Motivated by experiments on cell segregation, we present a two-species model of interacting particles, aiming at a quantitative description of this phenomenon. Under precise scaling hypothesis, we derive from the microscopic model a macroscopic one and we analyze it. In particular, we determine the range of parameters for which segregation is expected. We compare our analytical results and numerical simulations of the macroscopic model to direct simulations of the particles, and comment on possible links with experiments.
Citation: Julien Barré, Pierre Degond, Diane Peurichard, Ewelina Zatorska. Modelling pattern formation through differential repulsion[J]. Networks and Heterogeneous Media, 2020, 15(3): 307-352. doi: 10.3934/nhm.2020021
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Motivated by experiments on cell segregation, we present a two-species model of interacting particles, aiming at a quantitative description of this phenomenon. Under precise scaling hypothesis, we derive from the microscopic model a macroscopic one and we analyze it. In particular, we determine the range of parameters for which segregation is expected. We compare our analytical results and numerical simulations of the macroscopic model to direct simulations of the particles, and comment on possible links with experiments.
This paper is concerned with the finite element approximation of system of J = 2 quasi-variational inequalities QVIs with term sources and obstacles depending on solution: Find a vector U=(u1,u2)∈(H10(Ω))2 satisfying
ai(ui,v−ui)≥(fi(ui),v−ui);v∈H10(Ω) | (1.1) |
v,ui≤Mui;ui≥0. |
Where Ω is a bounded smooth domain of RN with N≥ 1, each ai(.,.) is a continuous elliptic bilinear form, (.,.) is the inner product in L2(Ω) and each fi is a regular, nonlinear functional depending on solutions. The obstacle M provide the coupling between the unknowns u1; u2
Mui=k+infμ≠iuμ; |
k is a positive number. We point out that in the case where fi are independent of the solution, the system (1.1) coincides with that introduced by Bensoussan and Lions in [1] which arises in the management of energy production problems.
It is easy to note that the structure of system (1.1) is analogous to that of the classical obstacle problem [2] where the term source and obstacle are depending upon the solution sought. The terminology QVI being chosen is a result of this remark.
Numerical analysis of system of quasi-variational inequalities where term sources not depending on solutions were achieved in several works, we refer to [3,4,5,6,7,8] for system of quasi-variational inequalities with coercive or noncoercive operators.
For results on systems related to evolutionary Hamilton-Jacobi-Bellman equation we refer to [9,10,11].
The main objective of this paper is to show that problem (1.1) can be properly approximated by a finite element method and an optimal L∞-error estimates is derived, which coincides with the optimal convergence order of elliptic variational inequalities of an obstacle type problem [12].
The approximation is carried out by first introducing a modified Bensoussan-Lions type iterative scheme depending on parameters which is shown to converge geometrically to the continuous solution. By a symmetrical approach, using the standard finite element method and a discrete maximum principle (DMP), the geometric convergence of the discrete modified Bensoussan-Lions type iterative scheme depending upon parameters is given as well. An L∞-error estimates is then established combining the geometric convergence of both the continuous and discrete iterative schemes and the known uniform error estimates in elliptic VIs.
It is worth mentioning that even the guiding idea of this paper rests on the algorithmic approach followed in many papers cited above, the treatment of the geometric convergence of both continuous and discrete schemes is totally different because of the nonlinear nature of terms sources. Also, it is used for the first time for a system of QVIs.
An outline of this paper is as follows: In section 2, we lay down some definitions and classical results related to variational inequalities and prove a Lipschitz continuous and discrete dependency with respect to the source term, the boundary condition and the obstacle. Section 3 discusses the continuous Bensoussan-Lions type iterative scheme and proves its geometrical convergence. In Section 4, we establish the finite element counter parts of the continuous system and the continuous Bensoussan-Lions type iterative scheme respectively and the geometrical convergence of the discrete scheme. Section 5 is devoted the L∞-error analysis of the method.
We are given functions aijk(x),aik(x),ai0(x),1≤i≤2 sufficiently smooth functions such that 1≤j,k≤N
∑1≤j,k≤Naijk(x)ξjξk≥α|ξ|2,ξ∈RN,α>0 |
ai0(x)≥βi>0,(x∈Ω) | (2.1) |
where βi is a positive constant. We define the bilinear forms: For all u,v∈H10(Ω)
ai(u,v)=∫Ω(∑1≤j,k≤Naijk(x)∂u∂xj∂v∂xk+∑Nk=1aik(x)∂u∂xkv+ai0(x)uv)dx | (2.2) |
We are given right-hand sides
fisuchthatfi∈L∞(Ω),fi≥f0>0, |
a nonlinear functional and Lipschitz continuous on R; that is
|fi(x)−fi(y)|≤ki|x−y|,∀x,y∈R, |
such that
αi=kiβi<1, | (2.3) |
where βi is a constant defined in (2.1). For W=(w1,w2)∈(L∞+(Ω))2 we introduce the norm
‖W‖∞=max1≤i≤2‖wi‖L∞(Ω). |
Let be Ω a bounded polyhedral domain of R2 or R3 with sufficiently smooth boundary ∂Ω. We consider the bilinear form of the same form of those defined in (2.2), the linear form
(f,v)=∫Ωf(x)v(x)dx, | (2.4) |
The right hand side
f∈L∞(Ω), | (2.5) |
the obstacle
ψ∈W2,∞(Ω)andψ≥0, | (2.6) |
the boundary condition g∈L∞(∂Ω) and the nonempty convex set
Kg={v∈H1(Ω)suchthatv=gon∂Ωandv≤ψonΩ}. | (2.7) |
We consider the variational inequality V.I.: Find u∈Kg such that
a(u,v−u)≥(f,v−u),∀v∈Kg. | (2.8) |
Proposition 1 Let (f,g,ψ); (˜f,˜g,˜ψ)be a pair of data and ζ=σ(f,g,ψ); ˜ζ=σ(˜f,˜g,˜ψ) the corresponding solution to (2.8). If f≤˜f in Ω, g≤˜g on ∂Ω and ψ≤˜ψ then, ζ≤˜ζ in Ω.
