# Null Controllability of a Nonlinear Age Structured Model for a Two-Sex Population

Authors: Amidou Traoré; Okana S. Sougué; Yacouba Simporé; Oumar Traoré
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

This paper is devoted to study the null controllability properties of a nonlinear age and two-sex population dynamics structured model without spatial structure. Here, the nonlinearity and the couplage are at the birth level. In this work, we consider two cases of null controllability problem. The first problem is related to the extinction of male and female subpopulation density. The second case concerns the null controllability of male or female subpopulation individuals. In both cases, ifA is the maximal age, a time interval of duration A after the extinction of males or females, one must get the total extinction of the population. Our method uses first an observability inequality related to the adjoint of an auxiliary system, a null controllability of the linear auxiliary system, and after Kakutani’s fixed-point theorem.


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## Body

## 1. Introduction

In practice, this study takes place in the fight against malaria. Malaria is a serious disease (in 2017, there were 219 million cases in the world [1]), and our work takes its importance in the strategy to fight against it.In fact, malaria is a vector-borne disease transmitted by an infective female anopheles mosquito. A malaria control strategy in Brazil or Burkina Faso consists of releasing genetically modified male mosquitoes (precisely sterile males) in the nature. This can reduce the reproduction of mosquitoes since females mate only once in their life cycle.In the theoretical framework, very few authors have studied control problems of a two-sex structured population dynamics model.The control problems of coupled systems of population dynamics models take an intense interest and are widely investigated in many papers. Among them, we have [2–4] and the references therein. In fact, in [2], the authors studied a coupled reaction-diffusion equations describing interaction between a prey population and a predator one. The goal of the above work is to look for a suitable control supported on a small spatial subdomain which guarantees the stabilization of the predator population to zero. In [4], the objective was different. More precisely, the authors consider an age-dependent prey-predator system and they prove the existence and uniqueness of an optimal control (called also “optimal effort”) which gives the maximal harvest via the study of the optimal harvesting problem associated with their coupled model. In [5], He and Ainseba study the null controllability of a butterfly population by acting on eggs, larvas, and female moths in a small age interval.In [3], the authors analyze the growth of a two-sex population with a fixed age-specific sex ratio without diffusion. The model is intended to give an insight into the dynamics of a population where the mating process takes place at random choice and the proportion between females and males is not influenced by environmental or social factors but only depends on a differential mortality or on a possible transition from one sex to the other (e.g., in sequential hermaphrodite species). Simpore and Traore study in [6] the null controllability of a nonlinear age, space, and two-sex structured population dynamics model. They first study an approximate null controllability result for an auxiliary cascade system and prove the null controllability of the nonlinear system by means of Schauder’s fixed-point theorem.Unlike the model treated in [6], we consider a nonlinear cascade system with two different fertility rates and without space variable. The fertility rate of the male λ and the fertility rate β of the female depend on the total population of the fertile males.Remark 1.
Here, we are talking about the null controllability as an extinction of the population, but we can see in the sense of exact controllability to the trajectories since there is an equivalence between the null controllability and exact controllability to the trajectories in the linear case.


## 2. Model and Main Results

In this paper, we study the null controllability of an infinite dimensional nonlinear coupled system describing the dynamics of a two-sex structured population. Letm,f be the solution of the following system:
(1)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=χΘ2vfinQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,Mfa,tdainQT,f0,t=γ∫0Aβa,Mfa,tdainQT,M=∫0Aλama,tdainQT,where T is a positive number, Q=0,A×0,T, Θ=0,a2×0,T, Θ1=a1,a2×0,T, and Θ2=b1,b2×0,T. Here, 0≤a1<a2≤A, 0≤b1<b2≤A, QA=0,A, and QT=0,T.We denote the density of males and females of agea at time t, respectively, by ma,t and fa,t. Moreover, μm and μf denote, respectively, the natural mortality rate of males and females. The control functions are vm and vf and depend on a and t. In addition, χΘ1 and χΘ2 are the characteristic functions of the support of the control vm and vf, respectively.We have denoted byβ the positive function describing the fertility rate that depends on a and also on
(2)M=∫0Aλama,tda,where λ is the fertility function of the male individuals. Thus, the densities of newborn male and female individuals at time t are given, respectively, by m0,t=1−γNt and f0,t=γNt where
(3)Nt=∫0Aβa,Mfa,tda.We assume that the fertility rateβ,λ and the mortality rate μf,μm satisfy the demographic properties:
(4)H1μma≥0,μfa≥0a.ea∈0,A,μm∈Lloc10,A,μf∈Lloc10,A,∫0Aμmada=+∞,∫0Aμfada=+∞,H2βa,p∈C0,A×ℝ,βa,p≥0foreverya,p∈0,A×ℝ.We further assume that the birth functionβ and the fertility function λ verify the following hypothesis:
(5)H3thereexistsb∈0,Asuchthatβa,p=0,∀a,p∈0,b×ℝ,thereexistsaconstantα+>0suchthat0≤β≤α+,∀a,p∈0,A×ℝ,βa,0=0,∀a∈0,A,H4λ∈C10,A,λa≥0foreverya∈0,A,λμm∈L10,A.The assumptionβa,0=0 for a∈0,A means that the birth rate is zero if there are no fertile male individuals.We can now state the main results. Ifa1,a2⊂b1,b2, we have the following theorem:Theorem 2.
Let us assume that the assumptionsH1−H4 hold true. If a1<b, for every time T>a1+A−a2 and for every m0,f0∈L2QA2, there exists vm,vf∈L2Θ1×L2Θ2 such that the associated solution m,f of system (1) verifies
(6)ma,T=fa,T=0a.ea∈0,A.Theorem 3.
Let us assume that the assumptionsH1−H4 hold true. We have the following:
(1)
Letvf=0. For any ϱ>0, for every time T>A−a2 and for every m0,f0∈L2QA2, there exists a control vm∈L2Θ such that the associated solution m,f of system (1) verifies(7)ma,T=0a.ea∈ϱ,A,where Θ=0,a2×0,T.
(2)
Letvm=0. For every time T>a1+A−a2 and for every m0,f0∈L2QA2, there exists a control vf∈L2Θ1 such that the associated solution m,f of system (1) verifies(8)fa,T=0a.ea∈0,A.Remark 4.
The first condition ofH3 is not necessary for Theorem 3-(1).We use the technique of [6, 7] combining final-state observability estimates with the use of characteristics to establish the observability inequalities necessary for the null controllability property of the auxiliary systems. Roughly, in our method, we first study the null controllability result for an auxiliary cascade system. Afterwards, we prove the null controllability result for system (1) by means of Kakutani’s fixed-point theorem.The remainder of this paper is as follows: in Section2, we describe the model and give the main results. Then, we study the existence and uniqueness of a positive solution for the model in Section 3. Section 4 is devoted to the proof of Theorem 2 and Theorem 3. Some illustrations of numerical simulations are given in Section 5.


