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Ways to improve biocides for metalworking fluid

  • Metalworking fluids (MWF) are mainly emulsions of oil in water containing additives such as corrosion inhibitors, emulsifiers, defoamers, and biocides. Microbial contamination of MWF is almost systematic, and some of their constituents serve as nutrients for contaminating microorganisms. Biocides for MWF are protection products used to counter microbial contaminations and growth. Ideally, a biocide for MWF should have the following non-exhaustive criteria: have a broad-spectrum activity, be usable at low concentrations, be compatible with the formulation and the physical-chemical properties of MWF, be stable over time, retain its effectiveness in the presence of soiling, have no corrosive action on metals, present no danger to humans and the environment, be inexpensive. The future lies in the development of new molecules with biocidal activity corresponding to these ideal specifications, but in the meantime, it is possible to improve the performance of existing molecules currently on the market. Different strategies for potentiation of the activity of existing biocides are possible. The compatibility of the potentiation strategies with their use in metal working fluids is discussed.

    Citation: Patrick Di Martino. Ways to improve biocides for metalworking fluid[J]. AIMS Microbiology, 2021, 7(1): 13-27. doi: 10.3934/microbiol.2021002

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  • Metalworking fluids (MWF) are mainly emulsions of oil in water containing additives such as corrosion inhibitors, emulsifiers, defoamers, and biocides. Microbial contamination of MWF is almost systematic, and some of their constituents serve as nutrients for contaminating microorganisms. Biocides for MWF are protection products used to counter microbial contaminations and growth. Ideally, a biocide for MWF should have the following non-exhaustive criteria: have a broad-spectrum activity, be usable at low concentrations, be compatible with the formulation and the physical-chemical properties of MWF, be stable over time, retain its effectiveness in the presence of soiling, have no corrosive action on metals, present no danger to humans and the environment, be inexpensive. The future lies in the development of new molecules with biocidal activity corresponding to these ideal specifications, but in the meantime, it is possible to improve the performance of existing molecules currently on the market. Different strategies for potentiation of the activity of existing biocides are possible. The compatibility of the potentiation strategies with their use in metal working fluids is discussed.


    Taxis is understood as the motion of an organism towards or away from an external stimulus. In particular when the stimulus is a chemical, it is called chemotaxis. It seems natural to address the problem of driving the motion of the organism by modifying or applying a chemical gradient. In this paper we have in mind a process in which chemotaxis takes place, the tumor angiogenesis. However, we could extend the same idea for other biological processes where the chemotaxis is involved.

    Tumor angiogenesis starts when, as a response to nutrient deprivation, cancer cells secrete a chemical factor known as Tumor Angiogenic Factors (TAF)s. (TAF)s diffuse in the extracellular matrix and activate endothelial cells which migrate, via chemotaxis, towards the source of TAF i.e., the tumor. When endothelial cells reach the tumor more nutrients are supplied to the tumor which grows further. See for instance [16] for additional details.

    It is clear that angiogenesis is a crucial step in the development of the tumor. Therefore, if we act in the (TAF)s molecules or if we modify the response of the endothelial cells to the (TAF)s molecules we may either reduce angiogenesis or avoid it.

    In this paper we consider two variables u, the density of endothelial cells and z the concentration of anti-angiogenic drug that modifies the sensitivity of a chemical gradient v (TAF) and the growth of the endothelial cells. We also assume that u, z and v are defined in a bounded domain ΩIR3. Here we are interested in the minimum of the functional defined by

    J(u,z)=a2Ωu(T)2+b2Ω×(0,T)z2, (1.1)

    where a and b are positive parameters and (u,z) is a solution of the nonlinear differential equation

    {utdiv(uV(u)α(z)v)=β(z,v)uu2inΩ×(0,T),nuV(u)α(z)nv=0onΩ×(0,T),u(0,x)=u0(x)inΩ. (1.2)

    and zUad where

    Uad:={zL2(Ω×(0,T)):z(x,t)0 for almost (x,t)Ω×(0,T)}.

    The functional has two terms; the first term refers to the amount of endothelial cells at the final stage and the second one can be seen as the cost of the drug over the time.

    Minimizing a functional (1.1), subject to a differential problem (1.2) and a convex constraint, zUad, is a problem of the optimal control theory. It is classical in this theory to call u state variable and to call z control variable, because it controls the state through the equation. The nonlinearity of (1.2) leads that the minimal problem is out of the convex framework. We will apply a generalization of the Lagrange multipliers theorem to obtain the optimality system which provides the necessary condition for the solution. This method, called Dubovitskii-Milyutin formalism (see [7]) has been mostly applied to ordinary differential equations (see [13]). In [6] it is applied to a linear partial differential problem with the feature that it is non well-posed. The optimal problem we study in this paper have the difficulty of the nonlinearity of the partial differential equation and the control constraint zUad which has empty interior in L2(Q).

    The formalism gives an optimality system constituted by two partial differential coupled equations, one is the state equation given in (1.2) and the other one is a linear equation for the adjoint variable, and a condition for the optimal control given by a projection operator. We will prove the uniqueness of solution of the optimality system for T small enough and this turns out the uniqueness of solution of the optimal problem.

    The chemotaxis is described in (1.2) by the term div(V(u)α(z)v). A similar description of the chemotaxis was introduced in [12] where there is an additional equation for v. Since, the minimal system proposed in [12] could generate singularities in finite time (see [11]) then to avoid the singularities it is proposed a model with a bounded drift term in [9]. In the case of angiogenesis, one of the first continuous models related to angiogenesis is introduced in [1]. A similar model to the one in the paper, without the variable z but with an additional equation for chemoattractant is given in [4]. The case of a therapy z satisfying an additional parabolic equation is considered in [17] and [5].