Proof. The proof is an adaptation of the proof of the monotonicity property of the solution of Ⅵ with nonlinear source term (see [13]). According to [14], ζ=max{ζ_} where {ζ_} is the set of all the subsolutions of ζ. Hence, ∀ζ_∈{ζ_}, ζ_ satisfies
a(ζ_,v)≤(f,v),∀v≥0withζ_≤ψandζ_≤g. |
By using the conditions f≤˜f in Ω, g≤˜g on ∂Ω and ψ≤˜ψ, we get
a(ζ_,v)≤(f,v)≤(˜f,v), |
with
ζ_≤ψ≤˜ψandζ_≤g≤˜gon∂Ω. |
Thus, ζ is a subsolution of ˜ζ=σ(˜f,˜g,˜ψ), that is ζ≤˜ζ in Ω.
This subsection is devoted to the establishment of a Lipschitz continuous dependence property of the solution with respect to the source term, the boundary condition and the obstacle by which we first, set out and demonstrate.
Proposition 2 Let (f,g,ψ); (˜f,˜g,˜ψ)be a pair of data andζ=σ(f,g,ψ); ˜ζ=σ(˜f,˜g,˜ψ) the corresponding solution to (2.8). Then, we have
‖ζ−˜ζ‖L∞(Ω)≤max{(1β)‖f−˜f‖L∞(Ω),‖g−˜g‖L∞(∂Ω),‖ψ−˜ψ‖L∞(Ω)}. | (2.9) |
Proof. The proof is an adaptation of the proof of a Lipschitz property of the solution of Ⅵ with nonlinear source term (see [13]). First, set
φ=max{(1β)‖f−˜f‖L∞(Ω),‖g−˜g‖L∞(∂Ω),‖ψ−˜ψ‖L∞(Ω)}. | (2.10) |
Then,
˜f≤f+‖f−˜f‖L∞(Ω) |
≤f+(1)‖f−˜f‖L∞(Ω) |
≤f+(a0(x)β)‖f−˜f‖L∞(Ω) |
≤f+a0(x)max{(1β)‖f−˜f‖L∞(Ω),‖g−˜g‖L∞(∂Ω),‖ψ−˜ψ‖L∞(Ω)}. |
So,
≤f+a0(x)φinΩ. | (2.11) |
Thus, for all 0<v,
(˜f,v)≤(f+a0(x)φ,v), |
with
˜ζ≤˜g≤g+φon∂Ω, |
˜ζ≤˜ψ≤ψ+φinΩ. |
So, according to the property ˜ζ is a subsolution ofσ(f+a0(x)φ,g+φ,ψ+φ)=σ(f,g,ψ)+φ, that is
˜ζ≤ζ+φin¯Ω |
or
˜ζ−ζ≤φin¯Ω. | (2.12) |
Similarly, interchanging the roles of the couples (f,g,ψ); (˜f,˜g,˜ψ), we obtain
ζ−˜ζ≤φin¯Ω, | (2.13) |
which completes the proof.
Let τh be a triangulation of Ω with meshsize h, Vh be the space of finite elements consisting of continuous piecewise linear functions v vanishing on ∂Ω and φs; s = 1, 2, …, m(h) be the basis functions of Vh.
The discrete counterpart of (2.8) consists of finding uh∈Kgh such that
a(uh,v−uh)≥(f,v−uh),∀v∈Kgh. | (2.14) |
Where
Kgh={v∈Vhsuchthatv=πhgon∂Ωandv≤rhψonΩ}, | (2.15) |
πh is an interpolation operator on ∂Ω and rhis the usual finite element restriction operator on Ω.
Theorem 3 (See [12] Under conditions (2.5) and (2.6), there exists a constant C independent of h such that
‖ζ−ζh‖L∞(Ω)≤Ch2|logh|2. | (2.16) |
Assuming that the DMP is satisfied, i.e. the matrix resulting from the finite element discretization is an M-matrix (see [15,16]), we prove the Lipschitz discrete dependence with respect to the boundary condition, the source term and the obstacle by a similar study to that undertaken previously for the Lipschitz continuous dependence property.
Proposition 4 Let (f,g,rhψ); (˜f,˜g,rh˜ψ)be a pair of data and ζh=σh(f,g,rhψ); ˜ζh=σh(˜f,˜g,rh˜ψ) the corresponding solution to (2.14). If f≤˜f in Ω, g≤˜g on ∂Ω and rhψ≤rh˜ψ then, ζh≤˜ζh in Ω.
Proof. The proof is similar to that of the continuous case.
The proposition below establishes a Lipschitz discrete dependence of the solution with respect to the data.
Proposition 5 Let the (d.m.p) holds. Then, we have
‖ζh−˜ζh‖L∞(Ω)≤max{(1β)‖f−˜f‖L∞(Ω),‖g−˜g‖L∞(∂Ω),‖rhψ−rh˜ψ‖L∞(Ω)} | (2.17) |
Proof. The proof is similar to that of the continuous case.
