1 − (ϕ, (v − u 2n+1 ))0 ≥ 0, ∀v ∈ V, ⎪ ⎪ ⎪ ⎪ ⎨ (2.20) u 2n+1 − Mu 2n −1 ≤ 0 Mu ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎩ 2n+1 u = u 2n on ∂1 , v = u 2n on ∂1 and (u 2n ) on 2 such that u 2n ∈ K ϕ solves ⎧ a(u 2n , v − u 2n ) − ( f, (v − u 2n ))2 − (ϕ, (v − u 2n ))0 ≥ 0, ∀v ∈ V, ⎪ ⎪ ⎪ ⎪ ⎨ u 2n − Mu 2n−2 ≤ 0, Mu 2n−2 ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎩ 2n u = u 2n−1 on ∂2 , v = u 2n−1 on ∂2 .

(2.21)

2.3 Discretization For i = 1, 2, let τ h i be a standard regular and quasi-uniform finite element triangulation in i ; h i (h 1 = h 2 = h) being the mesh size. We assume that the two triangulations are mutually independent on 1,2 in the sense that a triangle belonging to one triangulation does not necessarily belong to the other.

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Mohamed Haiour and Salah Boulaaras

hi Let V h i be the space of continuous piecewise

linear functions on τ which vanish on ¯ i , we define i ∩ ∂ j , i = j, i, j = 1, 2. For w ∈ C ∂ 

Vwh i = {vh ∈ V h i : vh = πh i (w) on i ∩ ∂ j ; vh = 0 in / 0 ; i = j, i, j = 1, 2},

(2.22)

where πh i denotes the interpolation operator on ∂i . We assume that the respective matrices resulting from the discretization of problems (2.20) and (2.21) are M-matrix [5]. We define the discrete counterparts of the continuous Schwarz sequences defined in ∈ V h 2n such that (2.20) and (2.21), respectively by u 2n+1 h uh

⎧ a(u 2n+1 , vh − u 2n+1 ) − ( f, (vh − u 2n+1 ))1 ⎪ h h h ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2n+1 h ⎪ ⎪ ⎨ −(ϕ, (v − u h ))0 ≥ 0, ∀vh ∈ V u 2n , h

⎪ ⎪ 2n ⎪ u 2n+1 = u 2n ⎪ h on ∂1 , vh = u h on ∂1 , h ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2n+1 ≤ rh Mu 2n−1 uh h h and u 2n h ∈ V(u 2n−1 ) such that h ⎧ 2n 2n a(u h , vh − u 2n ⎪ h ) − ( f, (vh − u h ))2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2n h ⎪ , ⎪ ⎨ −(ϕ, (v − u h ))0 ≥ 0, ∀v ∈ V(u 2n−1 ) h

⎪ ⎪ 2n−1 ⎪ ⎪ u 2n on ∂2 , vh = u 2n−1 on ∂2 , ⎪ h = uh h ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2n . u h ≤ rh Mu 2n−2 h

(2.23)

(2.24)

3. Error analysis This section is devoted to the proof of the main result of the present paper. We need to introduce an auxiliary sequence of discrete quasi-variational inequalities first and then prove the two fundamental theorems. For ζh0 = u 0h ∈ K ϕh , we define the sequences (ζh2n+1 ) such that ζh2n+1 ∈ V(uh 2n ) solves ⎧ a(ζh2n+1 , vh − ζh2n+1 ) − ( f, (vh − ζh2n+1 ))1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2n+1 ⎪ ⎪ ))0 ≥ 0, ∀vh ∈ V h 2n , ⎨ −(ϕ, (v − ζh (u h )

⎪ ⎪ 2n ⎪ ζh2n+1 = u 2n ⎪ h on ∂1 , vh = u h on ∂1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2n+1 ≤ rh Mu 2n−1 ζh h

(3.1)

Quasi-variational inequalities related to impulse control problem

489

and (ζh2n ) such that ζh2n ∈ V hu 2n−1 solves ( ) ⎧ a(ζh2n , vh − ζh2n ) − ( f, (vh − ζh2n ))2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2n h ⎪ ⎪ ⎨ −(ϕ, (v − ζh ))0 ≥ 0, ∀vh ∈ V 2n−1 , (u h

)

(3.2)