    The structure of the paper is as follows. In Section 2 we prove the existence of a unique weak solution of (1.2), global in time and positive, fixed z. The optimal control problem, the existence of the optimal control, its characterization and the uniqueness when T is small enough are studied in Section 3. In Section 4 we show some numerical simulations to illustrate the theoretical results.

    Let ΩIRN be a regular bounded domain and T>0 a fixed number. We will denote Q=Ω×(0,T) and Γ=Ω×(0,T). In this paper, we consider the problem

    {utdiv(uV(u)α(z)v)=β(z,v)uu2inQ,nuV(u)α(z)nv=0onΓ,u(0,x)=u0(x)inΩ. (2.1)

    Here, vW2,(Ω) is a known function, zL2(Q), and u0L2(Ω). The function V:[0,+)[0,+) verify V(0)=0 and VC2([0,))L([0,)). If it is needed we will extend this function by zero for negative values. The functions α:[0,+)[0,+) is in W1,([0,+)) and β:[0,+)×[0,+)[0,+) is in C2([0,))L([0,)). We would like to point out here that the main difficulty of the problem is that z is not defined on Ω.

    Let T>0 and X is a Banach space, we define the space Lp(0,T;X) of equivalence classes of measurable functions u:IX such that t(0,T)uX belongs to Lp(I) which is a Banach space for the norm

    uLp(X)={(T0upXdt)1/pif1p<Esssupt(0,T)u(t)Xifp=

    For instance, we will use

    uL2(L2)=(T0Ω|u(x)|2dxdt)1/2,
    uL2(H1)=(T0Ω|u(x)|2+|u(x)|2dxdt)1/2.

    The definition of a weak solution for the problem is

    Definition 2.1. We say that u is a weak solution of (1.2) if the following conditions are verified

    1) uW(0,T) where

    W(0,T):={uL2(0,T;H1(Ω)) such that utL2(0,T;(H1(Ω))) }.

    2) wC1([0,T]×ˉΩ):w(T,x)=0,xˉΩ, it holds

    T0ut,w(H1),H1+T0ΩuwT0ΩV(u)α(z)vw==T0Ωβ(z,v)uwT0Ωu2w+Ωu0(x)w(0,x). (2.2)

    Theorem 2.2. If u0L(Ω), u00, then there is a unique positive global weak solution to the problem (1.2) that also satisfies

    uL(0,T;L(Ω)).

    Proof. We denote by

    BR={uL2(0,T;L2(Ω)):uL2(L2)R},

    and we define the mapping S:BRL2(0,T;L2(Ω)) such that for each ˉuBR, S(ˉu)=u is the weak solution of the linear problem

    {ut=div(uV(ˉu)α(z)v)+β(z,v)uTk(ˉu)uinQ,nuV(ˉu)α(z)nv=0onΓ,u(0,x)=u0(x)inΩ. (2.3)

    where

    Tk(φ)={kifφ>k,φ+ifφk.

    We will prove the existence of a fixed point for S which is a weak solution of a truncated nonlinear problem. After that we will justify that the weak solution to the truncated problem is a solution of (1.2). At the end we will show the uniqueness of weak solution of (1.2).

    Step1_ First of all we show that the operator S is well defined. The existence and the uniqueness of the weak solution for the linear problem (2.3) follow from [14]. More precisely, by [14], Theorem 6.39] the problem (see [14], p. 136])

    {ut=div(u)+Ni=1ifi(x,t)+c0(x,t)uinQnu+Ni=1fiνi=0onΓu(0,x)=u0(x)inΩ. (2.4)

    (where fi=V(ˉu)α(z)iv and c0(x,t)=β(z,v)Tk(ˉu) in (2.3) and ν=(ν1,..,νN) is the outward normal to Ω), has a unique weak solution i.e. uL2(0,T;H1(Ω))L(0,T;L2(Ω)) such that

    T0Ω(uwt+|u||w|Ni=1fiiwc0(x,t)uw)dxdt=Ωu0(x)v(0,x)dx,

    for each wC1([0,T]×ˉΩ) such that w(T,x)=0,xΩ. Note that, if utL2(0,T;(H1(Ω))) then the previous definition it is exactly the one given in Definition (2.1). Therefore we should show that utL2(0,T;(H1(Ω)).

    ut,φ=Ωuφdxdt+ΩV(ˉu)α(z)vφdx+Ωc0(x,t)uφdxuL2φL2+V(ˉu)α(z)vL2φL2+c0(x,t)uL2φL2(uL2+V(ˉu)α(z)vL2+c0(x,t)uL2)φH1,

    where , stands for the duality pairing between (H1(Ω)) and H1(Ω). Therefore

    ut(H1(Ω))=sup0φH1(Ω)ut,φφH1uL2+V(ˉu)α(z)vL2+c0(x,t))uL2,

    and

    utL2(0,T;(H1(Ω)))=(T0ut2(H1(Ω))dt)1/2C (2.5)

    i.e., utL2(0,T;(H1(Ω))). Hence, S is well defined.

    In what follows we will apply the Schauder fixed point theorem to get the existence of a fixed point for S in BR.

    Step2_ We claim that there exists R>0 and T>0 such that S(BR)BR. In fact, multiplying the equation of (2.3) by u and integrating in Ω, it results

    12ddtΩu2=Ω|u|2+Ωα(z)V(ˉu)uv+Ωβ(z)u2ΩTk(ˉu)u212Ω|u|2+12Ωα(z)2V(ˉu)2|v|2+βΩu2βΩu2+C.