## 3. Well-Posedness Result

In this section, we study the existence of positive solution of the model. For this, we assume that the so-called demographic conditionsH1, H2, H3, and H4 are verified. Moreover, here, we suppose that
(9)H5βa,p=β1aβ2pforalla,p∈0,A×ℝ,thereexistsC>0suchthatβ2p−β2q≤Cp−qforallp,q∈ℝ,β1μf∈L10,Aholds true.Thus, we have the following result.Theorem 5.
Assume thatH1−H5 hold. For every m0,f0∈L20,A2 and vm,vf∈L2Q2, system (1) admits a unique solution m,f∈L20,A×0,T2 and we have the following estimations:
(10)mL20,A×0,T≤Kf0L20,T+m0L20,T+vmL2Q+vfL2Q,fL20,A×0,T≤Cm0L20,T+vfL2Q,where K and C are positive constants.
Moreover, suppose that(11)m0,f0≥0a.e0,A,vm,vf≥0a.eQ;then, m,f is also positive.Proof of Theorem 5.
Letp be fixed in L20,T, h and h′ be fixed in L2Q, and consider the following system:
(12)mt+ma+μmm=hinQ,ft+fa+μff=h′inQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,∫0Aλapa,tfa,tdainQT,f0,t=γ∫0Aβa,∫0Aλapa,tfa,tdainQT,
For everyf0∈L20,A and h′∈L2Q, the following system
(13)ft+fa+μff=h′inQ,fa,0=f0ainQA,f0,t=γ∫0Aβa,∫0Aλapa,tfa,tdainQTadmits a unique positive solution in L2Q, see [8, 9], and one has
(14)fL2Q2≤Cf0L20,A2+h′L2Q2,where C is a positive constant and independent of p because β∈L∞0,T×0,A.
Now,f and h′ are being known; the system
(15)mt+ma+μmm=hinQ,ma,0=m0ainQA,m0,t=1−γ∫0Aβa,∫0Aλapa,tfa,tdainQTadmits a unique positive system in L2Q, and we have the following estimation:
(16)mL2Q2≤Kf0L20,A2+m0L20,A2+hL2Q2+h′L2Q2,where K is a positive constant and independent of p because β∈L∞0,T×0,A.
Let us defineΦ:L+2Q⟶L+2Q, Φp=mp where mp is the unique solution of the system (15).
For anyp,q∈L+2Q, we set
(17)B1a,t=∫0Aλapt,ada,B2a,t=∫0Aλaqt,aa.e.t∈0,A×0,T,and w=mp−mqe−γ0t where γ0 is a positive parameter that will be chosen later; w is a solution of
(18)wt+wa+γ0+μmw=0inQ,wa,0=0inQA,w0,t=1−γe−γ0t×∫0Aβ2B1−β2B2β1afp+fp−fqβ2B2β1adainQT,
Multiplying (18) by w and integrating over 0,A×0,t, and using Young’s inequality, we get
(19)12wtL20,A2+∫0t∫0Aγ0+μmw2s,adads∫0t∫0Aβ2B1−β2B2β1afpda2ds+∫0t∫0Afp−fqβ2B2β1ada2ds≤C2λ∞2∫0t∫0Aps,ad−∫0Aqs,ada2∫0Aβ1afpsda2ds+∫0t∫0Afp−fqβ2B2β1ada2d≤C2Aλ∞2∫0t∫0Aps−qs2∫0Aβ1afpsda2ds+β1∞2β2∞2A∫0t∫0Afp−fq2ds.
Hence, for everyγ0>0, there is a constant C=max2C2Aλ∞2;2β1∞2β2∞2A such that
(20)wtL20,A2≤C∫0t∫0Aps−qs2∫0Aβ1afpsda2ds+∫0t∫0Afps−fqs2ds.
Now, setF=fp−fqe−δt where δ is a positive parameter that will be chosen later. Then, F solves the following auxiliary system:
(21)Ft+Fa+δ+μfF=0inQ,Fa,0=0inQA,w0,t=γ∫0Ae−γ0tβ2H1−β2H2β1afp+Fa,tβ2H2β1adainQT.
Similarly as above, we have(22)δ∫0t∫0AFa,s2ds≤C∫0t∫0Aps−qs2∫0Aβ1afpsda2ds+∫0t∫0AFa,s2ds.
Hence, there is a positive constantC′ such that
(23)∫0t∫0AFa,s2ds≤C′∫0t∫0Aps−qs2∫0Aβ1afpsda2ds.
SettingYt=∫0Aβ1afpdaa.ein0,T, Y solves the following system:
(24)Yt=∫0Aβ1ah′a,tda+∫0Aβ1′afa,tda−∫0Aμfaβ1afa,tdain0,T,Y0=∫0Aβ1af0ada.
Multiplying (24) by Y, integrating over 0,t, and using Young's inequality, we get
(25)Y2t≤Y20+∫0tY2sds+∫0t∫0Aβ1ah′a,sda+∫0Aβ1′afa,sda+∫0Aμfaβ1afa,sda2ds≤Y20+∫0tY2sd+3∫0t∫0Aβ1ah′a,sda2ds+3∫0t∫0Aβ1′afa,sda2ds+3∫0t∫0Aβ1aμfafa,sda2ds.
So,(26)Y2t≤∫0Aβ1af0ada2+∫0TY2tdt+3∫0T∫0Aβ1ah′a,tda2dt+3∫0T∫0Aβ1′afa,tda2dt+3∫0T∫0Aβ1aμfafa,tda2dt.
Let us setf~=e−λ0tf. Then, from (13), f~ satisfies the following system:
(27)f~t+f~a+λ0+μff~=e−λ0th′inQ,f~a,0=f0ainQA,f~0,t=γ∫0Aβa,∫0Aλapa,tf~a,tdainQT.
Multiplying the first equation of (27) by f~, integrating on Q, and using the inequality of Young, we get
(28)∫0T∫0Aλ0+μfaf~2a,tdadt≤12f0L20,A2+12h′L2Q2+12f~L2Q2+12∫0Tf~20,tdt.
Using the inequality of Cauchy-Schwarz and choosingλ0=3/2+α+2, we obtain
(29)∫0T∫0Aμfaf~2a,tdadt≤12f0L20,A2+h′L2Q2.
So,(30)∫0T∫0Aμfaf2a,tdadt≤e3+2β∞2T2f0L20,A2+h′L2Q2.
Using (26), (30), and against Young’s inequality, we have
(31)Y2t≤β1∞2Af0L20,A2+β1∞2AfL2Q2+3β1∞2Ah′L2Q2+3β′1∞2AfL2Q2+3Cβ1∞β1μfL10,Af0L20,A2+3Cβ1∞β1μfL10,Ah′L2Q2.
From (14), we have just proved the existence of a positive constant C such that
(32)Y2t≤Cf0L20,A2+h′L20,A2.
The estimate (32) means also that Y∈L∞0,T.
Combining (20), (23), and (32), we get the following estimate:
(33)Φp−ΦqtL20,A2≤σ∫0tps−qsL20,A2ds,where σ is a positive constant.
Let us define the metricd on L+2Q by setting
(34)dh1,h2=∫0T∥h1−h2t∥L20,A2exp−2σtdt1/2,forh1,h2∈L+2Q.
We have(35)dΦp,Φq2=∫0T∥Φp−Φqt∥L20,A2exp−2σtdt≤σ∫0Texp−2σt∫0t∥p−qs∥L20,A2dsdt.
Using the Fubini theorem, we conclude that(36)dΦp,Φq2=∫0T∥Φp−Φqt∥L20,A2exp−2σtdt≤∫0T∥p−qs∥L20,A2×∫sTσe−2σtdtds≤12dp,q2.
Then,Φ is a contraction on the complete metric space L+2Q into itself. Using Banach’s fixed-point theorem, we conclude the existence of a unique fixed-point m. Moreover, m is nonnegative. Hence, the unique couple m,f is the unique solution to our problem (1).


## 4. Null Controllability Results

For the sequel, the hypothesisH5 is not necessary. As a consequence, the uniqueness and the positivity of the solution of system (1) are not guaranteed.We first establish an observability inequality to show the controllability of a linear system. Then, by a fixed-point method, we show the controllability of the model.