We define the following fixed-point mapping
T:(L∞+(Ω))2→(L∞+(Ω))2 |
Z=(z1,z2)→TZ=ζ=(ζ1,ζ2). |
Where ζi∈H10(Ω)∩L∞(Ω) is a solution to the following variational inequality
ai(ζi,v−ζi)≥(fi(zi),v−ζi);v∈H10(Ω) | (3.1) |
v,ζi≤Mζi=k+zj;ζi≥0withi≠j. |
Thanks to [1,2], ζi is the unique solution to coercive variational inequality (3.1).
Remark 1 We remark that the solution U=(u1,u2) of the system (1.1) is the fixed point of the mapping T; that isTU=U.
Starting from U0=(u1,0,u2,0) where ui,0; i = 1; 2 is solution of the variational equation
ai(ui,0,v)=(fi(ui,0),v),∀v∈H10(Ω), |
and for all 0<wi<1; i=1,2 we define the sequences (u1,n+1) and (u2,n+1) such that u1,n+1 and u2,n+1 the components of the vector Un+1, solve the following elliptic variational inequalities respectively
(u1,n+1,v−u1,n+1)≥(w1f1(u1,n+1)+(1−w1)f1(u1,n),v−u1,n+1) | (3.2) |
v,u1,n+1≤Mu1,n+1=k+u2,n, | (3.3) |
a2(u2,n+1,v−u2,n+1)≥(w2f2(u2,n+1)+(1−w2)f2(u2,n),v−u2,n+1) | (3.4) |
v,u2,n+1≤Mu2,n+1=k+u1,n+1. | (3.5) |
Theorem 2 The sequences (u1,n+1) and (u2,n+1) converge geometrically to the solution U=(u1,u2) of the system (1.1); there exist a positive real ρ∈(0,1) which depends on αi and wi such that for all n≥0
‖Un+1−U‖∞≤ρn+1‖U0−U‖∞ | (3.6) |
where
ρ=max1≤i≤2α1(1−w1)1−α1w1<1. | (3.7) |
Proof. The proof will carry out by induction.
● We first deal with the case
‖u1−u1,0‖L∞(Ω)=max1≤i≤2‖ui−ui,0‖L∞(Ω). | (3.8) |
● Indeed for n = 0; using (1.1), (3.2), (3.3) and (2.9), we have
‖u1−u1,1‖L∞(Ω)≤max{(1β1)‖f1(u1)−(w1f1(u1,1)+(1−w1)f1(u1,0))‖L∞(Ω);‖u2−u2,0‖L∞(Ω)} |
≤max{(1β1)‖w1(f1(u1)−f1(u1,1))+(1−w1)(f1(u1)−f1(u1,0))‖L∞(Ω);‖u2−u2,0‖L∞(Ω)} |
≤max{(k1β1)(w1‖u1−u1,1‖L∞(Ω)+(1−w1)‖u1−u1,0‖L∞(Ω));‖u2−u2,0‖L∞(Ω)}. |
So,
‖u1−u1,1‖L∞(Ω)≤max{α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω);‖u2−u2,0‖L∞(Ω)} | (3.9) |
We distinguish two cases
max{α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω);‖u2−u2,0‖L∞(Ω)} |
=α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω) | (3.10) |
or
max{α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω);‖u2−u2,0‖L∞(Ω)} |
=‖u2−u2,0‖L∞(Ω) | (3.11) |
(3.9) in conjunction with case (3.10) implies
‖u1−u1,1‖L∞(Ω)≤α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω) | (3.12) |
with
‖u2−u2,0‖L∞(Ω)≤α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω), | (3.13) |
which implies
‖u1−u1,1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω). | (3.14) |
By replacing (3.14) in (3.13), we get
‖u2−u2,0‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω) |
≤ρmax1≤i≤2‖ui−ui,0‖L∞(Ω), |
which coincides with (3.8).
(3.9) in conjunction with (3.11) implies
‖u1−u1,1‖L∞(Ω)≤‖u2−u2,0‖L∞(Ω) | (3.15) |
with
α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω)≤‖u2−u2,0‖L∞(Ω). | (3.16) |
‖u2−u2,0‖L∞(Ω) is bounded below by both ‖u1−u1,1‖L∞(Ω)
and
α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω). |
So,
‖u1−u1,1‖L∞(Ω)≤α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω) |
or
α1w1‖u1−u1,1‖L∞(Ω)+α1(1−w1)‖u1−u1,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω). |
Then,
‖u1−u1,1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω) | (3.17) |
or
α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω). | (3.18) |
(3.15), (3.17) and (3.18) generate the following three possibilities
‖u1−u1,1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤‖u2−u2,0‖L∞(Ω)≤max1≤i≤2‖ui−ui,0‖L∞(Ω) |
or
‖u1−u1,1‖L∞(Ω)≤‖u2−u2,0‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤max1≤i≤2‖ui−ui,0‖L∞(Ω) |
or
α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω)≤‖u2−u2,0‖L∞(Ω)≤max1≤i≤2‖ui−ui,0‖L∞(Ω). |
All possibilities are true in the same time because they coincide with (3.8). So, there is either a contradiction and thus case (3.11) is impossible or case (3.11) is possible if and only if
‖u1−u1,1‖L∞(Ω)=α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω). |
Hence, both cases (3.10) and (3.11) imply (3.14).