    Thus for y(t)=Ωu2 we obtain the problem

    {y(t)ay(t)+b,y(0)=Ωu20,

    for some positive constants a,b. Consequently,

    0y(t)(y0+ba)eatba.

    We can choose any R>y(0)=u02 and determine T>0 such that

    y(t)=Ωu2(x,t)dxR,t(0,T), (2.6)

    and

    uL2(L2)=(T0Ω|u(x,t)|2dxdt)1/2R1/2T1/2R,

    if we take R>1 and T<1. So, for Ru02+1 there exists T<1 such that S(BR)BR.

    On the other hand, taking into account that

    y(t)+12Ω|u|2ay(t)+b,

    we can choose R, determine T and integrate the inequality on the interval (0,T) to obtain

    y(T)y(0)+T0Ω|u|2dxdtaT0ydt+bT0dt,
    T0Ω|u|2dxdty(0)+aT0Ωu2dxdt+bTu02+aR2+bT. (2.7)

    Consequently, uL2(H1) is bounded.

    Step3_ We claim that S is a continuous mapping. We will prove that

    S(ˉu1)S(ˉu2)L2(L2)Φ(k,ˉu1ˉu2L2(L2))ˉu1,ˉu2BR, (2.8)

    with Φ(k,s)0 when s0 for each fixed k, and T as in Step 2. If we denote u1=S(ˉu1) and u2=S(ˉu2), it holds

    {(u1)t=div(u1V(ˉu1)α(z)v)+β(z,v)u1Tk(ˉu1)u1,(u2)t=div(u2V(ˉu2)α(z)v)+β(z,v)u2Tk(ˉu2)u2,

    and taking w=u1u2=S(ˉu1)S(ˉu2), we have

    wt=Δwdiv((V(ˉu1)V(ˉu2)α(z)v)+β(z,v)w+(Tk(ˉu2)u2Tk(ˉu1)u1)=Δwdiv((V(ˉu1)V(ˉu2)α(z)v)+β(z,v)w+Tk(ˉu1)w++(Tk(ˉu2)Tk(ˉu1))u2.

    On multiplying the previous inequality by w and integrating on Ω we obtain

    12ddtΩw2=Ω|w|2Ω(V(ˉu1)V(ˉu2))α(z)vw++Ωβ(z,v)w2+ΩTk(ˉu1)w2+Ω(Tk(ˉu2)Tk(ˉu1))u2w. (2.9)

    Next, we provide a bound to the terms in the right hand side of (2.9)

    |Ω(V(ˉu1)V(ˉu2))α(z)vw|12Ω|w|2+12Ω(V(ˉu1)V(ˉu2))2α(z)2|v|212Ω|w|2+Cα2v2Ω|ˉu1ˉu2|2,
    Ωβ(z,v)w2βΩw2,
    ΩTk(ˉu1)w2kΩw2,
    Ω(Tk(ˉu2)Tk(ˉu1))u2w12Ωw2+12Ω(Tk(ˉu2)Tk(ˉu1))2u22.

    The Sobolev embedding H1(Ω)L6(Ω) together with the estimate u2L2(0,T;H1(Ω)) implies that u2L2(0,T;L6(Ω)). On the other hand if we apply the Hölder inequality with exponents 3 and 3/2 to the last term of the above inequality we obtain

    Ω(Tk(ˉu2)Tk(ˉu1))2u22u22L3/2(Tk(ˉu2)Tk(ˉu1))2L3=u22L3(Tk(ˉu2)Tk(ˉu1))2L6.

    Using the interpolation inequality,

    u22L3u2L2u2L6C(R)u2L6,

    and, thus

    Ω(Tk(ˉu2)Tk(ˉu1))2u22C(R)u2L6(Tk(ˉu2)Tk(ˉu1))2L6.

    Then (2.9) becomes

    12ddtΩw2C1ˉu1ˉu22L2+C2w2L2+C(R)u2L6(Tk(ˉu2)Tk(ˉu1))2L6. (2.10)

    We denote by y(t)=Ωw(x,t)2dx. Since y(0)=0 it follows from (2.10) that

    y(t)ectT0ecs(C1ˉu1ˉu22L2+C(R)u2L6(Tk(ˉu2)Tk(ˉu1))2L6)dsecT(C3ˉu1ˉu2L2(L2)+T0C(R)ecsu2L6(Tk(ˉu2)Tk(ˉu1))2L6ds). (2.11)

    By the Hölder inequality it holds

    T0C(R)ecsu2L6(Tk(ˉu2)Tk(ˉu1))2L6ds
    (T0u22L6)1/2(T0(Tk(ˉu2)Tk(ˉu1)2L6)3)1/3(T0(C(R)ecs)6ds)1/6.

    Since

    (T0(Tk(ˉu2)Tk(ˉu1)2L6)3)1/3=(T0Ω|Tk(ˉu2)Tk(ˉu1)|6dxds)1/3=
    =(T0Ω|Tk(ˉu2)Tk(ˉu1)|4|Tk(ˉu2)Tk(ˉu1)|2dxds)1/3k4/3(T0Tk(ˉu2)Tk(ˉu1)2L2ds)1/3k4/3(T0ˉu2ˉu12L2ds)1/3=k4/3ˉu2ˉu12/3L2(L2),

    it results from (2.11),

    y(t)C4ˉu2ˉu12L2(L2)+C5k4/3ˉu2ˉu12/3L2(L2). (2.12)

    Integrating on (0,T), we obtain

    S(ˉu1)S(ˉu2)L2(L2)T(C4ˉu2ˉu12L2(L2)+C5k4/3ˉu2ˉu12/3L2(L2)),

    which proves the desired continuity.