### 4.1. Null Controllability of an Auxiliary Coupled System

This section is devoted to the study of an auxiliary system obtained from system (1).Letp be a L2QT function, we define the auxiliary system given by
(37)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=χΘ2vfinQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tdainQT,f0,t=γ∫0Aβa,pfa,tdainQT.Letp be fixed in L2QT; for m0,f0∈L2QA2 and vm,vf∈L2Θ1×L2Θ2, system (37) admits a unique solution m,f∈L2Q2, see Section 3.The system above is null approximately controllable. Indeed, we have the following result:Theorem 6.
Let us assume that assumptionsH1−H2 hold. For every time T>a1+A−a2, for every κ,θ>0, and for every m0,f0∈L2QA2, there exists a control vκ,vθ such that the solutions m and f of system (37) verify
(38)∣m.,TL20,A≤κ,f.,TL20,A≤θ.The adjoint system of (37) is given by
(39)−nt−na+μmn=0inQ,−lt−la+μfl=1−γβa,pn0,t+γβa,pl0,tinQ,na,T=nTa,la,T=lTainQA,nA,t=0,lA,t=0inQT.The main idea in this part is to establish an observability inequality of (39) that will allow us to prove the approximate null controllability of (37).The basic idea for establishing this inequality is the estimation of nonlocal terms. For that, let us start by formulating a representation of the solution of the adjoint system by the method of characteristic and semigroup.For everynT,lT∈L2QA2, under the assumptions H1 and H2, coupled system (39) admits a unique solution n,l. Moreover, integrating along the characteristic lines, the solution n,l of (39) is as follows:
(40)nt=π1a+T−tπ1anTa+T−tifT−t≤A−a,0ifA−a<T−t,(41)lt=π2a+T−tπ2alTa+T−t+∫tTπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifT−t≤A−a,∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifA−a<T−t,where π1a=e−∫0aμmrdr and π2a=e−∫0aμfrdr.Suppose that the assumptionsH1, H2, H3, and H4 are fulfilled; then, we have the following result.Theorem 7.
Under the assumptions of Theorem2, if a1,a2⊂b1,b2, there exists a constant CT>0 independent of p such that the couple n,l solution of (39) verifies the following inequality:
(42)∫0An2a,0da+∫0Al2a,0da≤CT∫Θ1n2a,tdadt+∫Θ2l2a,tdadt.For the proof of Theorem7, we state the following estimations of the nonlocal terms.Proposition 8.
Under the assumptions of Theorem2, there exists C>0 such that
(43)∫0T−ηn20,tdt≤C∫0T∫a1a2n2a,tdadt,where a1<η<T.
In particular, for everyϱ>0, if a1=0 and nTa=0 a.e a∈0,ϱ, there is Cϱ,T>0 such that
(44)∫0Tn20,tdt≤Cϱ∫0T∫0a2n2a,tdadt.
Moreover, if the first condition ofH3 hold, we have the inequality
(45)∫0T−ηl20,tdt≤C∫Θ2l2a,tdadt,
for everyη such that b1<b and b1<η<T.Proof of Proposition 8.
The staten of (39) verifies
(46)−∂n∂t−∂n∂a+μmn=0in0,a2×0,T,na,T=nTin0,a2.
We denote byn~a,t=na,te−∫0aμmsds. Then, n~ satisfies
(47)∂n~∂t+∂n~∂a=0ina1,a2×0,T.
Proving the inequality(48)∫0T−ηn~20,tdt≤C∫0T∫a1a2n~2a,tdadtleads to getting inequality (43).
Indeed, we have(49)∫0T−ηn20,tdt=∫0T−ηn~20,tdt≤C∫0T∫a1a2n~2a,tdadt=C∫0T∫a1a2e−2∫0aμmsdsn2a,tdadt≤C∫0T∫a1a2n2a,tdadt.
We consider the following characteristic trajectoryγλ=T−λ,T+t−λ. If λ=T, the backward characteristic starts from 0,t. If T<a1, the trajectory γλ never reaches the observation region a1,a2. So, we choose T>a1.
Without loss of generality, let us assume thatη<a2<T.Step 1.
Estimation ofn0,t,t∈0,T−η(i)
Estimation fort∈0,T−a2
We denote by(50)wλ=n~T−λ,T+t−λ,λ∈T−a2,T.
Then,∂w/∂λ=0 for all λ∈T−a2,T. In particular, wλ is constant for all λ∈T−a2,T. Since η<a2, we have
(51)wt=1a2−η∫t−a2t−ηwλdλ.
Therefore,(52)wT=n~0,t=1a2−η∫T−a2T−ηn~T−λ,T+t−λdλ=1a2−η∫ηa2n~s,t+sds.
Using the fact that(53)∫t−a2t−ηwλdλ2≤a2−η∫t−a2t−ηw2λdλand integrating with respect to t over 0,T−a2, we get
(54)∫0T−a2n~20,tdt≤C∫a1a2∫0T−a2n~2s,t+sdtds=C∫a1a2∫sT−a2+sn~2s,ududs.
Finally,(55)∫0T−a2n~20,tdt≤C∫0T∫a1a2n~2a,tdadt.(ii)
Estimation fort∈T−a2,T−η
We define(56)wλ=n~T−λ,T+t−λ,λ∈T−a1,T.
Then,∂w/∂λ=0 for all λ∈T−a2,T. In particular, wλ is constant for all λ∈T−a2,T. Thus, we have
(57)wt=1η−a1∫t−ηt−a1wλdλ=1η−a1∫a1ηn~s,t+sds.
Integrating with respect tot over T−a2,T−η, we get
(58)∫T−a2T−ηn~20,tdt≤Cη,a1∫a1η∫T−a2T−ηn~s,t+sdtds.
Finally,(59)∫T−a2T−ηn~20,tdt≤Cη,a1∫0T∫a1a2n~a,tdadt.
Combining (55) and (59), we obtain
(60)∫0T−ηn~20,tdt≤Cη,a1,a2∫0T∫a1a2n~a,tdadt,leading to (43).
Remark thatlimη⟶a1+1/η−a1=+∞ then, limη⟶a1+Cη,a1,a2=+∞.
Suppose now thata1=0. From the above, we have for all ϱ>0, the existence of a constant depending on ϱ such that
(61)∫0T−ϱn20,tdt≤Cϱ,a2∫0T∫a1a2n2a,tdadt.
Moreover, ifnTa=0 in 0,ϱ, we have wt=0 in T−ϱ,T. Then, w≡0. So n0,t=0 in T−ϱ,T.
Finally, ifnTa=0 in 0,ϱ, we have the following inequality:
(62)∫0Tn20,tdt≤Cϱ,a2∫0T∫0a2n2a,tdadt.Step 2.
Estimation ofl0,t,t∈0,T−η
Consideringν=minb,b2, the state l of the adjoint system can be rewritten as
(63)−∂l∂t−∂l∂a+μfl=0in0,ν×0,T,la,T=lTin0,ν.
We denote byl~a,t=la,te−∫0aμfsds. Then, l~ satisfies
∂l~/∂t+∂l~/∂a=0inb1,ν×0,T.(174)
(i)
Estimation fort∈0,T−ν
Definingw as
(64)wλ=l~T−λ,T+t−λ,λ∈T−ν,T−b1,
then,wλ is constant for all λ∈T−ν,T−b1 and we have wt=1/ν−η∫t−νt−ηwλdλ. Likewise step 1, we obtain
(65)∫0T−νl~20,tdt≤Cη,ν∫Θ2l~2a,tdadt.(ii)
Estimation fort∈T−ν,T−η
We denote by(66)wλ=l~T−λ,T+t−λ,λ∈T−η,T−b1.
As above,wλ is constant for all λ∈T−η,T−b1 and we set wt=1/η−b1∫t−ηt−b1wλdλ. Then,
(67)∫T−νT−ηl~20,tdt≤Cη,b1∫Θ2l~2a,tdadt.Combining (65) and (67), the inequality (45) follows.Proposition 9.
Let us assume the assumptionsH1−H3.For every T>supa1,A−a2, there exists CT>0 such that the solution n,l of system (37) verifies the following inequality:
(68)∫0An2a,0da≤CT∫Θ1n2a,tdadt.Note that for everyT>supa1,A−a2, there exists a0∈a1,a2 such that na,0=0 for all a∈a0,A. This is a consequence of the following lemma.Lemma 10.
Let us suppose thatT>supa1,A−a2. Then, there exists a0∈a1,a2 such that T>A−a0>A−a for all a∈a0,A. Therefore, na,0=0 for all a∈a0,A.Proof of Lemma 10.
Suppose thatT>A−a2; then, there exists κ>0 (we choose κ such that κ<a2−a1) T>A−a2+κ. So T>A−a2−κ, and we denote a0=a2−κ. Then, T>A−a0>A−a for all a∈a0,A. Finally, from (40) for all a,t such that T−t>A−a, we get na,0=0 for all a∈a0,A.Proof of Proposition 9.
From the Lemma10, we have to prove the following inequality:
(69)∫0a0n2a,0da≤CT∫0T∫a1a2n2a,tdadt.
We setn~a,t=e−∫0aμmsdsna,t. Then, from the first equation of (39), n~ satisfies
(70)∂n~∂t+∂n~∂a=0in0,A×0,T.
Inequality (47) is a consequence of the following estimation:
(71)∫0a0n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.
Indeed, we have(72)∫0a0n2a,0da=∫0a0e2∫0aμmsdsn~2a,0da≤e2μmL10,a0∫0a0n~2a,0da≤Ce4μmL10,a0∫0T∫a1a2n2a,tdadt.
We consider in this proof the characteristicsγλ=a+λ,λ. For λ=0, the characteristics start from a,0. We have two cases.Case 1.
T<a2.
Two situations can arise:(i)
b0=a2−T<a1<a0, in this situation, we split the interval 0,a0 as(73)0,a0=0,b0∪b0,a1∪a1,a0(ii)
a1<b0<a0, in this situation, we split the interval 0,a0 as(74)0,a0=0,a1∪a1,a0Case 2.
T≥a2.
In this case, we split the interval0,a0 as
(75)0,a0=0,a1∪a1,a0.We make the proof for the second case. For the proof in the first case, see [6].Upper bound on0,a1:Fora∈0,a1, we set wλ=n~T+a−λ,T−λ,λ∈a1,T. We prove easily that w is a constant, see the proof of Proposition 8. Then, we set
(76)wt=1a2−a1∫t−a2t−a1wλdλ.Integrating with respect toa over 0,a1, we get
(77)∫0a1n~2a,0da≤C∫0a1∫a1a2n~2a+α,αdαda.Finally, we obtain(78)∫0a1n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.Upper bound ona1,a0:Fora∈a1,a0, we set wλ=n~T+a−λ,T−λ,λ∈a0,T and
(79)wt=1a2−a0∫t−a2t−a0wλdλ.Making as above and integrating with respect toa over a1,a0, it follows that
(80)∫a1a0n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.The inequalities (78) and (80) give the desired result.We also need the following estimate for the proof of Theorem7.Proposition 11.
Let us assume the assumptionsH1−H2, let b1<a0<b and T>b1.
Then, there existsCT>0 such that the solution l of (39) verifies
(81)∫0a0l2a,0da≤CT∫Θ2l2a,tdadt.Proof of Proposition 11.
We suppose thatβ=0 in 0,b, the function l verifies
(82)−∂l∂t−∂l∂a+μfl=0in0,b×0,T,la,T=lTin0,b.
Proceeding as in the proof of Proposition9, we get the desired result.For the proof of Theorem7, we start with the following lemma.Lemma 12.
Suppose thatT>A−a2+a1 and a1<b.
Then, there existsa0∈a1,b and κ>0 such that
(83)T>T−a1+κ>A−a0>A−aforalla∈a0,A.
Therefore,(84)la,0=∫0A−aπ2a+sπ2aβa+s,psl0,s+n0,sdsforalla∈a0,A.Proof of Lemma 12.
Notice that the solutionl of (39) is given by
(85)lt=π2a+T−tπ2alTa+T−t+∫tTπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifT−t≤A−a,∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifA−a<T−t.
Without loss of generality, we suppose thata2=b.
Suppose thatT>a1+A−a2⟺T−a1>A−a2. Then, there exists κ>0 (we choose κ such that 2κ<a2−a1) such that T−a1+κ>A−a2−κ. We denote by a0=a2−κ, and as A−a0>A−a for all a∈a0,A, then T>T−a1+κ>A−a0>A−a for all a∈a0,A.
Moreover, fora,t such that T−t>A−a, we have
(86)la,t=∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sds.
Fort=0 and a∈a0,A,T−0>A−a0>A−a, one has
(87)la,0=∫0A−aπ2a+sπ2aβa+s,ps1−γn0,s+γl0,sds.
Remark that asa1,a2⊂b1,b2, if a0∈a1,a2, then a0∈b1,b2.Now, we can prove Theorem7.Proof of Theorem 7.
Leta0 as in Lemma 12, we have
(88)∫0Al2a,0da=∫0a0l2a,0da+∫a0Al2a,0da.
Using the results of Lemma12, the assumption of β, and the regularity of π2a+s/π2a, we can prove the existence of a constant KT independent of p (β∈L∞0,T×0,A such that
(89)∫a0Al2a,0da≤KT∫0T−a1+κn20,tdt+∫0T−a1+κl20,tdt.
Moreover, we haveb1≤a1≤a1+κ. Using Proposition 8, it follows that
(90)∫0T−a1+κn20,tdt≤CT∫Θ1n2a,tdadt,∫0T−a1+κl20,t≤CT′∫Θ2l2a,tdadt.
Finally, adding the above to the results of Proposition9 and Proposition 11, we get
(91)∫0An2a,0da+∫0Al2a,0da≤CT∫Θ1n2a,tdadt+∫Θ2l2a,tdadt.Forε>0 and θ>0, we consider the functional Jε,θ defined by
(92)Jε,θvm,vf=12∫Θ1vm2dadt+12∫Θ2vf2dadt+12ε∫0Am2a,Tda+12ε∫0Af2a,Tda,where m,f is the solution of the following system:
(93)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=χΘ2vfinQ,ma,0=m0a,fa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tda,f0,t=γ∫0Aβa,pfa,tdainQT.Lemma 13.
The functionalJε,θ is continuous, strictly convex, and coercive. Consequently, Jε,θ reaches its minimum at a point vm,ε,vf,θ∈L2Θ1×L2Θ2. Setting mε,fθ the associated solution of (93) and nε,lθ the solution of (39) with
(94)nεa,T=−1εmεa,T,lθa,T=−1θfθa,T,we have
(95)χΘ1vm,ε=χΘ1nε,χΘ2vf,θ=χΘ2lθ.Moreover, there existCi>0,1≤i≤4, independent of ε and θ such that
(96)∫Θ1nε2a,tdadt≤C1∫0Am02ada+∫0Af02ada,∫0Amε2a,Tda≤εC2∫0Am02ada+∫0Af02ada,∫Θ2lθ2a,tdadt≤C3∫0Am02ada+∫0Af02ada,∫0Afθ2a,Tda≤θC4∫0Am02ada+∫0Af02ada.Proof of Lemma 13.
It is easy to check thatJε,θ is coercive, continuous, and strictly convex. Then, it admits a unique minimizer vε,vθ. The maximum principle gives
(97)χΘ1vm,ε=χΘ1nε,χΘ2vf,θ=χΘ2lθwhere the couple nε,lθ is the solution of the system
(98)−∂tnε−∂anε+μmnε=0inQ,−∂tlθ−∂alθ+μflθ=1−γβa,pnε0,t+γβa,plθ0,tinQ,nεa,T=−1εmεa,T,lθa,T=−1θfθa,TinQA,nεA,t=0,lθA,t=0inQT.
Multiplying the first and the second equation of (98) by, respectively, mε and fθ, integrating with respect to Q and using (97), we get
(99)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda=−∫0Am0anεa,0da−1−γ∫0T∫0Aβa,pfθa,tnε0,tdadt(100)∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda=−∫0Af0alθa,0da+1−γ∫0T∫0Aβa,pfθa,tnε0,tdadt.
Combining (99) and (100), we obtain
(101)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda=−∫0Am0anεa,0da−∫0Af0alθa,0da.
Using the inequality of Young, we have, for anyδ>0,
(102)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda≤δ2∫0Am02ada+12δ∫0Anε2a,0da+δ2∫0Af02ada+12δ∫0Alθ2a,0da.
Using observability inequality (42) and choosing δ=CT in the previous inequality, it follows that
(103)12∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+12∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda≤CT2∫0Am02ada+∫0Af02ada.
This gives the desired result necessary to the proof of the main one.Now, we consider the system(104)∂tmεp+∂amεp+μmmεp=χΘ1nεinQ,∂tfθp+∂afθp+μmfθp=χΘ2lθinQ,mεpa,0=m0a,fθpa,0=f0ainQA,mεp0,t=1−γ∫0Aβa,pfθpa,tda,fθp0,t=γ∫0Aβa,pfθpa,tdainQT,where nε,lθ is the solution of (98) that minimizes the functional Jε,θ. We have the following result:Lemma 14.
Under the assumptions of Theorem2, the solutions mε and fθ verify the following inequalities:
(105)∫0Amε2a,Tda+∫0T∫0A1+μmmε2a,tdadt≤C∫0Am02ada+∫0Af02ada,(106)∫0Afθ2a,Tda+∫0T∫0A1+μffθ2a,tdadt≤C∫0Am02ada+∫0Af02ada.Proof of Lemma 14.
We denote by(107)yε,zθ=e−λ0tmε,e−λ0tfθ.
The functionsyε and zθ verify
(108)∂tyε+∂ayε+λ0+μmyε=χΘ1e−λ0tnε,(109)∂tzθ+∂azθ+λ0+μfzθ=χΘ2e−λ0tlθ.
Multiplying equality (108) and equality (109) by, respectively, yε and zθ and integrating with respect to Q, we get
(110)12∫0Ayε2a,Tda+12∫0Tyε2A,tdt+∫0T∫0Aλ0+μmayε2a,tdadt=12∫0Ay02ada+1−γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ1e−λ0tnεyεdadt,(111)12∫0Azθ2a,Tda+12∫0Tzθ2A,tdt+∫0T∫0Aλ0+μfazθ2a,tdadt=12∫0Af02ada+γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ2e−λ0tlθzθdadt.
Using the Young inequality, Cauchy-Schwarz inequality, and the fact thatβ is L∞, we prove that
(112)1−γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ1e−λ0tnεyεdadt≤α+2zθL2Q2+12yεL2Q2+12nεL2Θ12,(113)γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ2e−λ0tlθzθdadt≤α+2zθL2Q2+12zθL2Q2+12lθL2Θ22.
Therefore, choosingλ0>α+2+3/2, we get
(114)12∫0Azθ2a,Tda+∫0T∫0A1+μfazθ2a,tdadt≤12f0QA2+lθL2Θ22.
Finally, applying the result of Lemma13 to the above inequality, it follows that
(115)12∫0Azθ2a,Tda+∫0T∫0A1+μfazθ2a,tdadt≤C∫0Af02ada+∫0Am02adaand then the inequality (106) holds.
Likewise, we have(116)12∫0Ayε2a,Tda+∫0T∫0A1+μmyε2a,tdadt≤12m0QA2+α+2zθL2Q2+12nεL2Θ12.
Using the above inequality, Lemma13, and inequality (115), we obtain
(117)∫0Ayε2a,Tda+∫0T∫0A1+μmyε2a,tdadt≤C∫0Af02ada+∫0Am02adaand then we get the desired result.Finally, from Lemma13 and Lemma 14, if ε,θ⟶0,0, we get
(118)χΘ1nε,χΘ2lθ⇀χΘ1vm,χΘ2vf,mε,fθ⇀m,f,with m,f solution of problem (37) and
(119)m.,T=f.,T=0a.ea∈0,A.We have now the necessary ingredients for the proof of Theorem2.


