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Research article

Green tea catechins and intracellular calcium dynamics in prostate cancer cells

  • Perturbations of internal Ca2+ ([Ca2+]i) homeostasis play a key role in several pathologies and in neoplastic transformation, where deregulated cell proliferation, together with the suppression of apoptosis, provides the condition for abnormal tissue growth and invasion. Green tea catechins have been shown to affect cancer development by interference with basic cellular functions, most of which are mediated by [Ca2+]i. Prostate cancer (PCa) is one of the most common malignancy in men in Western countries and the androgen-independent carcinoma is a lethal form for which there is still no effective therapy. Different evidences suggested that consumption of green tea may have beneficial effects against PCa. We have previously described how the main green tea flavonoid, (−)-epigallocatechin-3-gallate (EGCG), inhibited proliferation and induced dose-dependent peaks of [Ca2+]i in metastatic androgen-insensitive DU145 and PC3 PCa cells, by a mechanism that combined Ca2+ entry and Ca2+-induced Ca2+ release. In the present study, we studied the effect of green tea extract (GTE) on the same cell lines. Proliferation, measured by MTT assay, was inhibited by GTE with IC50 close to 60 µg/ml, a value that is higher than that expected by EGCG effect alone. [Ca2+]i, measured in real time by the fluorescent dye Fura-2, was transiently increased by GTE by a mechanism that resembled that described for EGCG, but was largely independent of external Ca2+. These observations suggested that other components, acting in synergy with EGCG, were involved in GTE effect, and confirmed the view that the alleged health benefits of green tea for PCa prevention may be related to [Ca2+]i deregulation in malignant cells. These results may be significant to understand the functional mechanisms by which flavonoids exert their beneficial or toxic actions.

    Citation: Carla Marchetti. Green tea catechins and intracellular calcium dynamics in prostate cancer cells[J]. AIMS Molecular Science, 2021, 8(1): 1-12. doi: 10.3934/molsci.2021001

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  • Perturbations of internal Ca2+ ([Ca2+]i) homeostasis play a key role in several pathologies and in neoplastic transformation, where deregulated cell proliferation, together with the suppression of apoptosis, provides the condition for abnormal tissue growth and invasion. Green tea catechins have been shown to affect cancer development by interference with basic cellular functions, most of which are mediated by [Ca2+]i. Prostate cancer (PCa) is one of the most common malignancy in men in Western countries and the androgen-independent carcinoma is a lethal form for which there is still no effective therapy. Different evidences suggested that consumption of green tea may have beneficial effects against PCa. We have previously described how the main green tea flavonoid, (−)-epigallocatechin-3-gallate (EGCG), inhibited proliferation and induced dose-dependent peaks of [Ca2+]i in metastatic androgen-insensitive DU145 and PC3 PCa cells, by a mechanism that combined Ca2+ entry and Ca2+-induced Ca2+ release. In the present study, we studied the effect of green tea extract (GTE) on the same cell lines. Proliferation, measured by MTT assay, was inhibited by GTE with IC50 close to 60 µg/ml, a value that is higher than that expected by EGCG effect alone. [Ca2+]i, measured in real time by the fluorescent dye Fura-2, was transiently increased by GTE by a mechanism that resembled that described for EGCG, but was largely independent of external Ca2+. These observations suggested that other components, acting in synergy with EGCG, were involved in GTE effect, and confirmed the view that the alleged health benefits of green tea for PCa prevention may be related to [Ca2+]i deregulation in malignant cells. These results may be significant to understand the functional mechanisms by which flavonoids exert their beneficial or toxic actions.


    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,vui)(fi(ui),vui);vH10(Ω) (1.1)
    v,uiMui;ui0.