    Step4_ We claim that S is a compact mapping in L2(Q). We know that for each ¯uBR S(¯u)=u is bounded in L2(0,T;H1(Ω)) and S(¯u)t=ut is bounded in L2(0,T;(H1(Ω))) (see (2.7) and (2.5)) then by the Lions-Aubin Lemma (see for instance [18]) S(BR) is embedded compactly in L2(0,T;L2(Ω)).

    Step5_ By the Schauder Theorem we have a solution to the problem

    {ut=div(uV(u)α(z)v)+β(z,v)uTk(u)uinQ,nuV(u)α(z)nv=0onΓ,u(0,x)=u0(x)inΩ. (2.13)

    We will prove that any solution of (2.13) is nonnegative. Let u=min(u,0). Taking u as a test function in the above equation we obtain

    d2dtΩ(u)20.

    Integrating on the space variable infer

    0Ωu(t)2Ω((u0))2=0

    Therefore, u(t)=0 a.e. in Ω.

    Step6_ We prove that a solution to (2.3) satisfies uL(0,T;L(Ω)) independently of the k, as a consequence a solution of (2.13) is a solution to (1.2) for k sufficiently large. We multiply (2.3) by pTk(u)p1 for p2, a cut off function that approximate to pup1 (see for instance [3], p. 1196]) to get

    ddtΩTk(u)p+4(p1)pΩ|Tk(u)p/2|2=pΩβ(z,v)uppΩTk(u)p+1++2ΩTk(u)p/21V(u)(up/2).

    Since V(0)=0 and V is a Lipschitz function then,

    ddtΩTk(u)p+2Ω|Tk(u)p/2|2pβΩTk(u)p+2ΩTk(u)p/21|V(u)V(0)||(Tk(u)p/2)|pβΩTk(u)p+2CΩTk(u)p/2|(Tk(u)p/2)|pβΩTk(u)p+2C(ΩTk(u)p)1/2(Ω|(Tk(u)p/2)|2)1/2

    Adding on both sides of the above inequality ΩTk(u)p we deduce that

    ddtΩTk(u)p+ΩTk(u)p+32Ω|(Tk(u)p/2)|2β(p+1+2C2β)ΩTk(u)p. (2.14)

    At this point we will provide a bound for Ωup. We claim that for any ϵ>0

    ΩTk(u)pϵΩ|Tk(u)p/2|2+C(ϵ)(ΩTk(u)p/2)2. (2.15)

    Next lines are devoted to the proof of the above inequality. Let ¯z=1|Ω|Ωz. We know that

    Ωz2=Ω(z¯z)2+1|Ω|(Ωz)2.

    On the other hand, by the Hölder inequality and the Poincare-Wintinger inequality we have

    Ω(z¯z)2z¯z3/23z¯z1/21=z¯z3/232z1/21ϵz¯z23+C(ϵ)(Ω|z|)2ϵΩ|z|2+C(ϵ)(Ω|z|)2.

    Therefore if we take z=Tk(u)p/2 the claim easily follows. By (2.15) we get from (2.14) that

    ddtΩTk(u)p+ΩTk(u)pC(p+C)(ΩTk(u)p/2)2.

    For any 0tT<Tmax, where Tmax stand for the maximal existence time, the above inequality asserts

    ΩTk(u)p(t)Ωup0et+C(p+C)sup0tT(ΩTk(u)p/2)2¯C(p+C)max{u0p,sup0tT(ΩTk(u)p/2)2}. (2.16)

    Let us define

    θ(p/2):=max{u0,sup0tT(ΩTk(u)p/2)2/p}.

    From (2.16), by a recursive procedure, we have

    (ΩTk(u)p(t))1/p(¯C(p+k))1/pθ(p/2)(¯C(p+C))1/p(¯C(p/2+C))2/pθ(p/4)

    In particular, for p=2j we deduce

    (ΩTk(u)2j(t))2j¯Csjji=0(2i+C)2iθ(1),

    where sj=ji=02i. Taking into account that

    θ(1)max{u0,sup0tTΩu}:=C1,

    Therefore we can take k+ to get

    (Ωu2j(t))2j¯Csjji=0(2i+C)2iC1

    Since

    i=02iC,ji=0(2i+C)2iC

    we can take j to conclude that

    u(t)C.

    Step7_ Uniqueness of solution for (1.2). Let be two solutions of (1.2), u1,u2, and we take w=u1u2. We can argue as in Step 3 to obtain

    12ddtw2L2aw2L2+Ω(u1+u2)w2 (2.17)

    for some a>0. By the boundedness of u1+u2 we easily get the result by the Gronwall Lemma.

    In this section we are interested in the minimum of the functional

    J:W(0,T)×L2(Q)IR+,

    defined by

    J(u,z)=a2Ωu(T)2+b2Qz2, (3.1)

    a and b are positive parameters and (u,z) has to be a solution of the nonlinear differential equation

    {ut=(uα(z)V(u)v)+β(z,v)uu2inQ,nuα(z)V(u)nv=0inΓ,u(x,0)=u0(x)inΩ, (3.2)

    and besides, we claim z that

    zL2(Q),  z0   for almost (x,t)Q   (a convex constraint). (3.3)

    The function u represents the endothelial cell (EC) density, the initial data u0L(Ω), u00 and not identically zero, means that the angiogenesis process has already begun, the TAF concentration is given by v and it is known in the equation and the concentration of the anti-angiogenic drug is represented by z which is the control function. Minimizing J means to find the appropriated therapy z such that the total amount of EC at the end of the treatment, this is T, is the smallest as possible applying the least amount of drug possible. The term Ωu(T)2 represents the total amount of EC in Ω at the final time T and the term Qz2 plays the total amount of drug in Ω for the time interval (0,T).