### 4.2. Proof of Theorem2

In this section, we established the existence of a fixed point for the preceding auxiliary problem. Indeed, we consider thatH3 holds and we suppose to simplify that λ0=λA=0. For each p∈L2QT, let us denote by Λp⊂L20,T the set of all ∫0Aλampda, where the couple mp,fp is the solution of the following system:
(120)∂tmp+∂amp+μmmp=χΘ1ninQ,∂tfp+∂afp+μmfp=χΘ2linQ,mpa,0=m0a,fpa,0=f0ainQA,mp0,t=1−γ∫0Aβa,pfpa,tda,fp0,t=γ∫0Aβa,pfpa,tdainQT,and np,lp the corresponding solution of the minimizer of Jε,θ with mpa,T=fpa,T=0 for almost every a∈0,A.We have the following result.Proposition 15.
Under the assumptions of Theorem2, for any p∈L2QT, the solution of problem (120) satisfies
(121)∣Yt∣+ddtYL20,T≤Cm0L2QA+f0L2QA,where Yt=∫0Aλampda and the constant C is independent of p, m0, and f0.Proof of Proposition 15.
LetYt=∫0Aλampda. It is easy to prove that Y is the solution of system
(122)∂tY+∫0Aμmaλampda=RtinQT,Y0=∫0Aλam0ada,where
(123)Rt=∫0Aλ′ampda+1−γλ0∫0Aβa,pfpda+∫0a2λanpda.
Using Lemma14 and the assumptions on β and λ, we infer that there exists K>0 such that
(124)RL2QT≤Km0L2QA+f0L2QA.
By using (122), the Young inequality and integrating on QT, we obtain
(125)∫0T∂tY2dt≤2∫0TRt2dt+2∫0T∫0Aμmaλamεpda2dt.
Moreover, the Cauchy-Schwarz inequality leads to(126)∫0T∫0Aμmaλamεpda2dt≤∫0Aμmaλada∫0T∫0Aμmaλamε2pdadt.
The inequality (105) and the fact that λ∈C0,A give
(127)∫0T∫0Aμmaλamε2pdadt≤K1m0L2QA2+f0L2QA2,where K1>0 is independent of p, ε, and θ. Moreover, as λμm∈L10,A, and using (124), it follows that
(128)ddtYL20,T≤Cm0L2QA+f0L2QA.
Now, letY~=e−λ0tY. Then, Y~ satisfies
(129)∂tY~+λ0Y~+e−λ0t∫0Aμmaλamεpda=e−λ0tRtinQT,Y~0=∫0Aλam0ada.
Multiplying the first equation of (129) by Y~, integrating on 0,t, and using successively Cauchy-Schwarz and Young inequalities, we deduce that
(130)Y~t2+λ0∫0tY~2dt≤Y~02+∫0tY~2dt+∫0T∫0Aμmaλamεpda2dt+RL2QT.
Using the above calculations and choosingλ0>2, we get
(131)Y~t2≤K2m0L2QA+f0L2QA.
The desired result comes from (128) and (131).It is obvious thatΛp is convex, and let
(132)W0,T=Y∈L∞0,T,YL∞0,T≤N;dYdtL20,T≤N,with N=Cm0L2QA+f0L2QA.We haveW0,T⊂W1,10,T. Moreover, the injection of W1,10,T into L20,T is compact, see [10], page 129. So W0,T is relatively compact in L20,T. From Proposition 15, we have ΛW0,T⊂W0,T, and we see that ΛW0,T is a relatively compact subset of L20,T. Let us now prove that Λ is uppersemicontinuous. This is equivalent to prove that for any closed subset K of L20,T, Λ−1K is closed in L20,T. Let pk∈Λ−1K such that pk converges towards p in L20,T. Then, pk is bounded, and for all k, there exists Pk∈K such that Pk∈Λpk. Therefore, from the definition of Λ, there exists mk,fk∈L20,T×0,A2 associated with nk,lk∈L2Θ1×L2Θ2 solution of (120) such that Pk=∫0Aλamkpkda and satisfying the inequalities of Lemma 13 and Lemma 14. Consequently, mk,fk and nk,lk are bounded, respectively, in L20,T×0,A2 and in L2Θ1×L2Θ2. Thus, there exists subsequences still denoted by mk,fk and nk,lk that converge weakly to m,f in L20,T×0,A2 and n,l in L2Θ1×L2Θ2, respectively. Using hypothesis H3, it follows that ∫0Aλamkpkda converges strongly to ∫0Aλampda in L20,T.Now, by a standard device, we see thatm,f associated with n,l are a solution of (120) and satisfy the inequalities of Lemma 13 and Lemma 14. This implies that P∈Λp.On the other hand, thanks to Proposition15, one can extract a subsequence also denoted by Pk that converges strongly towards the function P in L20,T. Since K is closed, we deduce that P∈K. Finally, we deduce that p∈Λ−1P.Applying the Kakutani fixed-point theorem [11] in the space L20,T to the mapping Λ, we infer that there is at least one Y∈W0,T such that Y∈ΛY. This completes the null controllability proof of the model (1).


### 4.3. Proof of Theorem3

#### 4.3.1. Proof of Theorem3-(1)

In this section, we always consider the following system:(133)mt+ma+μmm=χΘvminQ,ft+fa+μff=0inQ,ma,0=m0,fa,0=f0inQA,m0,t=1−γ∫0Aβa,pfda,f0,t=γ∫0Aβa,pfdainQT,for every p in L2QT. Under the assumptions of Theorem 3, the controllability problem that is to find vm∈L2Θ such that m,f solution of the system (133) verifies
(134)m.,T=0a∈ρ,Ais equivalent to the following observability inequality.Proposition 16.
Let us assume the assumptionsH1−H2−H3, for every T>A−a2 and for any ρ>0, if hTa=0a.e in 0,ϱ, there exists Cρ,T>0 such that the following inequality:
(135)∫0Ah2a,0da+∫0Ag2a,0da≤Cρ,T∫Θh2a,tdadtholds, where h,g is the solution of
(136)−ht−ha+μmh=0inQ,−gt−ga+μfg=1−γβa,pg0,t+γβa,ph0,tinQ,ha,T=hT,ga,T=0inQA,hA,t=0,gA,t=0inQT.For the proof of Proposition16, we state the following estimate.Proposition 17.
Under the assumptionsH1 and H2, there exists a constant C>0 such that the solution h,g of system (136) verifies
(137)∫0Ag2a,0da+∫0T∫0A1+μfg2a,tdadt≤C∫0Th20,tdt.
Moreover, we deduce forhT=0 a.e in 0,ϱ that there exists a constant Cϱ,T>0 such that
(138)∫0Ag2a,0da≤Cϱ,T∫0Th20,tdt≤Cϱ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 16.
Settingy=eλ0tg, the function y verifies
(139)−∂ty−∂ay+λ0+μfy=γβa,peλ0th0,t+1−γβa,py0,t.
Multiplying equality (139) by y and integrating on Q, we obtain
(140)12∫0Ay2a,0da+12∫0Ty20,tdt+∫0T∫0Aλ0+μfy2a,tdadt=∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt.
Using Young inequality and the condition onβ, we get
(141)∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt≤e2λ0Tα+22δh0,.L2QT2+δ2yL2Q2+α+22δy0,.L2QT2+δ2yL2Q2.
Choosingδ=α+2, we obtain
(142)12∫0Ay2a,0da+∫0T∫0Aλ0+μfy2a,tdadt≤e2λ0T2h0,.L2QT2+α+2yL2Q2.
Finally, choosingλ0>α+2+1, it follows that
(143)12∫0Ay2a,0da+∫0T∫0A1+μfy2a,tdadt≤e2λ0T2∫0Ah20,tdt.
So,(144)∫0Ay2a,0da≤e2λ0T∫0Ah20,tdt.
Finally, we get(145)∫0Ag2a,0da≤e2λ0T∫0Ah20,tdt.
Combining the above inequality and inequality (44) of Proposition 8, for hT=0 a.e in 0,ρ, we get
(146)∫0Ag2a,0da≤Cρ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 17.
We use the results of Proposition9 and Proposition 17. Indeed, by combining (68) and (146), the desired result is obtained.Now, letε>0 and ϱ>0. We consider the functional Jε defined by
(147)Jεvm=12∫0T∫a1a2vm2a,tdadt+12ε∫ρAm2a,Tda,where m,f is the solution of the following system:
(148)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=0inQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tdainQT,f0,t=γ∫0Aβa,pfa,tdainQT.We have the following lemma.Lemma 18.
The functionalJε is continuous, strictly convex, and coercive. Consequently, Jε reaches its minimum at one and has vm,ε=χΘhε and there exists positive constants C1,C2 independent of ε such that
(149)∫0T∫0a2hε2a,tdadt≤C1∫0Am02ada+∫0Af02ada,(150)∫ρAmε2a,Tda≤εC2∫0Am02ada+∫0Af02ada.Proof of Lemma 18.
The proof is similar to that of Lemma13.By makingε tending towards zero, we thus obtain that χΘhε⇀χΘvm and mε,fε⇀m,f, where m,f is the solution of system (148) that verifies
(151)m.,T=0a.einϱ,A.Finally, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear problem.