● Let us now discuss the second case
‖u2−u2,0‖L∞(Ω)=max1≤i≤2‖ui−ui,0‖L∞(Ω). | (3.19) |
(3.9) in conjunction with (3.10) implies (3.14) with
‖u2−u2,0‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω) |
≤ρmax1≤i≤2‖ui−ui,0‖L∞(Ω)<‖u2−u2,0‖L∞(Ω), |
which contradicts (3.19) which means that (3.10) is impossible. (3.9) in conjunction with (3.11) we get (3.17) and (3.18). So,
‖u1−u1,1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤max1≤i≤2‖ui−ui,0‖L∞(Ω) |
or
α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω)≤max1≤i≤2‖ui−ui,0‖L∞(Ω). |
We remark that both alternatives are true in same time because both coincide with (3.19) which implies that in case (3.11), we must have
‖u1−u1,1‖L∞(Ω)=α1(1−w1)1−α1w1‖u1−u1,0‖L∞(Ω). |
Hence, in both cases (3.8) and (3.19), we obtain (3.14). Hence,
‖u1−u1,1‖L∞(Ω)≤ρmax1≤i≤2‖ui−ui,0‖L∞(Ω). | (3.20) |
● As
U1=(u1,1,u2,1)andU=(u1,u2), |
we need to deal also with ‖u2−u2,1‖L∞(Ω), by following the same reasoning as that adopted for u1 and u1,1, we get
‖u2−u2,1‖L∞(Ω)≤max{α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω);‖u1−u1,1‖L∞(Ω)} | (3.21) |
Again we distinguish two possibilities
max{α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω);‖u1−u1,1‖L∞(Ω)} |
=α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω); | (3.22) |
or
max{α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω);‖u1−u1,1‖L∞(Ω)} |
=‖u1−u1,1‖L∞(Ω). | (3.23) |
(3.21) and (3.22) imply
‖u2−u2,1‖L∞(Ω)≤α2(1−w2)(1−α2w2)‖u2−u2,0‖L∞(Ω) | (3.24) |
with
‖u1−u1,1‖L∞(Ω)≤α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω). | (3.25) |
By substituting (3.24) in (3.25), we get
‖u1−u1,1‖L∞(Ω)≤α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω)≤ρmax1≤i≤2‖ui−ui,0‖L∞(Ω), |
which coincides with (3.20). (3.21) and (3.23) imply
‖u2−u2,1‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω), | (3.26) |
with
α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω). |
It is clear that ‖u1−u1,1‖L∞(Ω) is bounded below by both
‖u2−u2,1‖L∞(Ω) |
and
α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω), |
which leads us to distinguish the following possibilities
‖u2−u2,1‖L∞(Ω)≤α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω) |
or
α2w2‖u2−u2,1‖L∞(Ω)+α2(1−w2)‖u2−u2,0‖L∞(Ω)≤‖u2−u2,1‖L∞(Ω). |
Then,
‖u2−u2,1‖L∞(Ω)≤α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω) | (3.27) |
or
α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω)≤‖u2−u2,1‖L∞(Ω). | (3.28) |
Thus, (3.26)-(3.28) imply that the three following alternatives are required
‖u2−u2,1‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω)≤α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω) |
or
‖u2−u2,1‖L∞(Ω)≤α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω) |
or
α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω)≤‖u2−u2,1‖L∞(Ω)≤‖u1−u1,1‖L∞(Ω). |
It is clear that all alternatives coincide with (3.20). So, we must have
‖u2−u2,1‖L∞(Ω)=α2(1−w2)1−α2w2‖u2−u2,0‖L∞(Ω). |
Thus, in both cases (3.22) and (3.23) we obtain (3.24). Hence,
‖u2−u2,1‖L∞(Ω)≤ρmax1≤i≤2‖ui−ui,0‖L∞(Ω). | (3.29) |
(3.20) and (3.29) imply
‖U1−U‖∞≤ρ‖U0−U‖∞. |
● Let us assume that, for n≥0
‖ui−ui,n‖L∞(Ω)≤ρnmax1≤i≤2‖ui−ui,0‖L∞(Ω),i=1,2. | (3.30) |
● We prove
‖ui−ui,n+1‖L∞(Ω)≤ρn+1max1≤i≤2‖ui−ui,n‖L∞(Ω),i=1,2. | (3.31) |
By adopting the same arguments for (1.1), (3.2), (3.3) and (2.9) as that applied for the previous iterates, we get
‖u1−u1,n+1‖L∞(Ω)≤max{(1β1)‖f1(u1)−(w1f1(u1,n+1)+(1−w1)f1(u1,n))‖L∞(Ω);‖u2−u2,n‖L∞(Ω)} |
So,
‖u1−u1,n+1‖L∞(Ω)≤max{α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω);‖u2−u2,n‖L∞(Ω)} | (3.32) |
Also we distinguish two cases:
max{α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω);‖u2−u2,n‖L∞(Ω)} |
=α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω) | (3.33) |
or
max{α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω);‖u2−u2,n‖L∞(Ω)}=‖u2−u2,n‖L∞(Ω) | (3.34) |
(3.32) in conjunction with (3.33) implies
‖u1−u1,n+1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω), | (3.35) |
with
‖u2−u2,n‖L∞(Ω)≤α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω). | (3.36) |
By replacing (3.35) in (3.36) we get, according to (3.30); i = 1
‖u2−u2,n‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω)≤ρn+1max1≤i≤2‖ui−ui,0‖L∞(Ω) |
which matches with (3.30); i = 2. (3.32) in conjunction with (3.34) implies
‖u1−u1,n+1‖L∞(Ω)≤‖u2−u2,n‖L∞(Ω) | (3.37) |
with
α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω)≤‖u2−u2,n‖L∞(Ω). |
‖u2−u2,n‖L∞(Ω) is bounded below by both ‖u1−u1,n+1‖L∞(Ω)
and
α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω) |
So,
‖u1−u1,n+1‖L∞(Ω)≤α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω) |
or
α1w1‖u1−u1,n+1‖L∞(Ω)+α1(1−w1)‖u1−u1,n‖L∞(Ω)≤‖u1−u1,n+1‖L∞(Ω). |
Thus,
‖u1−u1,n+1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω) |
or
α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω)≤‖u1−u1,n+1‖L∞(Ω). |
By taking into account (3.37), we get
‖u1−u1,n+1‖L∞(Ω)≤‖u2−u2,n‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω) |
or
‖u1−u1,n+1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω)≤‖u2−u2,n‖L∞(Ω) |
or
α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω)≤‖u1−u1,n+1‖L∞(Ω)≤‖u2−u2,n‖L∞(Ω). |
Three possibilities are true because all coincide with (3.30). So, we necessarily get
‖u1−u1,n+1‖L∞(Ω)≤α1(1−w1)1−α1w1‖u1−u1,n‖L∞(Ω). |
Thus, both cases (3.33) and (3.34) imply (3.35). Hence, by using (3.30) we get (3.31) for i = 1. The proof for (3.31); i = 2 is obtain in similar way by using (3.31); i = 1 and (3.35) so, it will be omitted. The desired result (3.6) follows naturally from (3.31).