    The proof of the existence of solution of the optimal problem (3.1)–(3.3) is standard by a minimizing sequence. We will do the proof for the reader's convenience.

    Theorem 3.1. The optimal control problem (3.1)–(3.3) has a solution (ˆu,ˆz) in W(0,T)×L2(Q).

    Proof. Let be {un,zn} a minimizing sequence such that {J(un,zn)} is decreasing to ˆβ, which is the infimum of J(u,z) subject to (3.2) and (3.3). Then, {un(T)} is bounded in L2(Ω) and {zn} is bounded in L2(Q). So, there exist ˆzL2(Q) and a subsequence of {zn} which converges to ˆz in L2(Q)-weakly. Multiplying the differential equation by un and integrating by parts, we obtain

    12ddtun2L2+un2L2=Ωα(zn)V(un)vun+Ωβ(zn,v)u2nu3n. (3.4)

    Using that α,VL(IR), βL(IR2) and un0, we have

    12ddtun2L2C1+C2un2L2

    and so,

    un(t)2L2u02L2+C1T+t0C2un(s)2L2ds.

    Applying Gronwall's Lemma we deduce that {un} is bounded in L(0,T;L2(Ω)). Using this fact and returning to (3.4) we obtain that {un} is bounded in L2(0,T;H1(Ω)). This boundness gives that {(un)t} is bounded in L2(0,T;(H1(Ω))). Effectively, taking any wH1(Ω) and applying the duality product H1(Ω),H1(Ω) in the equation of (3.2) we have

    (un)t,w=Ωunw+ΩV(un)α(z)vw+Ωβ(z,v)unwΩu2nw.

    Since {un} is bounded in L(0,T;L2(Ω)), {un} is bounded in L2(0,T;H1(Ω)), applying the Hölder inequality, we get

    |(un)t,w|unL2wL2+C1wL2+C2wL2.

    Then, we obtain

    (un)t(H1(Ω))unL2+C.

    So,

    T0(un)t2(H1(Ω))C1+C2un2L2(0,T;H1(Ω)),

    which proves that {(un)t} is bounded in L2(0,T;(H1(Ω))).

    And this gives that {un} is bounded in W(0,T). By the Aubin-Lions Lemma (see for instance [15]) we can assure that there exist a subsequence of {un} and a function ˆu in W(0,T) such that (still denoting by un the subsequence) {un} converges to ˆu in L2(Q) strongly.

    The variational formulation of the problem (3.2) for the functions (un,zn) is

    T0(un)t,φ+Qunφ=Qα(zn)V(un)vφ+Qβ(zn,v)unφQu2nφ. (3.5)

    The functions α and β are convex and continuous (for β it is only necessary to be convex and continuous in the first variable), V is a continuous function in IR. Then, using the convergence of the sequences, {un} and {zn}, it is possible to pass to the limit in (3.5) and so, we obtain that (ˆu,ˆz) solves the problem (3.2). Besides, as {zn} weakly converges to ˆz in L2(Q) and zn0, we have that ˆz0. This proves that (ˆu,ˆz) belongs to the admissible set of the optimal control problem. The functional J is continuous and convex, so

    J(un,zn)J(ˆu,ˆz),

    which gives that J(ˆu,ˆz)=ˆβ and then, (ˆu,ˆz) is a solution of the problem (3.1)–(3.3).

    Remark 1. The solution of the optimal problem might not be unique.

    In order to obtain the first order necessary conditions of the optimal control problem, we enunciate the Duvobitskii-Milyutin theorem (see [7]), which is a generalization of the Lagrange's Multipliers theorem.

    Theorem 3.2. Let X be a normed space. Assume that the functional J:XIR has a local minimum with constraints Z=n+1i=1ZiX at a point x0Z. Assume that J is regularly decreasing at x0, with decreasing (and convex) cone DC0; the inequality constraints Zi, 1in, are regular at x0, with feasible (and convex) cones FCi, 1in; the equality constraint Zn+1 is also regular at x0, with tangent (and convex) cone TCn+1. Then, there exist continuous linear functionals f0DC0, fiFCi for 1in and fn+1TCn+1 (we denote by the corresponding dual cone), not all identically zero, such that they satisfy the Euler-Lagrange equation:

    f0+ni=1fi+fn+1=0in X.

    The constraint (3.3) must be considered as an equality constraint, together with (3.2), because the convex U={zL2(Q): z0} is a set with empty interior in L2(Q). So, applying the Duvobitskii-Milyutin theorem to (3.1)–(3.3) there exist two functionals, f0(DC), f(TC) such that

    f0+f=0  W(0,T)×L2(Q),

    and they are not simultaneously identically zero.

    The identification of the cones. Following [2], we are going to define the cones and their dual ones in a point. The functional f0(DC) where (DC) is the dual decreasing cone in (ˆu,ˆz), is given by

    f0=λJ(ˆu,ˆz)  with  λ0.

    The computation of the tangent cone and its dual one is technically more complicated because there are two constraints, (3.2) and (3.3), considered like equalities generating a tangent cone. Each of these constraints generates, by itself, a tangent cone. We denote Z1 the constraint set given by (3.2) and TC1 the tangent cone associated to Z1 in (ˆu,ˆz). By Lyusternik's theorem (see [2]), this cone is the set

    TC1={(u,z)W(0,T)×L2(Q): solution the problem
    {utΔu+(α(ˆz)V(ˆu)uv)β(ˆz,v)u+2ˆuu++(α(ˆz)zV(ˆu)v)1β(ˆz,v)zˆu=0,nuα(ˆz)V(ˆu)unvα(ˆz)zV(ˆu)nv=0,u(0)=0. (3.6)

    And (TC1) is

    (TC1)={h,(u,z)=0  (u,z)TC1}.