⎪ ⎪ ⎪ ζh2n = u 2n−1 on ∂2 , vh = u 2n−1 on ∂2 , ⎪ h h ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2n . ζh ≤ rh Mu 2n−2 h 3.1 Convergence proof via the maximum principle We introduce the sets ⎧ 2n−1 2n h 2n 2n ⎪ on ∂2 , ⎨ u h ∈ V(u 2n−1 ) : −u h ≤ f, u h = u h h 2n T = ⎪ ⎩ 2n−2 u 2n on 2 and u 2n h ≤ r h Mu h h = 0 in / 0 and T

2n+1

=

⎫ ⎪ ⎬ ⎪ ⎭

⎧ 2n+1 ⎫ ⎪ ∈ V h 2n : −u 2n+1 ≤ f, u 2n+1 = u 2n ⎨ uh ⎬ h on ∂1 , ⎪ h h (u ) h

⎪ ⎩

u 2n+1 ≤ rh Mu 2n−1 on 1 and u 2n+1 = 0 in / 0 . h h h

⎪ ⎭

.

2n+1 Lemma 3 [12]. If A is the M-matrix and u 2n ) is the solution (2.23), (resp. h (resp. u h 2n+1 2n 2n (2.24)), then u h (resp. ∇u h ) is the minimal of T (resp. T 2n+1 ).

Theorem 4. Let u h be a solution of (2.6). Then the iterative sequence {u 2n h } (resp. 2n+1 2n+1 2n 2n 2n+1 ∈ T ) and u h ≤ u 2n+2 ≤ {u h }) is monotone; that is, u h ∈ T (resp. u h h 2n 0 uh ≤ · · · ≤ uh . Proof. We take u 0h = u h | 2 such that −u 0h = f . We know that if u 0h ≤ rh Mu h then (−u 0h − f ) |2 ≤ 0, i.e. (∇u 0h , ∇(vh − u 0h ))2 − ( f, (vh − u 0h ))2 − (ϕ, (v − u 0h ))0 ≥ 0. Therefore u 0h ∈ T 0 . From Lemma 2 we know that u 2h is the minimal element of T 0 . So u 2h ≤ rh Mu 0h and yields u 2h ≤ u 0h . By induction, for index n we obtain 2n−2 u 2n ≤ · · · ≤ u 2h ≤ u 0h = u h . h ≤ uh



We know that if u 3h ≤ rh Mu 1h , then −u 3h − f |1 ≤ 0, i.e. (∇u 3h , ∇(vh − u 3h ))1 − ( f, (vh − u 3h ))1 − (ϕ, (v − u 0h ))0 ≥ 0.

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Mohamed Haiour and Salah Boulaaras

Therefore u 3h ∈ T 3 . From Lemma 4 we know that u 3h is the minimal element of T 3 which yields u 3h ≤ u 1h . By induction, for index n we obtain ≤ u 2n−1 ≤ · · · ≤ u 1h . u 2n+1 h h  Lemma 4. If A = (ai j )i, j={1...N } is the M-matrix, then there exists two constants k1 , k2 , i.e., k1 = sup {wh (x), x ∈ γ2 } ∈ (0, 1), and k2 = sup {wh (x), x ∈ γ1 } ∈ (0, 1) such that sup|u h − u 2n+1 | ≤ k1 sup|u h − u 2n h | h

(3.3)

sup|u h − u 2n+1 | ≤ k2 sup|u h − u 2n h |. h

(3.4)

γ1

γ1

and γ2

γ2

Proof. Setting M1 = supγ1 |u 2n+1 − u h | and M = supγ1 |u h − u 2n h |, we may suppose that h M1 = 0. We prove M1 < M.

(3.5)

If (3.3) is not true then there exists xi0 ∈ γ1 such that |u 2n+1 (xi0 ) − u(xi0 )| = M1 ≥ M. Hence, we have (noting aii > 0, ai j ≤ 0 for i = j because A is the M-matrix) 0≥

N N   aii0 (u 2n+1 − u ) ≥ aii0 ≥ 0. h h i=1

i=1

We know by Theorem 3 that u 2n+1 ≥ u h which implies that h  aii0 |u 2n+1 − u h | − M1 = 0. h i =i 0

Therefore − u h | = M1 , |u 2n+1 h

if aii0 = 0.