#### 4.3.2. Proof of Theorem3-(2)

Letp∈L2QT, under the assumptions of Theorem 3, the following controllability problem finds vf∈L2Θ such that the solution of the system
(152)ft+fa+μff=χΘ2vfinQ,fa,0=f0ainQA,f0,t=γ∫0Aβa,pfa,tdainQTverifies
(153)f.,T=0a.ein0,Ais equivalent to the following observability inequality.Proposition 19.
Let us assume true the assumptionsH1−H2−H3. For any T>a1+A−a2, there exists CT>0 such that
(154)∫0Ag2a,0da≤CT∫Θ2g2a,tdadt,where g is a solution of the system
(155)−gt−ga+μfg=γβa,pg0,tinQ,ga,T=gTinQA,gA,t=0inQT.Proof of Proposition 19.
Using inequality (45) of Proposition 8, the result of Proposition 11, and the representation of the solution of the system (155), we get the desired result.To conclude, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear system. We omit all details because the extension is straightforward.


## 4.1. Null Controllability of an Auxiliary Coupled System

This section is devoted to the study of an auxiliary system obtained from system (1).Letp be a L2QT function, we define the auxiliary system given by
(37)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=χΘ2vfinQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tdainQT,f0,t=γ∫0Aβa,pfa,tdainQT.Letp be fixed in L2QT; for m0,f0∈L2QA2 and vm,vf∈L2Θ1×L2Θ2, system (37) admits a unique solution m,f∈L2Q2, see Section 3.The system above is null approximately controllable. Indeed, we have the following result:Theorem 6.
Let us assume that assumptionsH1−H2 hold. For every time T>a1+A−a2, for every κ,θ>0, and for every m0,f0∈L2QA2, there exists a control vκ,vθ such that the solutions m and f of system (37) verify
(38)∣m.,TL20,A≤κ,f.,TL20,A≤θ.The adjoint system of (37) is given by
(39)−nt−na+μmn=0inQ,−lt−la+μfl=1−γβa,pn0,t+γβa,pl0,tinQ,na,T=nTa,la,T=lTainQA,nA,t=0,lA,t=0inQT.The main idea in this part is to establish an observability inequality of (39) that will allow us to prove the approximate null controllability of (37).The basic idea for establishing this inequality is the estimation of nonlocal terms. For that, let us start by formulating a representation of the solution of the adjoint system by the method of characteristic and semigroup.For everynT,lT∈L2QA2, under the assumptions H1 and H2, coupled system (39) admits a unique solution n,l. Moreover, integrating along the characteristic lines, the solution n,l of (39) is as follows:
(40)nt=π1a+T−tπ1anTa+T−tifT−t≤A−a,0ifA−a<T−t,(41)lt=π2a+T−tπ2alTa+T−t+∫tTπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifT−t≤A−a,∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifA−a<T−t,where π1a=e−∫0aμmrdr and π2a=e−∫0aμfrdr.Suppose that the assumptionsH1, H2, H3, and H4 are fulfilled; then, we have the following result.Theorem 7.
Under the assumptions of Theorem2, if a1,a2⊂b1,b2, there exists a constant CT>0 independent of p such that the couple n,l solution of (39) verifies the following inequality:
(42)∫0An2a,0da+∫0Al2a,0da≤CT∫Θ1n2a,tdadt+∫Θ2l2a,tdadt.For the proof of Theorem7, we state the following estimations of the nonlocal terms.Proposition 8.
Under the assumptions of Theorem2, there exists C>0 such that
(43)∫0T−ηn20,tdt≤C∫0T∫a1a2n2a,tdadt,where a1<η<T.
In particular, for everyϱ>0, if a1=0 and nTa=0 a.e a∈0,ϱ, there is Cϱ,T>0 such that
(44)∫0Tn20,tdt≤Cϱ∫0T∫0a2n2a,tdadt.
Moreover, if the first condition ofH3 hold, we have the inequality
(45)∫0T−ηl20,tdt≤C∫Θ2l2a,tdadt,
for everyη such that b1<b and b1<η<T.Proof of Proposition 8.
The staten of (39) verifies
(46)−∂n∂t−∂n∂a+μmn=0in0,a2×0,T,na,T=nTin0,a2.
We denote byn~a,t=na,te−∫0aμmsds. Then, n~ satisfies
(47)∂n~∂t+∂n~∂a=0ina1,a2×0,T.
Proving the inequality(48)∫0T−ηn~20,tdt≤C∫0T∫a1a2n~2a,tdadtleads to getting inequality (43).
Indeed, we have(49)∫0T−ηn20,tdt=∫0T−ηn~20,tdt≤C∫0T∫a1a2n~2a,tdadt=C∫0T∫a1a2e−2∫0aμmsdsn2a,tdadt≤C∫0T∫a1a2n2a,tdadt.
We consider the following characteristic trajectoryγλ=T−λ,T+t−λ. If λ=T, the backward characteristic starts from 0,t. If T<a1, the trajectory γλ never reaches the observation region a1,a2. So, we choose T>a1.
Without loss of generality, let us assume thatη<a2<T.Step 1.
Estimation ofn0,t,t∈0,T−η(i)
Estimation fort∈0,T−a2
We denote by(50)wλ=n~T−λ,T+t−λ,λ∈T−a2,T.
Then,∂w/∂λ=0 for all λ∈T−a2,T. In particular, wλ is constant for all λ∈T−a2,T. Since η<a2, we have
(51)wt=1a2−η∫t−a2t−ηwλdλ.
Therefore,(52)wT=n~0,t=1a2−η∫T−a2T−ηn~T−λ,T+t−λdλ=1a2−η∫ηa2n~s,t+sds.
Using the fact that(53)∫t−a2t−ηwλdλ2≤a2−η∫t−a2t−ηw2λdλand integrating with respect to t over 0,T−a2, we get
(54)∫0T−a2n~20,tdt≤C∫a1a2∫0T−a2n~2s,t+sdtds=C∫a1a2∫sT−a2+sn~2s,ududs.
Finally,(55)∫0T−a2n~20,tdt≤C∫0T∫a1a2n~2a,tdadt.(ii)
Estimation fort∈T−a2,T−η
We define(56)wλ=n~T−λ,T+t−λ,λ∈T−a1,T.
Then,∂w/∂λ=0 for all λ∈T−a2,T. In particular, wλ is constant for all λ∈T−a2,T. Thus, we have
(57)wt=1η−a1∫t−ηt−a1wλdλ=1η−a1∫a1ηn~s,t+sds.
Integrating with respect tot over T−a2,T−η, we get
(58)∫T−a2T−ηn~20,tdt≤Cη,a1∫a1η∫T−a2T−ηn~s,t+sdtds.
Finally,(59)∫T−a2T−ηn~20,tdt≤Cη,a1∫0T∫a1a2n~a,tdadt.
Combining (55) and (59), we obtain
(60)∫0T−ηn~20,tdt≤Cη,a1,a2∫0T∫a1a2n~a,tdadt,leading to (43).
Remark thatlimη⟶a1+1/η−a1=+∞ then, limη⟶a1+Cη,a1,a2=+∞.
Suppose now thata1=0. From the above, we have for all ϱ>0, the existence of a constant depending on ϱ such that
(61)∫0T−ϱn20,tdt≤Cϱ,a2∫0T∫a1a2n2a,tdadt.
Moreover, ifnTa=0 in 0,ϱ, we have wt=0 in T−ϱ,T. Then, w≡0. So n0,t=0 in T−ϱ,T.
Finally, ifnTa=0 in 0,ϱ, we have the following inequality:
(62)∫0Tn20,tdt≤Cϱ,a2∫0T∫0a2n2a,tdadt.Step 2.
Estimation ofl0,t,t∈0,T−η
Consideringν=minb,b2, the state l of the adjoint system can be rewritten as
(63)−∂l∂t−∂l∂a+μfl=0in0,ν×0,T,la,T=lTin0,ν.
We denote byl~a,t=la,te−∫0aμfsds. Then, l~ satisfies
∂l~/∂t+∂l~/∂a=0inb1,ν×0,T.(174)
(i)
Estimation fort∈0,T−ν
Definingw as
(64)wλ=l~T−λ,T+t−λ,λ∈T−ν,T−b1,
then,wλ is constant for all λ∈T−ν,T−b1 and we have wt=1/ν−η∫t−νt−ηwλdλ. Likewise step 1, we obtain
(65)∫0T−νl~20,tdt≤Cη,ν∫Θ2l~2a,tdadt.(ii)
Estimation fort∈T−ν,T−η
We denote by(66)wλ=l~T−λ,T+t−λ,λ∈T−η,T−b1.
As above,wλ is constant for all λ∈T−η,T−b1 and we set wt=1/η−b1∫t−ηt−b1wλdλ. Then,
(67)∫T−νT−ηl~20,tdt≤Cη,b1∫Θ2l~2a,tdadt.Combining (65) and (67), the inequality (45) follows.Proposition 9.
Let us assume the assumptionsH1−H3.For every T>supa1,A−a2, there exists CT>0 such that the solution n,l of system (37) verifies the following inequality:
(68)∫0An2a,0da≤CT∫Θ1n2a,tdadt.Note that for everyT>supa1,A−a2, there exists a0∈a1,a2 such that na,0=0 for all a∈a0,A. This is a consequence of the following lemma.Lemma 10.
Let us suppose thatT>supa1,A−a2. Then, there exists a0∈a1,a2 such that T>A−a0>A−a for all a∈a0,A. Therefore, na,0=0 for all a∈a0,A.Proof of Lemma 10.
Suppose thatT>A−a2; then, there exists κ>0 (we choose κ such that κ<a2−a1) T>A−a2+κ. So T>A−a2−κ, and we denote a0=a2−κ. Then, T>A−a0>A−a for all a∈a0,A. Finally, from (40) for all a,t such that T−t>A−a, we get na,0=0 for all a∈a0,A.Proof of Proposition 9.
From the Lemma10, we have to prove the following inequality:
(69)∫0a0n2a,0da≤CT∫0T∫a1a2n2a,tdadt.
We setn~a,t=e−∫0aμmsdsna,t. Then, from the first equation of (39), n~ satisfies
(70)∂n~∂t+∂n~∂a=0in0,A×0,T.
Inequality (47) is a consequence of the following estimation:
(71)∫0a0n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.
Indeed, we have(72)∫0a0n2a,0da=∫0a0e2∫0aμmsdsn~2a,0da≤e2μmL10,a0∫0a0n~2a,0da≤Ce4μmL10,a0∫0T∫a1a2n2a,tdadt.