    If we consider the convex constraint (3.3) and we call Z2 to this set, the dual tangent cone to Z2 in the point (ˆu,ˆz), denoted by (TC2), is

    (TC2)={(0,g):  g,zˆzL2(Q),L2(Q)0  z0}.

    The question is if we could assure that the dual cone to the cone associated to the set Z1Z2 in (ˆu,ˆz), denoted by (TC), would be equal to (TC1)+(TC2). We can always assure that (TC1)+(TC2) is included in (TC), but the converse inclusion will be true if we prove that (TC1) and (TC2) are a system of the same sense cones (see [19], Definition 2.2.1]). The definition of a system of the same sense cones (SSS) is the following (see [19]):

    Definition 3.3. Let {Ci}ni=1 a system of cones in a normed space X. It is a (SSS) if for every M>0 there exist M1,...Mn such that, for each x=ni=1xi, xiCi, the inequality xM implies the inequalities xiMi,  i=1,...,n.

    This definition is not handle to determine if a system of cones is or not a (SSS). We enunciate a theorem that characterizes a (SSS) constituted by two cones when one of them is given by a linear continuous operator (see [19]).

    Theorem 3.4. Let E=X×Y be the product space of two normed spaces and C1,C2 the following cones:

    C1={(x,y)E:  x=Ay},

    where A is a linear continuous operator from Y to X, and

    C2=X×~C2,

    and ~C2 is a cone in Y. If we denote by (Ci) the dual cone to Ci, then

    (C1)={(x,y)E,  y=Ax}

    and

    (C2)={(0,y)E,  y~C2}.

    Besides, the system {(C1),(C2)} is a (SSS).

    In our case, C1=TC1 and C2=TC2. The tangent cone associated to (3.2), TC1, is given by a linear continuous operator, the operator of the differential equation (3.6). By this theorem, we have that (TC)=(TC1)+(TC2).

    In the following theorem we obtain the optimality system of the optimal control problem (3.1)–(3.3).

    Theorem 3.5. If (ˆu,ˆz) is a solution of the optimal control problem (3.1)–(3.3), then there exists pW(0,T) such that (ˆu,ˆz,p) satisfies

    {ˆutΔˆu=(α(ˆz)V(ˆu)v)+β(ˆz,v)ˆuˆu2inΩ×(0,T),nˆuα(ˆz)V(ˆu)nv=0inΩ×(0,T),ˆu(x,0)=u0(x)inΩ,ptΔp=α(ˆz)V(ˆu)vp+β(ˆz,v)p2ˆupinΩ×(0,T),np=0inΩ×(0,T),p(T)=ˆu(T)inΩ,ˆz=[ab(α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup)]+.  (3.7)

    Proof. The Euler-Lagrange equation

    The Euler-Lagrange equation, (E-L), is

    f0+f1+f2=0  in  W(0,T)×L2(Q),

    with f0(DC), f1(TC1) and f2(TC2). We know that

    f0=λJ(ˆu,ˆz),  λ0
    f1,(u,z)=0  (u,z)  solution of (3.6)
    f2=(0,˜f2)  with  <˜f2,zˆz>≥0  z0.

    We will see that λ cannot be zero.

    Suppose that λ=0. Then, the (E-L) equation is

    f1+f2=0.

    Applying this equation to any (u,z)TC1 we have f2,(u,z)=0 and so, ˜f2,z=0 for every z such that there exists u verifying (u,z)TC1. Since for every zL2(Q) there exists this u, we have that ˜f20 and then, f20 and f10, which is impossible. So, λ0, we can rescale λ=1 and the (E-L) equation is

    J(ˆu,ˆz)+f1+(0,˜f2)=0  in  W(0,T)×L2(Q).

    We have that

    J(ˆu,ˆz)(u,z)=(˜f2,z)  (u,z)TC1,

    this is

    aΩˆu(T)u(T)+bQˆzz=(˜f2,z)  (u,z)TC1. (3.8)

    In the following, we describe a standard procedure in order to rewrite the first term in (3.8) using the optimal control ˆz. This is done defining an adjoint problem. We multiply the differential equation of (3.6) by a function p, which will be the solution of the adjoint problem, and we integrate by parts. Taking account, we obtain

    u,ptΔpα(ˆz)V(ˆu)vpβ(ˆz,v)p+2ˆup+Ωu(T)p(T)
    z,α(ˆz)V(ˆu)vpz,1β(ˆz,v)ˆup+T0Ωunp=0.

    We define the adjoint problem

    {ptΔp=α(ˆz)V(ˆu)vp+β(ˆz,v)p2ˆupnp=0p(T)=ˆu(T). (3.9)

    Then

    Ωu(T)ˆu(T)z,α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup=0,

    and replacing this equality in (3.8) we obtain

    az,α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup+bQˆzz=(˜f2,z),

    for every zL2(Q) because, as we have already said, for every zL2(Q) there is a function u verifying (u,z)TC1. So, we have obtained the functional ˜f2:

    ˜f2=a(α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup)+bˆz.

    Since (˜f2,zˆz)0 for every z0, we have that

    (a(α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup)+bˆz,zˆz)0  z0,

    and this is equivalent to

    ˆz=PU(ab(α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup)),

    where U is the convex set {zL2(Q):  z0} and P is the projection operator of L2(Q) on U.