This section, we will handle the discrete problem by a perfect symmetry in the treatment of that the continuous one. Indeed, we define the discrete system of QVIs: Find a vector Uh=(u1h,u2h)∈(Vh)2 such that
{ai(uih,v−uih)≥(fi(uih),v−uih);v∈Vhv,uih≤rh(Muih)=rh(k+ujh);i≠j.uih≥0anduih=πhgon∂Ω. | (4.1) |
The related discrete fixed-point mapping
Th:(Vh)2→(Vh)2 |
Zh=(z1h,z2h)→ThZh=ζh=(ζ1h,ζ2h), |
where ζih∈Vh is the unique solution to the following discrete variational inequality
ai(ζih,v−ζih)≥(fi(zih),v−ζih);v∈Vh | (4.2) |
v,ζih≤rh(Mζih)=rh(k+zjh);ζih≥0withi≠jandζih=πhgon∂Ω. |
Remark 1 We remark that the solution Uh=(u1h,u2h) of the system (4.1) is the fixed point of the mapping Th; that is ThUh=Uh.
Starting from U0h=(u1,0h,u2,0h) where ui,0h=rhui,0;i=1,2 is the discrete analog of ui,0 then,
‖ui,0−ui,0h‖L∞(Ω)≤Ch2|logh|2. | (4.3) |
For all 0<wi<1;i=1,2 we define the discrete sequences (u1,n+1h) and (u2,n+1h) such that u1,n+1h and u2,n+1h components of the vector Un+1h solve discrete elliptic variational inequalities
a1(u1,n+1h,v−u1,n+1h)≥(w1f1(u1,n+1h)+(1−w1)f1(u1,nh),v−u1,n+1h) | (4.4) |
v,u1,n+1h≤rh(Mu1,n+1h)=rh(k+u2,nh), | (4.5) |
a2(u2,n+1h,v−u2,n+1h)≥(w2f2(u2,n+1h)+(1−w2)f2(u2,nh),v−u2,n+1h) | (4.6) |
v,u2,n+1h≤rh(Mu2,n+1h)=rh(k+u1,n+1h). | (4.7) |
Theorem 2 The discrete sequences (u1,n+1h) and (u2,n+1h) converge geometrically to the discrete solution Uh=(u1h,u2h) of the system (4.1); there exist a positive real ρ∈(0,1) defined in (3.7) such that for all n≥0
‖Un+1h−Uh‖∞≤ρn+1‖U0h−Uh‖∞. | (4.8) |
Proof. The proof is similar to that of the continuous case.
This section is devoted to the proof of the main result of this paper. For that purpose we need to introduce an auxiliary system.
Let wi,0h=ui,0h;i=1,2 be an initialization. For all 0<wi<1;i=1,2 we define the discrete sequences (w1,n+1h) and (w2,n+1h) such that w1,n+1h and w2,n+1h solve coercive variational inequalities
a1(w1,n+1h,v−w1,n+1h)≥(w1f1(u1,n+1)+(1−w1)f1(u1,n),v−w1,n+1h) | (5.1) |
v,w1,n+1h≤rh(Mu1,n+1)=rh(k+u2,n), | (5.2) |
a2(w2,n+1h,v−w2,n+1h)≥(w2f2(u2,n+1)+(1−w2)f2(u2,n),v−w2,n+1h) | (5.3) |
v,w2,n+1h≤rh(Mu2,n+1)=rh(k+u1,n+1). | (5.4) |
It is clear that wi,n+1h;i=1,2 components of the vector Wn+1h are finite element approximation of ui,n+1 defined in (3.2)–(3.4). Thus, making use of (2.16); we get
‖wi,n+1h−ui,n+1‖L∞(Ω)≤Ch2|log|2;i=1,2andn≥0. | (5.5) |
The algorithmic approach used in the present paper rests on the following crucial lemma, where the error estimate between the nth iterate Un and its discrete counter parts Un+1h is established.
Lemma 1 Let (Un+1) and (Un+1h) be the vectors whose components are sequences defined in (3.2)–(3.5) and (4.4)–(4.7) respectively. Then,
‖Un+1−Un+1h‖∞≤(γ(1−ρn+11−ρ)+ρn+1)maxn≥0‖Un−Wnh‖∞. | (5.6) |
Where
γ=max1≤i≤2{1(1−αiwi)}. | (5.7) |
Proof. The proof of the lemma rests on the discrete Lipschitz continuous dependency with respect to source term and obstacle and will carry out by induction.