We consider in this proof the characteristicsγλ=a+λ,λ. For λ=0, the characteristics start from a,0. We have two cases.Case 1.
T<a2.
Two situations can arise:(i)
b0=a2−T<a1<a0, in this situation, we split the interval 0,a0 as(73)0,a0=0,b0∪b0,a1∪a1,a0(ii)
a1<b0<a0, in this situation, we split the interval 0,a0 as(74)0,a0=0,a1∪a1,a0Case 2.
T≥a2.
In this case, we split the interval0,a0 as
(75)0,a0=0,a1∪a1,a0.We make the proof for the second case. For the proof in the first case, see [6].Upper bound on0,a1:Fora∈0,a1, we set wλ=n~T+a−λ,T−λ,λ∈a1,T. We prove easily that w is a constant, see the proof of Proposition 8. Then, we set
(76)wt=1a2−a1∫t−a2t−a1wλdλ.Integrating with respect toa over 0,a1, we get
(77)∫0a1n~2a,0da≤C∫0a1∫a1a2n~2a+α,αdαda.Finally, we obtain(78)∫0a1n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.Upper bound ona1,a0:Fora∈a1,a0, we set wλ=n~T+a−λ,T−λ,λ∈a0,T and
(79)wt=1a2−a0∫t−a2t−a0wλdλ.Making as above and integrating with respect toa over a1,a0, it follows that
(80)∫a1a0n~2a,0da≤C∫0T∫a1a2n~2a,tdadt.The inequalities (78) and (80) give the desired result.We also need the following estimate for the proof of Theorem7.Proposition 11.
Let us assume the assumptionsH1−H2, let b1<a0<b and T>b1.
Then, there existsCT>0 such that the solution l of (39) verifies
(81)∫0a0l2a,0da≤CT∫Θ2l2a,tdadt.Proof of Proposition 11.
We suppose thatβ=0 in 0,b, the function l verifies
(82)−∂l∂t−∂l∂a+μfl=0in0,b×0,T,la,T=lTin0,b.
Proceeding as in the proof of Proposition9, we get the desired result.For the proof of Theorem7, we start with the following lemma.Lemma 12.
Suppose thatT>A−a2+a1 and a1<b.
Then, there existsa0∈a1,b and κ>0 such that
(83)T>T−a1+κ>A−a0>A−aforalla∈a0,A.
Therefore,(84)la,0=∫0A−aπ2a+sπ2aβa+s,psl0,s+n0,sdsforalla∈a0,A.Proof of Lemma 12.
Notice that the solutionl of (39) is given by
(85)lt=π2a+T−tπ2alTa+T−t+∫tTπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifT−t≤A−a,∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sdsifA−a<T−t.
Without loss of generality, we suppose thata2=b.
Suppose thatT>a1+A−a2⟺T−a1>A−a2. Then, there exists κ>0 (we choose κ such that 2κ<a2−a1) such that T−a1+κ>A−a2−κ. We denote by a0=a2−κ, and as A−a0>A−a for all a∈a0,A, then T>T−a1+κ>A−a0>A−a for all a∈a0,A.
Moreover, fora,t such that T−t>A−a, we have
(86)la,t=∫tt+A−aπ2a+s−tπ2aβa+s−t,ps1−γn0,s+γl0,sds.
Fort=0 and a∈a0,A,T−0>A−a0>A−a, one has
(87)la,0=∫0A−aπ2a+sπ2aβa+s,ps1−γn0,s+γl0,sds.
Remark that asa1,a2⊂b1,b2, if a0∈a1,a2, then a0∈b1,b2.Now, we can prove Theorem7.Proof of Theorem 7.
Leta0 as in Lemma 12, we have
(88)∫0Al2a,0da=∫0a0l2a,0da+∫a0Al2a,0da.
Using the results of Lemma12, the assumption of β, and the regularity of π2a+s/π2a, we can prove the existence of a constant KT independent of p (β∈L∞0,T×0,A such that
(89)∫a0Al2a,0da≤KT∫0T−a1+κn20,tdt+∫0T−a1+κl20,tdt.
Moreover, we haveb1≤a1≤a1+κ. Using Proposition 8, it follows that
(90)∫0T−a1+κn20,tdt≤CT∫Θ1n2a,tdadt,∫0T−a1+κl20,t≤CT′∫Θ2l2a,tdadt.
Finally, adding the above to the results of Proposition9 and Proposition 11, we get
(91)∫0An2a,0da+∫0Al2a,0da≤CT∫Θ1n2a,tdadt+∫Θ2l2a,tdadt.Forε>0 and θ>0, we consider the functional Jε,θ defined by
(92)Jε,θvm,vf=12∫Θ1vm2dadt+12∫Θ2vf2dadt+12ε∫0Am2a,Tda+12ε∫0Af2a,Tda,where m,f is the solution of the following system:
(93)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=χΘ2vfinQ,ma,0=m0a,fa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tda,f0,t=γ∫0Aβa,pfa,tdainQT.Lemma 13.
The functionalJε,θ is continuous, strictly convex, and coercive. Consequently, Jε,θ reaches its minimum at a point vm,ε,vf,θ∈L2Θ1×L2Θ2. Setting mε,fθ the associated solution of (93) and nε,lθ the solution of (39) with
(94)nεa,T=−1εmεa,T,lθa,T=−1θfθa,T,we have
(95)χΘ1vm,ε=χΘ1nε,χΘ2vf,θ=χΘ2lθ.Moreover, there existCi>0,1≤i≤4, independent of ε and θ such that
(96)∫Θ1nε2a,tdadt≤C1∫0Am02ada+∫0Af02ada,∫0Amε2a,Tda≤εC2∫0Am02ada+∫0Af02ada,∫Θ2lθ2a,tdadt≤C3∫0Am02ada+∫0Af02ada,∫0Afθ2a,Tda≤θC4∫0Am02ada+∫0Af02ada.Proof of Lemma 13.
It is easy to check thatJε,θ is coercive, continuous, and strictly convex. Then, it admits a unique minimizer vε,vθ. The maximum principle gives
(97)χΘ1vm,ε=χΘ1nε,χΘ2vf,θ=χΘ2lθwhere the couple nε,lθ is the solution of the system
(98)−∂tnε−∂anε+μmnε=0inQ,−∂tlθ−∂alθ+μflθ=1−γβa,pnε0,t+γβa,plθ0,tinQ,nεa,T=−1εmεa,T,lθa,T=−1θfθa,TinQA,nεA,t=0,lθA,t=0inQT.
Multiplying the first and the second equation of (98) by, respectively, mε and fθ, integrating with respect to Q and using (97), we get
(99)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda=−∫0Am0anεa,0da−1−γ∫0T∫0Aβa,pfθa,tnε0,tdadt(100)∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda=−∫0Af0alθa,0da+1−γ∫0T∫0Aβa,pfθa,tnε0,tdadt.
Combining (99) and (100), we obtain
(101)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda=−∫0Am0anεa,0da−∫0Af0alθa,0da.
Using the inequality of Young, we have, for anyδ>0,
(102)∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda≤δ2∫0Am02ada+12δ∫0Anε2a,0da+δ2∫0Af02ada+12δ∫0Alθ2a,0da.
Using observability inequality (42) and choosing δ=CT in the previous inequality, it follows that
(103)12∫Θ1nε2a,tdadt+1ε∫0Amε2a,Tda+12∫Θ2lθ2a,tdadt+1θ∫0Alθ2a,Tda≤CT2∫0Am02ada+∫0Af02ada.
This gives the desired result necessary to the proof of the main one.Now, we consider the system(104)∂tmεp+∂amεp+μmmεp=χΘ1nεinQ,∂tfθp+∂afθp+μmfθp=χΘ2lθinQ,mεpa,0=m0a,fθpa,0=f0ainQA,mεp0,t=1−γ∫0Aβa,pfθpa,tda,fθp0,t=γ∫0Aβa,pfθpa,tdainQT,where nε,lθ is the solution of (98) that minimizes the functional Jε,θ. We have the following result:Lemma 14.
Under the assumptions of Theorem2, the solutions mε and fθ verify the following inequalities:
(105)∫0Amε2a,Tda+∫0T∫0A1+μmmε2a,tdadt≤C∫0Am02ada+∫0Af02ada,(106)∫0Afθ2a,Tda+∫0T∫0A1+μffθ2a,tdadt≤C∫0Am02ada+∫0Af02ada.Proof of Lemma 14.
We denote by(107)yε,zθ=e−λ0tmε,e−λ0tfθ.
The functionsyε and zθ verify
(108)∂tyε+∂ayε+λ0+μmyε=χΘ1e−λ0tnε,(109)∂tzθ+∂azθ+λ0+μfzθ=χΘ2e−λ0tlθ.
Multiplying equality (108) and equality (109) by, respectively, yε and zθ and integrating with respect to Q, we get
(110)12∫0Ayε2a,Tda+12∫0Tyε2A,tdt+∫0T∫0Aλ0+μmayε2a,tdadt=12∫0Ay02ada+1−γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ1e−λ0tnεyεdadt,(111)12∫0Azθ2a,Tda+12∫0Tzθ2A,tdt+∫0T∫0Aλ0+μfazθ2a,tdadt=12∫0Af02ada+γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ2e−λ0tlθzθdadt.
Using the Young inequality, Cauchy-Schwarz inequality, and the fact thatβ is L∞, we prove that
(112)1−γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ1e−λ0tnεyεdadt≤α+2zθL2Q2+12yεL2Q2+12nεL2Θ12,(113)γ2∫0T∫0Aβa,pzθda2dt+∫0T∫0AχΘ2e−λ0tlθzθdadt≤α+2zθL2Q2+12zθL2Q2+12lθL2Θ22.
Therefore, choosingλ0>α+2+3/2, we get
(114)12∫0Azθ2a,Tda+∫0T∫0A1+μfazθ2a,tdadt≤12f0QA2+lθL2Θ22.
Finally, applying the result of Lemma13 to the above inequality, it follows that
(115)12∫0Azθ2a,Tda+∫0T∫0A1+μfazθ2a,tdadt≤C∫0Af02ada+∫0Am02adaand then the inequality (106) holds.
Likewise, we have(116)12∫0Ayε2a,Tda+∫0T∫0A1+μmyε2a,tdadt≤12m0QA2+α+2zθL2Q2+12nεL2Θ12.
Using the above inequality, Lemma13, and inequality (115), we obtain
(117)∫0Ayε2a,Tda+∫0T∫0A1+μmyε2a,tdadt≤C∫0Af02ada+∫0Am02adaand then we get the desired result.Finally, from Lemma13 and Lemma 14, if ε,θ⟶0,0, we get
(118)χΘ1nε,χΘ2lθ⇀χΘ1vm,χΘ2vf,mε,fθ⇀m,f,with m,f solution of problem (37) and
(119)m.,T=f.,T=0a.ea∈0,A.We have now the necessary ingredients for the proof of Theorem2.