    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+infμiuμ;

    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),1i2 sufficiently smooth functions such that 1j,kN

    1j,kNaijk(x)ξjξkα|ξ|2,ξRN,α>0
    ai0(x)βi>0,(xΩ) (2.1)

    where βi is a positive constant. We define the bilinear forms: For all u,vH10(Ω)

    ai(u,v)=Ω(1j,kNaijk(x)uxjvxk+Nk=1aik(x)uxkv+ai0(x)uv)dx (2.2)

    We are given right-hand sides

    fisuchthatfiL(Ω),fif0>0,

    a nonlinear functional and Lipschitz continuous on R; that is

    |fi(x)fi(y)|ki|xy|,x,yR

    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=max1i2wiL(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

    fL(Ω), (2.5)

    the obstacle

    ψW2,(Ω)andψ0, (2.6)

    the boundary condition gL(Ω) and the nonempty convex set

    Kg={vH1(Ω)suchthatv=gonΩandvψonΩ}. (2.7)

    We consider the variational inequality V.I.: Find uKg such that

    a(u,vu)(f,vu),vKg. (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),v0withζ_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˜gonΩ.

    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˜ζL(Ω)max{(1β)f˜fL(Ω),g˜gL(Ω),ψ˜ψL(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˜fL(Ω),g˜gL(Ω),ψ˜ψL(Ω)}. (2.10)

    Then,

    ˜ff+f˜fL(Ω)
    f+(1)f˜fL(Ω)
    f+(a0(x)β)f˜fL(Ω)
    f+a0(x)max{(1β)f˜fL(Ω),g˜gL(Ω),ψ˜ψL(Ω)}.

    So,

    f+a0(x)φinΩ. (2.11)

    Thus, for all 0<v,

    (˜f,v)(f+a0(x)φ,v),

    with

    ˜ζ˜gg+φonΩ,
    ˜ζ˜ψψ+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

    ˜ζζ+φin¯Ω

    or

    ˜ζζφin¯Ω. (2.12)

    Similarly, interchanging the roles of the couples (f,g,ψ); (˜f,˜g,˜ψ), we obtain

    ζ˜ζφin¯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 uhKgh such that

    a(uh,vuh)(f,vuh),vKgh. (2.14)

    Where

    Kgh={vVhsuchthatv=πhgonΩandvrhψonΩ}, (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

    ζζhL(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 rhψrh˜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˜ζhL(Ω)max{(1β)f˜fL(Ω),g˜gL(Ω),rhψrh˜ψL(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 ζiH10(Ω)L(Ω) is a solution to the following variational inequality

    ai(ζi,vζi)(fi(zi),vζi);vH10(Ω) (3.1)
    v,ζiMζi=k+zj;ζi0withij.

    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),vH10(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,vu1,n+1)(w1f1(u1,n+1)+(1w1)f1(u1,n),vu1,n+1) (3.2)
    v,u1,n+1Mu1,n+1=k+u2,n, (3.3)
    a2(u2,n+1,vu2,n+1)(w2f2(u2,n+1)+(1w2)f2(u2,n),vu2,n+1) (3.4)
    v,u2,n+1Mu2,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 n0

    Un+1Uρn+1U0U (3.6)

    where

    ρ=max1i2α1(1w1)1α1w1<1. (3.7)

    Proof. The proof will carry out by induction.

    ● We first deal with the case

    u1u1,0L(Ω)=max1i2uiui,0L(Ω). (3.8)

    ● Indeed for n = 0; using (1.1), (3.2), (3.3) and (2.9), we have

    u1u1,1L(Ω)max{(1β1)f1(u1)(w1f1(u1,1)+(1w1)f1(u1,0))L(Ω);u2u2,0L(Ω)}
    max{(1β1)w1(f1(u1)f1(u1,1))+(1w1)(f1(u1)f1(u1,0))L(Ω);u2u2,0L(Ω)}
    max{(k1β1)(w1u1u1,1L(Ω)+(1w1)u1u1,0L(Ω));u2u2,0L(Ω)}.

    So,

    u1u1,1L(Ω)max{α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω);u2u2,0L(Ω)} (3.9)

    We distinguish two cases

    max{α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω);u2u2,0L(Ω)}
    =α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω) (3.10)

    or

    max{α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω);u2u2,0L(Ω)}
    =u2u2,0L(Ω) (3.11)

    (3.9) in conjunction with case (3.10) implies

    u1u1,1L(Ω)α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω) (3.12)

    with

    u2u2,0L(Ω)α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω) (3.13)

    which implies

    u1u1,1L(Ω)α1(1w1)1α1w1u1u1,0L(Ω). (3.14)

    By replacing (3.14) in (3.13), we get

    u2u2,0L(Ω)α1(1w1)1α1w1u1u1,0L(Ω)
    ρmax1i2uiui,0L(Ω),

    which coincides with (3.8).