● For n = 0, we have
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω)+‖w1,1h−u1,1h‖L∞(Ω). |
(5.1), (5.2), (4.4), (4.5) and (2.17) imply
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω) |
+max{(1β1)‖f1(u1,1)−(w1f1(u1,1h)+(1−w1)f1(u1,0h))‖L∞(Ω);‖rh(k+u2,0)−rh(k+u2,0h)‖L∞(Ω)} |
So,
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω) |
+max{(k1β1)w1‖u1,1−u1,1h‖L∞(Ω)+(k1β1)(1−w1)‖u1,0−u1,0h‖L∞(Ω);‖rh(k+u2,0)−rh(k+u2,0h)‖L∞(Ω)}. |
Therefore,
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω) | (5.8) |
+max{α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω);‖u2,0−u2,0h‖L∞(Ω)}. |
We distinguish two cases
max{α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω);‖u2,0−u2,0h‖L∞(Ω)} |
=α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω) | (5.9) |
or
max{α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω);‖u2,0−u2,0h‖L∞(Ω)}=‖u2,0−u2,0h‖L∞(Ω) | (5.10) |
(5.8) in conjunction with (5.9) imply
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω)+α1w1‖u1,1−u1,1h‖L∞(Ω) |
+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω) |
with
‖u2,0−u2,0h‖L∞(Ω)≤α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω). | (5.11) |
So,
(1−α1w1)‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω), |
with (5.11). Then,
‖u1,1−u1,1h‖L∞(Ω)≤1(1−α1w1)‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,0−u1,0h‖L∞(Ω). | (5.12) |
By replacing (5.12) in (5.11) we obtain
‖u2,0−u2,0h‖L∞(Ω)≤α1w1(1−α1w1)‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,0−u1,0h‖L∞(Ω). |
According to (5.5) and (4.3) we get,
‖u2,0−u2,0h‖L∞(Ω)≤α1(1−α1w1)Ch2|logh|2, |
which coincides with (4.3).
(5.8) and (5.10) imply
‖u1,1−u1,1h‖L∞(Ω)≤‖u1,1−w1,1h‖L∞(Ω)+‖u2,0−u2,0h‖L∞(Ω) | (5.13) |
with
α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω)≤‖u2,0−u2,0h‖L∞(Ω). |
Then, multiplying (5.13) by α1w1 and adding α1(1−w1)‖u1,0−u1,0h‖L∞(Ω), we obtain
α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω)≤α1w1‖u1,1−w1,1h‖L∞(Ω)+α1w1‖u2,0−u2,0h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω). |
We note that
α1w1‖u1,1−u1,1h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω) |
is bounded by both
α1w1‖u1,1−w1,1h‖L∞(Ω)+α1w1‖u2,0−u2,0h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω) |
and
‖u2,0−u2,0h‖L∞(Ω). |
So,
α1w1‖u1,1−w1,1h‖L∞(Ω)+α1w1‖u2,0−u2,0h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω)≤‖u2,0−u2,0h‖L∞(Ω) |
or
‖u2,0−u2,0h‖L∞(Ω)≤α1w1‖u1,1−w1,1h‖L∞(Ω)+α1w1‖u2,0−u2,0h‖L∞(Ω)+α1(1−w1)‖u1,0−u1,0h‖L∞(Ω). |
Therefore, according to (5.5) and (4.3), we get
α1w1(1−α1w1)‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,0−u1,0h‖L∞(Ω)≤‖u2,0−u2,0h‖L∞(Ω)≤Ch2|logh|2 |
or
‖u2,0−u2,0h‖L∞(Ω)≤α1w1(1−α1w1)‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,0−u1,0h‖L∞(Ω)≤α1(1−α1w1)Ch2|logh|2. |
So, the last two alternatives are true at the same time because both coincide with (4.3). We necessarily deduce that
‖u2,0−u2,0h‖L∞(Ω)=α1w1(1−α1w1)‖u1,1−w1,1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,0−u1,0h‖L∞(Ω). | (5.14) |
By replacing (5.14) in (5.13); we get (5.12). Hence, in both cases (5.9) and (5.10); we can write
‖u1,1−u1,1h‖L∞(Ω)≤max1≤i≤2{1(1−αiwi)}max1≤i≤2‖ui,1−wi,1h‖L∞(Ω) |
+max1≤i≤2{αi(1−wi)(1−αiwi)}max1≤i≤2‖ui,0−ui,0h‖L∞(Ω). |
Thus,
‖u1,1−u1,1h‖L∞(Ω)≤(γ+ρ)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). | (5.15) |
● In a similar way, that is by following the same steps as for u1,1 and u1,1h, u2,1 and u2,1h satisfy
‖u2,1−u2,1h‖L∞(Ω)≤‖u2,1−w2,1h‖L∞(Ω)+‖w2,1h−u2,1h‖L∞(Ω). |
So,
‖u2,1−u2,1h‖L∞(Ω)≤‖u2,1−w2,1h‖L∞(Ω) |
+max{α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω);‖u1,1−u1,1h‖L∞(Ω)}. | (5.16) |
We distinguish also two cases
max{α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω);‖u1,1−u1,1h‖L∞(Ω)}=α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω) | (5.17) |
or
max{α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω);‖u1,1−u1,1h‖L∞(Ω)}=‖u1,1−u1,1h‖L∞(Ω). | (5.18) |
(5.16) in conjunction with case (5.17); we get
‖u2,1−u2,1h‖L∞(Ω)≤‖u2,1−w2,1h‖L∞(Ω)+α2w2‖u2,1−u2,1h‖L∞(Ω)
+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω) |
with