## 4.2. Proof of Theorem2

In this section, we established the existence of a fixed point for the preceding auxiliary problem. Indeed, we consider thatH3 holds and we suppose to simplify that λ0=λA=0. For each p∈L2QT, let us denote by Λp⊂L20,T the set of all ∫0Aλampda, where the couple mp,fp is the solution of the following system:
(120)∂tmp+∂amp+μmmp=χΘ1ninQ,∂tfp+∂afp+μmfp=χΘ2linQ,mpa,0=m0a,fpa,0=f0ainQA,mp0,t=1−γ∫0Aβa,pfpa,tda,fp0,t=γ∫0Aβa,pfpa,tdainQT,and np,lp the corresponding solution of the minimizer of Jε,θ with mpa,T=fpa,T=0 for almost every a∈0,A.We have the following result.Proposition 15.
Under the assumptions of Theorem2, for any p∈L2QT, the solution of problem (120) satisfies
(121)∣Yt∣+ddtYL20,T≤Cm0L2QA+f0L2QA,where Yt=∫0Aλampda and the constant C is independent of p, m0, and f0.Proof of Proposition 15.
LetYt=∫0Aλampda. It is easy to prove that Y is the solution of system
(122)∂tY+∫0Aμmaλampda=RtinQT,Y0=∫0Aλam0ada,where
(123)Rt=∫0Aλ′ampda+1−γλ0∫0Aβa,pfpda+∫0a2λanpda.
Using Lemma14 and the assumptions on β and λ, we infer that there exists K>0 such that
(124)RL2QT≤Km0L2QA+f0L2QA.
By using (122), the Young inequality and integrating on QT, we obtain
(125)∫0T∂tY2dt≤2∫0TRt2dt+2∫0T∫0Aμmaλamεpda2dt.
Moreover, the Cauchy-Schwarz inequality leads to(126)∫0T∫0Aμmaλamεpda2dt≤∫0Aμmaλada∫0T∫0Aμmaλamε2pdadt.
The inequality (105) and the fact that λ∈C0,A give
(127)∫0T∫0Aμmaλamε2pdadt≤K1m0L2QA2+f0L2QA2,where K1>0 is independent of p, ε, and θ. Moreover, as λμm∈L10,A, and using (124), it follows that
(128)ddtYL20,T≤Cm0L2QA+f0L2QA.
Now, letY~=e−λ0tY. Then, Y~ satisfies
(129)∂tY~+λ0Y~+e−λ0t∫0Aμmaλamεpda=e−λ0tRtinQT,Y~0=∫0Aλam0ada.
Multiplying the first equation of (129) by Y~, integrating on 0,t, and using successively Cauchy-Schwarz and Young inequalities, we deduce that
(130)Y~t2+λ0∫0tY~2dt≤Y~02+∫0tY~2dt+∫0T∫0Aμmaλamεpda2dt+RL2QT.
Using the above calculations and choosingλ0>2, we get
(131)Y~t2≤K2m0L2QA+f0L2QA.
The desired result comes from (128) and (131).It is obvious thatΛp is convex, and let
(132)W0,T=Y∈L∞0,T,YL∞0,T≤N;dYdtL20,T≤N,with N=Cm0L2QA+f0L2QA.We haveW0,T⊂W1,10,T. Moreover, the injection of W1,10,T into L20,T is compact, see [10], page 129. So W0,T is relatively compact in L20,T. From Proposition 15, we have ΛW0,T⊂W0,T, and we see that ΛW0,T is a relatively compact subset of L20,T. Let us now prove that Λ is uppersemicontinuous. This is equivalent to prove that for any closed subset K of L20,T, Λ−1K is closed in L20,T. Let pk∈Λ−1K such that pk converges towards p in L20,T. Then, pk is bounded, and for all k, there exists Pk∈K such that Pk∈Λpk. Therefore, from the definition of Λ, there exists mk,fk∈L20,T×0,A2 associated with nk,lk∈L2Θ1×L2Θ2 solution of (120) such that Pk=∫0Aλamkpkda and satisfying the inequalities of Lemma 13 and Lemma 14. Consequently, mk,fk and nk,lk are bounded, respectively, in L20,T×0,A2 and in L2Θ1×L2Θ2. Thus, there exists subsequences still denoted by mk,fk and nk,lk that converge weakly to m,f in L20,T×0,A2 and n,l in L2Θ1×L2Θ2, respectively. Using hypothesis H3, it follows that ∫0Aλamkpkda converges strongly to ∫0Aλampda in L20,T.Now, by a standard device, we see thatm,f associated with n,l are a solution of (120) and satisfy the inequalities of Lemma 13 and Lemma 14. This implies that P∈Λp.On the other hand, thanks to Proposition15, one can extract a subsequence also denoted by Pk that converges strongly towards the function P in L20,T. Since K is closed, we deduce that P∈K. Finally, we deduce that p∈Λ−1P.Applying the Kakutani fixed-point theorem [11] in the space L20,T to the mapping Λ, we infer that there is at least one Y∈W0,T such that Y∈ΛY. This completes the null controllability proof of the model (1).


## 4.3. Proof of Theorem3

### 4.3.1. Proof of Theorem3-(1)

In this section, we always consider the following system:(133)mt+ma+μmm=χΘvminQ,ft+fa+μff=0inQ,ma,0=m0,fa,0=f0inQA,m0,t=1−γ∫0Aβa,pfda,f0,t=γ∫0Aβa,pfdainQT,for every p in L2QT. Under the assumptions of Theorem 3, the controllability problem that is to find vm∈L2Θ such that m,f solution of the system (133) verifies
(134)m.,T=0a∈ρ,Ais equivalent to the following observability inequality.Proposition 16.
Let us assume the assumptionsH1−H2−H3, for every T>A−a2 and for any ρ>0, if hTa=0a.e in 0,ϱ, there exists Cρ,T>0 such that the following inequality:
(135)∫0Ah2a,0da+∫0Ag2a,0da≤Cρ,T∫Θh2a,tdadtholds, where h,g is the solution of
(136)−ht−ha+μmh=0inQ,−gt−ga+μfg=1−γβa,pg0,t+γβa,ph0,tinQ,ha,T=hT,ga,T=0inQA,hA,t=0,gA,t=0inQT.For the proof of Proposition16, we state the following estimate.Proposition 17.
Under the assumptionsH1 and H2, there exists a constant C>0 such that the solution h,g of system (136) verifies
(137)∫0Ag2a,0da+∫0T∫0A1+μfg2a,tdadt≤C∫0Th20,tdt.
Moreover, we deduce forhT=0 a.e in 0,ϱ that there exists a constant Cϱ,T>0 such that
(138)∫0Ag2a,0da≤Cϱ,T∫0Th20,tdt≤Cϱ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 16.
Settingy=eλ0tg, the function y verifies
(139)−∂ty−∂ay+λ0+μfy=γβa,peλ0th0,t+1−γβa,py0,t.
Multiplying equality (139) by y and integrating on Q, we obtain
(140)12∫0Ay2a,0da+12∫0Ty20,tdt+∫0T∫0Aλ0+μfy2a,tdadt=∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt.
Using Young inequality and the condition onβ, we get
(141)∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt≤e2λ0Tα+22δh0,.L2QT2+δ2yL2Q2+α+22δy0,.L2QT2+δ2yL2Q2.
Choosingδ=α+2, we obtain
(142)12∫0Ay2a,0da+∫0T∫0Aλ0+μfy2a,tdadt≤e2λ0T2h0,.L2QT2+α+2yL2Q2.
Finally, choosingλ0>α+2+1, it follows that
(143)12∫0Ay2a,0da+∫0T∫0A1+μfy2a,tdadt≤e2λ0T2∫0Ah20,tdt.
So,(144)∫0Ay2a,0da≤e2λ0T∫0Ah20,tdt.
Finally, we get(145)∫0Ag2a,0da≤e2λ0T∫0Ah20,tdt.
Combining the above inequality and inequality (44) of Proposition 8, for hT=0 a.e in 0,ρ, we get
(146)∫0Ag2a,0da≤Cρ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 17.
We use the results of Proposition9 and Proposition 17. Indeed, by combining (68) and (146), the desired result is obtained.Now, letε>0 and ϱ>0. We consider the functional Jε defined by
(147)Jεvm=12∫0T∫a1a2vm2a,tdadt+12ε∫ρAm2a,Tda,where m,f is the solution of the following system:
(148)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=0inQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tdainQT,f0,t=γ∫0Aβa,pfa,tdainQT.We have the following lemma.Lemma 18.
The functionalJε is continuous, strictly convex, and coercive. Consequently, Jε reaches its minimum at one and has vm,ε=χΘhε and there exists positive constants C1,C2 independent of ε such that
(149)∫0T∫0a2hε2a,tdadt≤C1∫0Am02ada+∫0Af02ada,(150)∫ρAmε2a,Tda≤εC2∫0Am02ada+∫0Af02ada.Proof of Lemma 18.
The proof is similar to that of Lemma13.By makingε tending towards zero, we thus obtain that χΘhε⇀χΘvm and mε,fε⇀m,f, where m,f is the solution of system (148) that verifies
(151)m.,T=0a.einϱ,A.Finally, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear problem.


### 4.3.2. Proof of Theorem3-(2)

Letp∈L2QT, under the assumptions of Theorem 3, the following controllability problem finds vf∈L2Θ such that the solution of the system
(152)ft+fa+μff=χΘ2vfinQ,fa,0=f0ainQA,f0,t=γ∫0Aβa,pfa,tdainQTverifies
(153)f.,T=0a.ein0,Ais equivalent to the following observability inequality.Proposition 19.
Let us assume true the assumptionsH1−H2−H3. For any T>a1+A−a2, there exists CT>0 such that
(154)∫0Ag2a,0da≤CT∫Θ2g2a,tdadt,where g is a solution of the system
(155)−gt−ga+μfg=γβa,pg0,tinQ,ga,T=gTinQA,gA,t=0inQT.Proof of Proposition 19.
Using inequality (45) of Proposition 8, the result of Proposition 11, and the representation of the solution of the system (155), we get the desired result.To conclude, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear system. We omit all details because the extension is straightforward.