    (3.9) in conjunction with (3.11) implies

    u1u1,1L(Ω)u2u2,0L(Ω) (3.15)

    with

    α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω)u2u2,0L(Ω). (3.16)

    u2u2,0L(Ω) is bounded below by both u1u1,1L(Ω)

    and

    α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω).

    So,

    u1u1,1L(Ω)α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω)

    or

    α1w1u1u1,1L(Ω)+α1(1w1)u1u1,0L(Ω)u1u1,1L(Ω).

    Then,

    u1u1,1L(Ω)α1(1w1)1α1w1u1u1,0L(Ω) (3.17)

    or

    α1(1w1)1α1w1u1u1,0L(Ω)u1u1,1L(Ω). (3.18)

    (3.15), (3.17) and (3.18) generate the following three possibilities

    u1u1,1L(Ω)α1(1w1)1α1w1u1u1,0L(Ω)u2u2,0L(Ω)max1i2uiui,0L(Ω)

    or

    u1u1,1L(Ω)u2u2,0L(Ω)α1(1w1)1α1w1u1u1,0L(Ω)max1i2uiui,0L(Ω)

    or

    α1(1w1)1α1w1u1u1,0L(Ω)u1u1,1L(Ω)u2u2,0L(Ω)max1i2uiui,0L(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

    u1u1,1L(Ω)=α1(1w1)1α1w1u1u1,0L(Ω).

    Hence, both cases (3.10) and (3.11) imply (3.14).

    ● Let us now discuss the second case

    u2u2,0L(Ω)=max1i2uiui,0L(Ω). (3.19)

    (3.9) in conjunction with (3.10) implies (3.14) with

    u2u2,0L(Ω)α1(1w1)1α1w1u1u1,0L(Ω)
    ρmax1i2uiui,0L(Ω)<u2u2,0L(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,

    u1u1,1L(Ω)α1(1w1)1α1w1u1u1,0L(Ω)max1i2uiui,0L(Ω)

    or

    α1(1w1)1α1w1u1u1,0L(Ω)u1u1,1L(Ω)max1i2uiui,0L(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

    u1u1,1L(Ω)=α1(1w1)1α1w1u1u1,0L(Ω).

    Hence, in both cases (3.8) and (3.19), we obtain (3.14). Hence,

    u1u1,1L(Ω)ρmax1i2uiui,0L(Ω). (3.20)

    ● As

    U1=(u1,1,u2,1)andU=(u1,u2),

    we need to deal also with u2u2,1L(Ω), by following the same reasoning as that adopted for u1 and u1,1, we get

    u2u2,1L(Ω)max{α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω);u1u1,1L(Ω)} (3.21)

    Again we distinguish two possibilities

    max{α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω);u1u1,1L(Ω)}
    =α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω); (3.22)

    or

    max{α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω);u1u1,1L(Ω)}
    =u1u1,1L(Ω). (3.23)

    (3.21) and (3.22) imply

    u2u2,1L(Ω)α2(1w2)(1α2w2)u2u2,0L(Ω) (3.24)

    with

    u1u1,1L(Ω)α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω). (3.25)

    By substituting (3.24) in (3.25), we get

    u1u1,1L(Ω)α2(1w2)1α2w2u2u2,0L(Ω)ρmax1i2uiui,0L(Ω),

    which coincides with (3.20). (3.21) and (3.23) imply

    u2u2,1L(Ω)u1u1,1L(Ω) (3.26)

    with

    α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω)u1u1,1L(Ω).

    It is clear that u1u1,1L(Ω) is bounded below by both

    u2u2,1L(Ω)

    and

    α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω)

    which leads us to distinguish the following possibilities

    u2u2,1L(Ω)α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω)

    or

    α2w2u2u2,1L(Ω)+α2(1w2)u2u2,0L(Ω)u2u2,1L(Ω).