‖u1,1−u1,1h‖L∞(Ω)≤α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω). | (5.19) |
So,
‖u2,1−u2,1h‖L∞(Ω)≤1(1−α2w2)‖u2,1−w2,1h‖L∞(Ω)+α2(1−w2)(1−α2w2)‖u2,0−u2,0h‖L∞(Ω) | (5.20) |
with, according to (5.20)
‖u1,1−u1,1h‖L∞(Ω)≤1(1−α2w2)‖u2,1−w2,1h‖L∞(Ω)+α2(1−w2)(1−α2w2)‖u2,0−u2,0h‖L∞(Ω). |
Then,
‖u1,1−u1,1h‖L∞(Ω)≤max1≤i≤2{1(1−αiwi)}‖ui,1−wi,1h‖L∞(Ω)+max1≤i≤2{αi(1−wi)(1−αiwi)}‖ui,0−ui,0h‖L∞(Ω). |
Therefore,
‖u1,1−u1,1h‖L∞(Ω)≤(γ+ρ)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω), |
which coincides with (5.15). The conjunction of (5.16) with case (5.18), implies
‖u2,1−u2,1h‖L∞(Ω)≤‖u2,1−w2,1h‖L∞(Ω)+‖u1,1−u1,1h‖L∞(Ω) | (5.21) |
with
α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω)≤‖u1,1−u1,1h‖L∞(Ω). |
Then, by multiplying (5.21) by α2w2 and addingα2(1−w2)‖u2,0−u2,0h‖L∞(Ω), we obtain that the term α2w2‖u2,1−u2,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω)is bounded by both
α2w2‖u2,1−w2,1h‖L∞(Ω)+α2w2‖u1,1−u1,1h‖L∞(Ω)+α2(1−w2)‖u2,0−u2,0h‖L∞(Ω) |
and
‖u1,1−u1,1h‖L∞(Ω). |
So, we distinguish again, the two following alternatives
α2w2(1−α2w2)‖u2,1−w2,1h‖L∞(Ω)+α2(1−w2)(1−α2w2)‖u2,0−u2,0h‖L∞(Ω)≤‖u1,1−u1,1h‖L∞(Ω)≤(γ+ρ)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω) |
or
‖u1,1−u1,1h‖L∞(Ω)≤α2w2(1−α2w2)‖u2,1−w2,1h‖L∞(Ω)+α2(1−w2)(1−α2w2)‖u2,0−u2,0h‖L∞(Ω)≤(γ+ρ)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). |
We remark that both alternatives coincide with (5.15), which implies that case (5.18) is possible if and only if
‖u1,1−u1,1h‖L∞(Ω)=α2w2(1−α2w2)‖u2,1−w2,1h‖L∞(Ω)+α2(1−w2)(1−α2w2)‖u2,0−u2,0h‖L∞(Ω). | (5.22) |
By substituting (5.22) in (5.21), we get (5.20). Hence, in both cases (5.17) and (5.18), we get
‖u2,1−u2,1h‖L∞(Ω)≤max1≤i≤2{1(1−αiwi)}max1≤i≤2‖ui,1−wi,1h‖L∞(Ω)+max1≤i≤2{αi(1−wi)(1−αiwi)}max1≤i≤2‖ui,0−ui,0h‖L∞(Ω). |
Thus,
‖u2,1−u2,1h‖L∞(Ω)≤(γ+ρ)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). | (5.23) |
(5.15) and (5.23) imply
‖U1−U1h‖∞≤(γ+ρ)maxn≥0‖Un−Wnh‖∞. |
● Let us assume that for n≥0 and i = 1, 2
‖ui,n−ui,nh‖L∞(Ω)≤(γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). | (5.24) |
● And prove for i = 1, 2
‖ui,n+1−ui,n+1h‖L∞(Ω)≤(γ(1+ρ+⋯+ρn)+ρn+1)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). | (5.25) |
We operate in the same way as in iterate n = 0. Let us begin with case i = 1 in (5.25)
‖u1,n+1−u1,n+1h‖L∞(Ω)≤‖u1,n+1−w1,n+1h‖L∞(Ω)+‖w1,n+1h−u1,n+1h‖L∞(Ω). |
So, by applying (2.17), we get
‖u1,n+1−u1,n+1h‖L∞(Ω)≤‖u1,n+1−w1,n+1h‖L∞(Ω) |
+max{α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω);‖u2,n−u2,nh‖L∞(Ω)} | (5.26) |
We distinguish again two cases
max{α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω);‖u2,n−u2,nh‖L∞(Ω)} |
=α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω) | (5.27) |
or
max{α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω);‖u2,n−u2,nh‖L∞(Ω)} |
=‖u2,n−u2,nh‖L∞(Ω). | (5.28) |
(5.26) in conjunction with case (5.27) implies
‖u1,n+1−u1,n+1h‖L∞(Ω)≤‖u1,n+1−w1,n+1h‖L∞(Ω)+α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω) |
and
‖u2,n−u2,nh‖L∞(Ω)≤α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω). |
Then,
‖u1,n+1−u1,n+1h‖L∞(Ω)≤1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω) | (5.29) |
with, according to (5.29)
‖u2,n−u2,nh‖L∞(Ω)≤α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω). |
(5.24) implies
‖u1,n+1−u1,n+1h‖L∞(Ω)≤1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)((γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω)) |
with
‖u2,n−u2,nh‖L∞(Ω)≤α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω) |
+α1(1−w1)(1−α1w1)((γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω)). |
Thus,
‖u1,n+1−u1,n+1h‖L∞(Ω)≤γ‖u1,n+1−w1,n+1h‖L∞(Ω)+ρ((γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω)) |
and as α1w1<1
‖u2,n−u2,nh‖L∞(Ω) |
≤γ‖u1,n+1−w1,n+1h‖L∞(Ω)+ρ((γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω)). |
Hence,
‖u1,n+1−u1,n+1h‖L∞(Ω)≤(γ(1+ρ+⋯+ρn)+ρn+1)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω) |
and
‖u2,n−u2,nh‖L∞(Ω)≤(γ(1+ρ+⋯+ρn)+ρn+1)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). |
which corresponds with (5.24) for i = 2: Inequality (5.26) with (5.28) imply