    In this case, because of the particular convex set we have, the operator P is well-known. It is

    PUg=g+=max{g,0},

    hence,

    ˆz=[ab(α(ˆz)V(ˆu)vp+1β(ˆz,v)ˆup)]+. (3.10)

    We have just obtained the optimality system, given by the theorem.

    Next, we are going to prove the uniqueness of solution of (3.7) for T small enough. To this aim it will be useful to obtain an uniform estimate in time for p on the interval [0,T]. In order to get this estimate we will suppose additional restrictions on α and v. In particular, we require that α=α0 is a constant function and nv=0. In this case the optimality system is given by the following equations:

    utΔu=(α0V(u)v)+β(z,v)uu2inΩ×(0,T), (3.11)
    nu=0onΩ×(0,T), (3.12)
    u(x,0)=u0(x)inΩ, (3.13)
    ptΔp=α0V(u)vp+β(z,v)p2upinΩ×(0,T), (3.14)
    np=0onΩ×(0,T), (3.15)
    p(T)=u(T)inΩ, (3.16)
    z=[ab(1β(z,v)up)]+.  (3.17)

    Lemma 3.6. We have that u(T)W1,3(Ω)C.

    Proof. Using the variations of constants formula in (3.11)–(3.13) we get

    u(t)=etAu0+t0e(ts)Ah(u,v)ds, (3.18)

    where A=Δ+I with domain {ψW2,3(Ω):nu=0} and

    h(u,v):=(α0V(u)v)+(β(z,v)+1)uu2.

    Let Xβ, β(0,1) the fractional powers of X. By [8], Theorem 1.6.1] we have that

    XβW1,3,

    for any β(1/2,1). Taking the norm W1,3 in (3.18) and applying the above embedding in the right hand side we infer

    u(T)W1,3(Ω)etAu0Xβ+T0e(ts)AhXβ.

    By [8], Theorem 1.4.3] we deduce

    u(T)W1,3(Ω)Cβeδttβu0L3+T0eδ(ts)(ts)βh(u,v)L3

    Let us notice that by the regularity of β, V and v and the embedding W1,3(Ω)L6(Ω) we have

    h(u,v)L3CuW1,3(Ω).

    As a consequence, we can conclude by the singular Gronwall Lemma (see [8], Section 1.2.1]).

    In what follows we will show the regularity for the backward linear parabolic equation (3.14)–(3.16). The backward equation can be written in a forward way by the change of variable ¯p(t)=p(Tt). Therefore, the problem is

    {¯ptΔ¯p=α0V(u)v¯p+β(z,v)¯p2u¯pin Ω×(0,T),n¯p=0on Ω×(0,T),¯p(0)=u(T)in Ω. (3.19)

    Lemma 3.7. We have that maxt[0,T]p(t)W1,3(Ω)C.

    Proof. As in the previous Lemma we apply the variation of constants formula for ¯p to get

    ¯p(t)=etAu(T)+t0e(ts)Af(u,v,¯p)ds,

    where

    f(u,v,¯p):=α0V(u)v¯p+(β(z,v)+1)¯p2u¯p.

    We take W1,3(Ω) in the variations of constants formula for p and we apply [8], Theorem 1.4.3, Theorem 1.6.1] and ([10], p. 59]) to infer

    ¯p(t)W1,3(Ω)Cu(T)W1,3(Ω)+T0(ts)βeδ(ts)f(u,v,¯p)L3ds

    with β(1/2,1). Since f(u,v,¯p)L3C¯pW1,3(Ω) then by the Gronwall Lemma (see [20]) we can conclude the result.

    Theorem 3.8. Let be α=α0 a constant function, the TAF concentration v satisfies that nv=0 on Γ, the function β is twice differentiable with respect to the first variable and it verifies that 211β is bounded and its norm in L is small. Then, the optimality system (3.7) has a unique solution when the time T is small enough.

    Proof. Using the hypothesis α=α0, the optimality system is given by these equations:

    z=[ab(1β(z,v)up)]+.  (3.20)

    We call T to the following operator:

    T:L2(L2)L2(L2)
    zT(z)=[ab1β(z,v)up]+,

    where L2(L2) is the space L2(0,T;L2(Ω)). The function u is obtained by the data z:

    T1:L2(L2)W(0,T)
    zu, the solution of (3.11), (3.12), (3.13)

    Having z and u, the function p can be solved:

    T2:L2(L2)×W(0,T)W(0,T)
    (z,T1(z))p, the solution of (3.14), (3.15), (3.16)

    With this notation, the operator T is

    T(z)=[ab1β(z,v)T1(z)T2(z,T1(z))]+.

    The existence of an optimal control is the existence of a fixed point of T. This is guaranteed by the Theorem 3.1. We want to prove that T is contractive. Then, there will be a unique fixed point, ˆz, and we can say that the optimality system has a unique solution and so, the optimal control problem has a unique optimal control.

    Let be z1,z2L2(L2).

    T(z1)T(z2)L2(L2)ab1β(z1,v)u1p11β(z2,v)u2p2L2(L2),

    where we have called ui=T1(zi) and pi=T2(zi,T1(zi)), i=1,2. Adding and subtracting the appropriate terms and using the triangular inequality we have

    T(z1)T(z2)L2(L2)ab[211βL(L)z1z2L2(L2)u1L(L)p1L(L)+
    +1βL(L)u1u2L2(L2)p1L(L)+1βL(L)p1p2L2(L2)u2L(L)].

    By Theorem 2.2, it is known that uL(L), where u is a solution of (3.11), (3.12) and (3.13), and pL(L), p is a solution of (3.14), (3.15) and (3.16), are bounded.