## 4.3.1. Proof of Theorem3-(1)

In this section, we always consider the following system:(133)mt+ma+μmm=χΘvminQ,ft+fa+μff=0inQ,ma,0=m0,fa,0=f0inQA,m0,t=1−γ∫0Aβa,pfda,f0,t=γ∫0Aβa,pfdainQT,for every p in L2QT. Under the assumptions of Theorem 3, the controllability problem that is to find vm∈L2Θ such that m,f solution of the system (133) verifies
(134)m.,T=0a∈ρ,Ais equivalent to the following observability inequality.Proposition 16.
Let us assume the assumptionsH1−H2−H3, for every T>A−a2 and for any ρ>0, if hTa=0a.e in 0,ϱ, there exists Cρ,T>0 such that the following inequality:
(135)∫0Ah2a,0da+∫0Ag2a,0da≤Cρ,T∫Θh2a,tdadtholds, where h,g is the solution of
(136)−ht−ha+μmh=0inQ,−gt−ga+μfg=1−γβa,pg0,t+γβa,ph0,tinQ,ha,T=hT,ga,T=0inQA,hA,t=0,gA,t=0inQT.For the proof of Proposition16, we state the following estimate.Proposition 17.
Under the assumptionsH1 and H2, there exists a constant C>0 such that the solution h,g of system (136) verifies
(137)∫0Ag2a,0da+∫0T∫0A1+μfg2a,tdadt≤C∫0Th20,tdt.
Moreover, we deduce forhT=0 a.e in 0,ϱ that there exists a constant Cϱ,T>0 such that
(138)∫0Ag2a,0da≤Cϱ,T∫0Th20,tdt≤Cϱ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 16.
Settingy=eλ0tg, the function y verifies
(139)−∂ty−∂ay+λ0+μfy=γβa,peλ0th0,t+1−γβa,py0,t.
Multiplying equality (139) by y and integrating on Q, we obtain
(140)12∫0Ay2a,0da+12∫0Ty20,tdt+∫0T∫0Aλ0+μfy2a,tdadt=∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt.
Using Young inequality and the condition onβ, we get
(141)∫0T∫0Aγβa,peλ0th0,t+1−γβa,py0,tydadt≤e2λ0Tα+22δh0,.L2QT2+δ2yL2Q2+α+22δy0,.L2QT2+δ2yL2Q2.
Choosingδ=α+2, we obtain
(142)12∫0Ay2a,0da+∫0T∫0Aλ0+μfy2a,tdadt≤e2λ0T2h0,.L2QT2+α+2yL2Q2.
Finally, choosingλ0>α+2+1, it follows that
(143)12∫0Ay2a,0da+∫0T∫0A1+μfy2a,tdadt≤e2λ0T2∫0Ah20,tdt.
So,(144)∫0Ay2a,0da≤e2λ0T∫0Ah20,tdt.
Finally, we get(145)∫0Ag2a,0da≤e2λ0T∫0Ah20,tdt.
Combining the above inequality and inequality (44) of Proposition 8, for hT=0 a.e in 0,ρ, we get
(146)∫0Ag2a,0da≤Cρ,T∫0T∫0a2h2a,tdadt.Proof of Proposition 17.
We use the results of Proposition9 and Proposition 17. Indeed, by combining (68) and (146), the desired result is obtained.Now, letε>0 and ϱ>0. We consider the functional Jε defined by
(147)Jεvm=12∫0T∫a1a2vm2a,tdadt+12ε∫ρAm2a,Tda,where m,f is the solution of the following system:
(148)mt+ma+μmm=χΘ1vminQ,ft+fa+μff=0inQ,ma,0=m0afa,0=f0ainQA,m0,t=1−γ∫0Aβa,pfa,tdainQT,f0,t=γ∫0Aβa,pfa,tdainQT.We have the following lemma.Lemma 18.
The functionalJε is continuous, strictly convex, and coercive. Consequently, Jε reaches its minimum at one and has vm,ε=χΘhε and there exists positive constants C1,C2 independent of ε such that
(149)∫0T∫0a2hε2a,tdadt≤C1∫0Am02ada+∫0Af02ada,(150)∫ρAmε2a,Tda≤εC2∫0Am02ada+∫0Af02ada.Proof of Lemma 18.
The proof is similar to that of Lemma13.By makingε tending towards zero, we thus obtain that χΘhε⇀χΘvm and mε,fε⇀m,f, where m,f is the solution of system (148) that verifies
(151)m.,T=0a.einϱ,A.Finally, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear problem.


## 4.3.2. Proof of Theorem3-(2)

Letp∈L2QT, under the assumptions of Theorem 3, the following controllability problem finds vf∈L2Θ such that the solution of the system
(152)ft+fa+μff=χΘ2vfinQ,fa,0=f0ainQA,f0,t=γ∫0Aβa,pfa,tdainQTverifies
(153)f.,T=0a.ein0,Ais equivalent to the following observability inequality.Proposition 19.
Let us assume true the assumptionsH1−H2−H3. For any T>a1+A−a2, there exists CT>0 such that
(154)∫0Ag2a,0da≤CT∫Θ2g2a,tdadt,where g is a solution of the system
(155)−gt−ga+μfg=γβa,pg0,tinQ,ga,T=gTinQA,gA,t=0inQT.Proof of Proposition 19.
Using inequality (45) of Proposition 8, the result of Proposition 11, and the representation of the solution of the system (155), we get the desired result.To conclude, a similar functionΛ is defined and a similar procedure is followed to get the null controllability for the nonlinear system. We omit all details because the extension is straightforward.


## 5. Numerical Illustrations

In this part, the idea is to highlight the numerical simulation of the nonlinear problem.The first parts are to reduce the PDE to the finite dimensional system of the form(156)U̇l=AlUl+Pl+BlYl,where Al, Bl, and Pl are matrices, and
(157)Ult=VltWltis the finite dimensional state vector. Here, vector Pl is the contribution of the nonlinear part, which comes from births; Bl is the control matrix, and Yl is the control vector.So let us consider the following system:(158)∂m∂t+∂am+μmm=vmin0,A×0,T,∂f∂t+∂af+μff=vfin0,A×0,T,m0,t=1−γ∫0Aλmda∫0Aβfdain0,T,f0,t=γ∫0Aλmda∫0Aβfdain0,T,ma,0=m0in0,A,fa,0=f0in0,A,To solve (158), the age discretization is performed with the finite difference method on a rectangular grid on 0,A. For a given rectangular grid T with vertex aj,1≤j≤n and uniform step size (without loss of generality) Δa in a−direction, we denote by the diameter of the grid. The finite difference approximation of the aging term is given by
(159)∂y∂aaj,t=yaj,t−yaj−1,tΔa.Letmjt be the approximation of maj,t and Vlt=mjt1≤j≤n and let fjt be the approximation of faj,t and Wlt=fjt1≤j≤n. We denote by Al the matrix of the mortality approximation and the aging approximation.We approximate∫0Aβafa,tda by the rectangle method. In fact,
(160)∫0Aβafa,tda=An∑j=1n−1βajfaj,t+β0f0,t+βAfA,t2,(161)∫0Aλama,tda=A/n∑j=1n−1λajmaj,t+λ0m0,t+λAmA,t2.Then,(162)f0=γA2n2∑k=1n−1βakfak,t∑j=1n−1λajmaj,t.But as(163)∂f1t∂a=f1t−f0tΔa,then
(164)∂f1t∂a=−γA2/n2∑k=1n−1βakfak,t∑j=1n−1λajmaj,t+f1tΔa,(165)m0=1−γA2n2∑k=1n−1βakfak,t∑j=1n−1λajmaj,t.Remark 20.
f0 and m0 are the approximations of births.Now, as(166)∂m1t∂a=m1t−m0tΔa,then
(167)∂m1t∂a=−1−γA2/n2∑k=1n−1βakfak,t∑j=1n−1λajmaj,t+m1tΔa.The corresponding discrete system is given for a continuous initial solutiony0 by
(168)dUldt=AlUl+Pl+BlYl,Ul0=y0aj1≤j≤n.Example 1.
For the simulation, we takeA=2, Δa=1/20, T=0.5, and n=20.
Moreover, we choose the initial condition(169)m0a=10e−1/A−a,f0a=15e−1/A−a.The fertilityβ is given by
(170)βa=e−0.5210a−4Γ5and the mortality rate μ (see [12, 13]) by
(171)μma=μfa=110A−a,λa=1.Remark 21.
Note thatlima⟶Aμma=lima⟶Aμfa=+∞. Therefore, the probability of survival e−∫0aμsds at maximum age A of male and female individuals is zero.Thus, the following numerical results were obtained.When the initial density of males or females is zero, the evolution of the population of males or females remains zero, see Figures1 and 2.Figure 1
Evolution of the uncontrolled system froma=t=0 to A=t=2.Figure 2
Evolution of the uncontrolled system froma=t=0 to A=t=2.We notice that numerically, we have the positivity of the state and the initial conditions, see Figures3 and 4. Also, when we tend towards A, the density of each population tends towards zero. Therefore, this fact reflects the reality.Figure 3
Initial conditions.Figure 4
Evolution of the uncontrolled system froma=t=0 to A=2 and t=0.5.When the initial density of male population or female population is zero, we have no males or females during the evolution of the population, see Figures1 and 2. This fact shows the importance of fertility function.Moreover, in the absence of males or females, we obtain the total extinction of females or males at timet=A=2 confirming the reality.We use ODE 45 for all the simulations of the system.Example 2.
We construct the control problem, which consists in minimizing the function, and we choose the classical Hum function(172)JεU,Y=1ε∑i=12∫0AUi2da+∑i=12∫0T∫0AYi2dadt.The approximate null controllability becomes the minimization of the functionalJε, where U,Y verifies the system
(173)dUldt=AlUl+Pl+BlYl.In this example, we use the data of Example1.For the final timeT=0.5, the control states and the controlled states are given below.We notice that the uncontrolled solutions are not null at the corresponding final time (see Figure4) while the controlled states are zero at the corresponding final time (see Figure 5). However, we could not take a positivity constraint in our simulations (see Figure 6).Figure 5
Evolution of the controlled system froma=t=0 to A=2 and t=0.5.Figure 6
Evolution of the control states froma=t=0 to A=2 and t=0.5.The CaSadi toolbox is used to simulate the control system (by the minimization of the function: Hum method).



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