‖u1,n+1−u1,n+1h‖L∞(Ω)≤‖u1,n+1−w1,n+1h‖L∞(Ω)+‖u2,n−u2,nh‖L∞(Ω) | (5.30) |
and
α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω)≤‖u2,n−u2,nh‖L∞(Ω). |
By multiplying (5.30) by α1w1 and adding the term α1(1−w1)‖u1,n−u1,nh‖L∞(Ω), we get that the term
α1w1‖u1,n+1−u1,n+1h‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω) |
is bounded by the following two terms
α1w1‖u1,n+1−w1,n+1h‖L∞(Ω)+α1w1‖u2,n−u2,nh‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω) |
and
‖u2,n−u2,nh‖L∞(Ω). |
So, we need to distinguish the followings possibilities
α1w1‖u1,n+1−w1,n+1h‖L∞(Ω)+α1w1‖u2,n−u2,nh‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω)≤‖u2,n−u2,nh‖L∞(Ω) |
or
‖u2,n−u2,nh‖L∞(Ω)≤α1w1‖u1,n+1−w1,n+1h‖L∞(Ω)+α1w1‖u2,n−u2,nh‖L∞(Ω)+α1(1−w1)‖u1,n−u1,nh‖L∞(Ω), |
which implies
α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω)≤‖u2,n−u2,nh‖L∞(Ω) |
or
‖u2,n−u2,nh‖L∞(Ω)≤α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω). |
By using (5.24), we can write
α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω)≤‖u2,n−u2,nh‖L∞(Ω)≤(γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω), |
or
‖u2,n−u2,nh‖L∞(Ω)≤α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω)≤(γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). |
Only the last alternative is true because it matches with (5.24) for i = 2. So, in (5.28) we get
‖u2,n−u2,nh‖L∞(Ω)≤α1w1(1−α1w1)‖u1,n+1−w1,n+1h‖L∞(Ω)+α1(1−w1)(1−α1w1)‖u1,n−u1,nh‖L∞(Ω). | (5.31) |
By replacing (5.31) in (5.30); we get (5.29). Hence, in both cases (5.27) and (5.28), we obtain
‖u1,n+1−u1,n+1h‖L∞(Ω)≤max1≤i≤2{1(1−αiwi)}max1≤i≤2‖ui,n+1−wi,n+1h‖L∞(Ω)+max1≤i≤2{αi(1−wi)(1−αiwi)}max1≤i≤2‖ui,n−ui,nh‖L∞(Ω). |
So,
‖u1,n+1−u1,n+1h‖L∞(Ω)≤γmaxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω)+ρ(γ(1+ρ+⋯+ρn−1)+ρn)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). |
Therefore,
‖u1,n+1−u1,n+1h‖L∞(Ω)≤(γ(1+ρ+⋯+ρn−1+ρn)+ρn+1)maxn≥0max1≤i≤2‖ui,n−wi,nh‖L∞(Ω). | (5.32) |
By using the last inequality (5.32) and by adopting the same reasoning we prove (5.25); i = 2, therefore, we get (5.6).
Theorem 2 Let U and Uh be the solution of systems (1.1) and (4.8), respectively. Then, there exists a constant C independent of h such that
‖U−Uh‖∞≤γ1−ρh2|logh|2. |
Proof. Making use of (3.6), (5.6) and (4.8), we have
‖U−Uh‖∞≤‖U−Un+1‖∞+‖Un+1−Un+1h‖∞+‖Un+1h−Uh‖∞ |
≤ρn+1‖U−U0‖∞+(γ(1−ρn+11−ρ)+ρn+1)maxn≥0‖Un−Wnh‖∞+ρn+1‖Uh−U0h‖∞. |
As n→+∞ and by using (5.5) we get (5.33).
In this work an optimal convergence order is derived for a class of system of two elliptic quasi-variational inequalities where terms sources and obstacles depend upon the solution, where the continuous and discrete Lipschitz dependence with respect to the terms sources, boundary condition and obstacles' played a leading role in obtaining the main result of this paper. As (1.1) plays a key role in solving Hamilton-Jacobi-Bellman equation the results obtained in this paper can give an optimal error estimate for HJB equation also even for J≥2. The approach used and the results obtained in this paper (optimal convergence order) remain valid when we deal with systems of J≥2 quasi-variational inequalities with terms sources depends on solution and the obstacles i independent of the solution, that is systems of the form; Find a vector U=(u1,…,uJ)∈(H10(Ω))J satisfying
{ai(ui,v−ui)≥(fi(ui),v−ui);v∈H10(Ω)v,ui≤ψi;ui≥0andi=1,…,J. |
The author states that no funding source or sponsor has participated in the realization of this work.
All authors declare no conflicts of interest in this paper.
Scheme of the predictions of the linear stability analysis by acting on the interspecies repulsion force
Form of the potential
Values of
(I): Values of
Microscopic simulations for Cases 1-4 for parameters described in Table 1.
Macroscopic simulations for Cases 1-4 for parameters described in table 1 for the final time of simulations equal to
(I) Simulations of the microscopic and macroscopic models for
Quantifiers computed on the simulation images as functions of the logarithm of the simulation time for