    Then,

    u2u2,1L(Ω)α2(1w2)1α2w2u2u2,0L(Ω) (3.27)

    or

    α2(1w2)1α2w2u2u2,0L(Ω)u2u2,1L(Ω). (3.28)

    Thus, (3.26)-(3.28) imply that the three following alternatives are required

    u2u2,1L(Ω)u1u1,1L(Ω)α2(1w2)1α2w2u2u2,0L(Ω)

    or

    u2u2,1L(Ω)α2(1w2)1α2w2u2u2,0L(Ω)u1u1,1L(Ω)

    or

    α2(1w2)1α2w2u2u2,0L(Ω)u2u2,1L(Ω)u1u1,1L(Ω).

    It is clear that all alternatives coincide with (3.20). So, we must have

    u2u2,1L(Ω)=α2(1w2)1α2w2u2u2,0L(Ω).

    Thus, in both cases (3.22) and (3.23) we obtain (3.24). Hence,

    u2u2,1L(Ω)ρmax1i2uiui,0L(Ω). (3.29)

    (3.20) and (3.29) imply

    U1UρU0U.

    ● Let us assume that, for n0

    uiui,nL(Ω)ρnmax1i2uiui,0L(Ω),i=1,2. (3.30)

    ● We prove

    uiui,n+1L(Ω)ρn+1max1i2uiui,nL(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

    u1u1,n+1L(Ω)max{(1β1)f1(u1)(w1f1(u1,n+1)+(1w1)f1(u1,n))L(Ω);u2u2,nL(Ω)}

    So,

    u1u1,n+1L(Ω)max{α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω);u2u2,nL(Ω)} (3.32)

    Also we distinguish two cases:

    max{α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω);u2u2,nL(Ω)}
    =α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω) (3.33)

    or

    max{α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω);u2u2,nL(Ω)}=u2u2,nL(Ω) (3.34)

    (3.32) in conjunction with (3.33) implies

    u1u1,n+1L(Ω)α1(1w1)1α1w1u1u1,nL(Ω), (3.35)

    with

    u2u2,nL(Ω)α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω). (3.36)

    By replacing (3.35) in (3.36) we get, according to (3.30); i = 1

    u2u2,nL(Ω)α1(1w1)1α1w1u1u1,nL(Ω)ρn+1max1i2uiui,0L(Ω)

    which matches with (3.30); i = 2. (3.32) in conjunction with (3.34) implies

    u1u1,n+1L(Ω)u2u2,nL(Ω) (3.37)

    with

    α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω)u2u2,nL(Ω).

    u2u2,nL(Ω) is bounded below by both u1u1,n+1L(Ω)

    and

    α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω)

    So,

    u1u1,n+1L(Ω)α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω)

    or

    α1w1u1u1,n+1L(Ω)+α1(1w1)u1u1,nL(Ω)u1u1,n+1L(Ω).

    Thus,

    u1u1,n+1L(Ω)α1(1w1)1α1w1u1u1,nL(Ω)

    or

    α1(1w1)1α1w1u1u1,nL(Ω)u1u1,n+1L(Ω).

    By taking into account (3.37), we get

    u1u1,n+1L(Ω)u2u2,nL(Ω)α1(1w1)1α1w1u1u1,nL(Ω)

    or

    u1u1,n+1L(Ω)α1(1w1)1α1w1u1u1,nL(Ω)u2u2,nL(Ω)

    or

    α1(1w1)1α1w1u1u1,nL(Ω)u1u1,n+1L(Ω)u2u2,nL(Ω).

    Three possibilities are true because all coincide with (3.30). So, we necessarily get

    u1u1,n+1L(Ω)α1(1w1)1α1w1u1u1,nL(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,vuih)(fi(uih),vuih);vVhv,uihrh(Muih)=rh(k+ujh);ij.uih0anduih=πhgonΩ. (4.1)

    The related discrete fixed-point mapping

    Th:(Vh)2(Vh)2
    Zh=(z1h,z2h)ThZh=ζh=(ζ1h,ζ2h),

    where ζihVh is the unique solution to the following discrete variational inequality

    ai(ζih,vζih)(fi(zih),vζih);vVh (4.2)
    v,ζihrh(Mζih)=rh(k+zjh);ζih0withijandζih=πhgonΩ.