(3.6)

Since (ai j )i, j={1...N } = A is irreducible, there exists xi1 , xi2 , . . . , xis ∈ γ2 such that ai0 i1 , ai1 i2 , . . . , ais lk = 0.



xi1 − u h xi1 | = M1 Similarly, we get We know by (3.6) that |u 2n+1 h







xis − u h xis | |u 2n+1 xi2 − u xi2 | = · · · = |u 2n+1 h



= |u 2n+1 x l k − u h x l k | = M1 . h Hence we have 0≥

N N   ali |u 2n+1 − u | ≥ ali M1 > 0, h h i=1

i=1

Quasi-variational inequalities related to impulse control problem which is a contradiction with (3.6). The proof for (3.4) case is similar.

491 

Remark 2. The demonstration of Lemma 4 is an adaptation of the one in [12] given for the problem of variational inequality. This lemma remains true for the problem introduced in this paper. The main convergence result is given by the following theorem. Theorem 5. The sequences (u 2n+1 ), (u 2n h ), n ≥ 0 produced by the Schwarz alternating h method converge geometrically to the solution u of the obstacle problem (2.3). More precisely, there exist k1 , k2 ∈ (0, 1) which depend on (1 , γ2 ) and (2 , γ1 ) such that for all n ≥ 0, | ≤ k1n k2n sup|u h − u 0h | sup|u h − u 2n+1 h

(3.7)

n n−1 0 sup|u h − u 2n h | ≤ k1 k2 sup|u h − u h |.

(3.8)

γ1

¯1 

and γ2

¯2 

Proof. Under Lemma 4, we have |u h − u 2n+1 | ≤ wh (x) sup|u h − u 2n h |. h γ1

Hence sup|u h − u 2n+1 | ≤ k1 sup|u h − u 2n h |. h γ1

Thus

sup|u h − u 2n+1 | h γ1

γ1

≤

k1n+1 k2n sup|u h γ1

− u 0h |

and we also have 2n−1 |. |u h − u 2n h | ≤ w (x) sup|u h − u h γ2

Hence 2n−1 |, sup|u h − u 2n h | ≤ k2 sup|u h − u h γ2

γ2

that is to say, n n 0 sup|u h − u 2n h | ≤ k1 k2 sup|u h − u h |. γ2

γ2

Equations (3.7) and (3.8) follow from the maximum principle which yields | = sup|u h − u 2n+1 | = sup|u h − u 2n sup|u h − u 2n+1 h | for n ≥ 0 h h ¯1 

γ1

γ1

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Mohamed Haiour and Salah Boulaaras

and sup|u h − u 2n+1 | = sup|u h − u 2n+1 | = sup|u h − u 2n h | for n ≥ 0. h h ¯2 

γ2

γ2

Hence sup|u h − u 2n+1 | ≤ k1n k2n sup|u h − u 0h | h ¯1 

γ1

and n n−1 0 sup|u h − u 2n h | ≤ k1 k2 sup|u h − u h |. ¯2 

γ2

 3.2 Error estimate for the quasi-variational inequalities Theorem 6. Let u be a solution of problem (2.3). Then there exists a constant C independent of both h and n such that

u − u 2n+1

L ∞ (¯ 1 ) ≤ Ch 2 | log h|3 h

(3.9)

and 2 3

u − u 2n ¯ 2 ) ≤ Ch | log h| . h L ∞ (

(3.10)

Proof. Setting k = k1 = k2 , and using Theorems 2 and 4 we have

u − u 2n+1

L ∞ (¯ 1 ) ≤ u − u h L ∞ (¯ 1 ) + u h − u 2n+1

L ∞ (¯ 1 ) h h ≤ u − u h L ∞ (¯ 1 ) + k 2n u h − u 0h L ∞ (γ1 ) ≤ Ch 2 | log h|2 + k 2n u h − u 0h L ∞ (γ1 ) ≤ Ch 2 | log h|2 + k 2n ( u − u h L ∞ (γ1 ) + u − u 0h L ∞ (γ1 ) ) ≤ Ch 2 | log h|2 + Ch 2 k 2n |log h|2 and also setting k 2n ≤ | log h|, we get

L ∞ (¯ 1 ) ≤ Ch 2 | log h|3 .

u − u 2n+1 h The proof for the (3.10) case is similar.



Quasi-variational inequalities related to impulse control problem

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