    Let w=u1u2. Writing the equation for w, multiplying by w and integrating in Ω we obtain the following estimate:

    12ddtw2L2+12w2L212Ωα20(V(u1)V(u2))2|v|2+Ωβ(z1,v)w2+
    +Ω(β(z1,v)β(z2,v))u2wΩ(u1+u2)w2.

    Applying the medium value theorem, V, β, 1β and u2L(Q) are bounded and the Young's inequality, we have

    ddtw2L2Aw2L2+Bz1z22L2,   A,B>0.

    Then,

    w(t)2L2BeAtz1z22L2(L2). (3.21)

    Therefore,

    u1u22L2(L2)BA(eAT1)z1z22L2(L2).

    We do a similar reasoning with p1p2. Let η=p1p2. Writing the equation for η, multiplying it by η and integrating in Ω, we get

    ddtΩη2+Ω|η|2=α0ΩV(u1)vηη+α0Ω(V(u1)V(u2))vp2η++Ωβ(z1,v)η2+Ω(β(z1,v)β(z2,v))p2η2Ωu1η22Ωwp2η. (3.22)

    We multiply the previous equality by 1 and we apply the Young inequality and the Hölder inequality to obtain

    ddtΩη2+Ω|η|2ϵΩ|η|2+C(ϵ)Ωη2++(u1u2)vL2p2L3ηL6+Cz1z22L2+CΩw2.

    Hence, the Sobolev embedding and (3.21) entails

    ddtΩη2C(ϵ)Ωη2+C(ϵ)BeATz1z2L2(L2)+Cz1z2L2.

    Next, we solve the differential equation to get

    Ωη2eC(ϵ)tu1(T)u2(T)L2(Ω)+ec(ϵ)tC(ϵ)BeATz1z2L2(L2)C(T)++eC(ϵ)tt0eC(ϵ)sCz1z22L2.

    We apply (3.21) for t=T to obtain

    Ωη2eC(ϵ)tC(T)z1z2L2(L2).

    Therefore, after integration on the interval [0,T] we have

    ηL2(L2)C(T)eC(ϵ)T1C(ϵ)z1z2L2(L2).

    After all of these inequalities we get the following equation

    T(z1)T(z2)L2(L2)C(C(T)+211βLu1Lp1L)z1z2L2(L2),

    where C(T) goes to zero as T goes to zero. If we assume that 211βL is small enough we can say that the operator T is contractive if T is small enough. Therefore, there exists a unique optimal control if T is small.

    We are going to solve the optimality system in some particular cases. The program has been done in FreeFEM, using P1-Lagrange finit element method and Euler method to solve the problem of u and p. In the case of u, because of the nonlinearity of the equation, we have used the Newton's method, and for the problem of p we have to solve a backward equation. The optimality system is a recursive equation, by the hypothesis of Theorem 3.8, it is contractive for T small enough, so, we have chosen z, we get u and p, we obtain a new z and we repeat until the difference between this one and the previous one satisfies the stop test.

    The domain Ω is plotted in the following figure:

    We have chosen the following data:

    The coefficient for the chemotaxis term

    α0=50,

    the time step,

    dt=0.1

    and a small final time, T,

    T=0.5.

    We solve this problem to get v

    Δv+v=g,  Ωvn=0,  Ω

    with

    g=10.01+x2+y2.

    The graphic of the TAF concentration is shown in Figure 1.

    The initial density of endothelial cells is

    u0=x2+y2+100,

    and it is drawn in Figure 2:

    Figure 1.  TAF concentration.
    Figure 2.  Density of ECs at the initial time.

    The function V which appears in the chemotaxis term is given by

    V(u)=u1+u2,

    and the function β which is the growth rate of the drug z, is

    β(z,v)=kexp(z)v1+v2,

    we have taken k=1. Finally, the coefficients, a and b are equal to one, that means that we optimize u(T) and z with the same weights. At the final time, the density of ECs is shown in Figure 3:

    And the concentration of z at the final time is plotted in Figure 4.

    The functional and norms are

    J(ˆu,ˆz)=65.1,   u(T)L2(Ω)=6.27,   zL2(Q)=9.53

    When the chemotaxis factor, α0, is smaller, the drug concentration is smaller too, it is necesary less chemical agent to control the angiogenesis process. We take

    α0=5,

    instead of 50 and the rest of the data the same as before. The density of ECs at the final time is shown in Figure 5.

    In this case, the value of J on the optimal and the norms of u(T) and z are

    J(ˆu,ˆz)=66.46,   u(T)L2(Ω)=5.7,   zL2(Q)=10,

    and the concentration of drug at the final times is drawn in Figure 6. If we prioritize to minimize the term of u(T), taking

    a=10,   b=1

    and we choose a growth function β smaller than the previous one, taking

    k=0.1

    the functions u(T) and z(T) decrease notably, as we can see in Figure 7 and Figure 8.

    Now, the functional and the norms are

    J(ˆu,ˆz)=70,   u(T)L2(Ω)=3,76,   zL2(Q)=0,46.
    Figure 3.  Density of ECs at the final time.
    Figure 4.  Concentration of the drug at the final time.
    Figure 5.  Density of ECs at the final time with α0 = 5.
    Figure 6.  Concentration of the drug at the final time with α0=5.
    Figure 7.  Density of ECs at the final time with α0=5 and k=0.1.
    Figure 8.  Concentration of the drug at the final time with α0=5 and k=0.1.

    The authors declare no conflict of interest.


    Acknowledgments



    This work was supported in part by a grant from the company CONDAT.

    Conflict of interest



    The author declares no conflicts of interest in this article.

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