    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,0ui,0hL(Ω)Ch2|logh|2. (4.3)

    For all 0<wi<1;i=1,2 we define the discrete sequences (u1,n+1h) and such that and components of the vector solve discrete elliptic variational inequalities

    (4.4)
    (4.5)
    (4.6)
    (4.7)

    Theorem 2 The discrete sequences and converge geometrically to the discrete solution of the system (4.1); there exist a positive real defined in (3.7) such that for all

    (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 be an initialization. For all we define the discrete sequences and such that and solve coercive variational inequalities

    (5.1)
    (5.2)
    (5.3)
    (5.4)

    It is clear that components of the vector are finite element approximation of defined in (3.2)–(3.4). Thus, making use of (2.16); we get

    (5.5)

    The algorithmic approach used in the present paper rests on the following crucial lemma, where the error estimate between the nth iterate and its discrete counter parts is established.

    Lemma 1 Let and be the vectors whose components are sequences defined in (3.2)–(3.5) and (4.4)–(4.7) respectively. Then,

    (5.6)

    Where

    (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

    (5.1), (5.2), (4.4), (4.5) and (2.17) imply

    So,

    Therefore,

    (5.8)

    We distinguish two cases

    (5.9)

    or

    (5.10)

    (5.8) in conjunction with (5.9) imply

    with

    (5.11)

    So,

    with (5.11). Then,

    (5.12)

    By replacing (5.12) in (5.11) we obtain

    According to (5.5) and (4.3) we get,

    which coincides with (4.3).

    (5.8) and (5.10) imply

    (5.13)

    with

    Then, multiplying (5.13) by and adding we obtain

    We note that

    is bounded by both

    and

    So,

    or

    Therefore, according to (5.5) and (4.3), we get

    or

    So, the last two alternatives are true at the same time because both coincide with (4.3). We necessarily deduce that

    (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

    Thus,

    (5.15)

    ● In a similar way, that is by following the same steps as for and , and satisfy

    So,

    (5.16)

    We distinguish also two cases

    (5.17)

    or

    (5.18)

    (5.16) in conjunction with case (5.17); we get

    +

    with

    (5.19)

    So,

    (5.20)

    with, according to (5.20)

    Then,

    Therefore,

    which coincides with (5.15). The conjunction of (5.16) with case (5.18), implies

    (5.21)

    with

    Then, by multiplying (5.21) by and adding, we obtain that the term is bounded by both

    and

    So, we distinguish again, the two following alternatives

    or

    We remark that both alternatives coincide with (5.15), which implies that case (5.18) is possible if and only if

    (5.22)

    By substituting (5.22) in (5.21), we get (5.20). Hence, in both cases (5.17) and (5.18), we get

    Thus,

    (5.23)

    (5.15) and (5.23) imply

    ● Let us assume that for and i = 1, 2

    (5.24)

    ● And prove for i = 1, 2

    (5.25)

    We operate in the same way as in iterate n = 0. Let us begin with case i = 1 in (5.25)

    So, by applying (2.17), we get

    (5.26)

    We distinguish again two cases

    (5.27)

    or

    (5.28)

    (5.26) in conjunction with case (5.27) implies

    and

    Then,

    (5.29)

    with, according to (5.29)

    (5.24) implies

    with

    Thus,

    and as

    Hence,

    and

    which corresponds with (5.24) for i = 2: Inequality (5.26) with (5.28) imply

    (5.30)

    and

    By multiplying (5.30) by and adding the term we get that the term

    is bounded by the following two terms

    and

    So, we need to distinguish the followings possibilities

    or

    which implies

    or

    By using (5.24), we can write

    or

    Only the last alternative is true because it matches with (5.24) for i = 2. So, in (5.28) we get

    (5.31)

    By replacing (5.31) in (5.30); we get (5.29). Hence, in both cases (5.27) and (5.28), we obtain

    So,

    Therefore,

    (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 and be the solution of systems (1.1) and (4.8), respectively. Then, there exists a constant C independent of h such that

    Proof. Making use of (3.6), (5.6) and (4.8), we have

    As 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 . The approach used and the results obtained in this paper (optimal convergence order) remain valid when we deal with systems of 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 satisfying

    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.



    Funding sources



    This research was supported by National Research Council (CNR Italy) and did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

    Conflict of interest



    The author declares no conflicts of interest in this paper.

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