New results on the Stackelberg-Nash exact control of linear parabolic equations

New results on the Stackelberg-Nash exact control of linear parabolic equations F. D. Araruna∗ , E. Fern´ andez-Cara† , S. Guerrero‡and M. C. Santos§ Abstract This paper is concerned with Stackelberg-Nash strategies to control parabolic equations. We have one control, the leader, that is responsible for a null controllability property; additionally, we have a couple of controls, called the followers, that provides a Nash equilibrium for two cost functionals. This is a classical situation in many fields of science and, in mathematics, leads to a lot of interesting questions and open problems and possesses many applications. In the main result, we prove the existence of a leader such that the corresponding controlled system is driven to zero. This way, we improve some questions that were left open in previous works. 1 Introduction There are plenty of situations where several controls are required in order to drive a system to one or more objectives. Usually, if we assign different roles to the controls, we speak of hierarchic control. In the case of a system governed by a PDE, this concept was introduced by J.-L. Lions (see [15, 16], where some techniques are presented). These works motivated the study of the subject and a lot of other results appeared; see for instance [4, 5, 12, 19, 20]. All these previous works combine the multicriteria optimization concepts and arguments and approximate controllability. In the context of null controllability, few is known; see [1] for some first results. In this paper, we solve a question that was left open in [1]. The solution requires some careful computations based on new Carleman estimates. Let us be more precise. Let Ω ⊂Rnbe a bounded domain whose boundary Γ is regular enough. Let T > 0 be given and define Q:= Ω×(0, T), with lateral boundary Σ := ∂Ω×(0, T ). In the sequel, we will denote by Ca generic positive constant which may differ from line to line. Sometimes, we will write C(Ω), C(Ω, T), etc. to indicate the data on which Cdepends. The usual norm and scalar product in L2(Ω) will be respectively denoted by k·k and (·,·). Let us consider the linear system    yt−∆y+a(x, t)y=f1O+v11O1+v21O2in Q, y= 0 on Σ, y(·,0) = y0in Ω, (1) where y=y(x, t) is the state, a∈L∞(Q) and y0=y0(x) is prescribed. In (1), the set O ⊂ Ω is the main control domain and O1,O2⊂Ω are the secondary control domains (all of them are supposed to be small); 1O, 1O1and 1O2are the characteristic functions of O,O1and O2, respectively; the controls are the leader f=f(x, t) and the followers v1=v1(x, t) and v2=v2(x, t). ∗Dpto. de Matem´atica, Universidade Federal da Para´ıba, 58051-900, Jo˜ao Pessoa - PB, Brazil. E-mail: [email protected]. Partially supported by INCTMat, CAPES, CNPq (Brazil) and MathAmSud COSIP. †Dpto. EDAN and IMUS, University of Sevilla, Aptdo. 1160, 41080 Sevilla, Spain ([email protected]). Partially supported by grant MTM2010-15592, DGI-MICINN (Spain) and CAPES (Brazil). ‡Laboratoire Jacques-Louis Lions, Universit´e Pierre et Marie Curie, 75252 Paris C´edex 05, France. E-mail: [email protected]. Partially supported by MathAmSud COSIP. §Dpto. de Matem´atica, Universidade Federal da Pernambuco, 50670-901, Recife - PE, Brazil. E-mail: [email protected]. Partially supported by CAPES (Brazil) and MathAmSud COSIP. 1 Let O1,d,O2,d ⊂Ω be open sets, representing observation domains for the followers. We will consider the (secondary) functionals Ji(f;v1, v2) := αi 2ZZOi,d×(0,T ) |y−yi,d|2dx dt +µ 2ZZOi×(0,T ) |vi|2dx dt, i = 1,2, and the main functional J(f) := 1 2ZZO×(0,T ) |f|2dx dt, where the αi>0 and µ > 0 are constants and the yi,d =yi,d(x, t) are given functions. The structure of the control process can be described as follows: 1. For each leader f, the followers v1and v2intend to be a Nash equilibrium for the costs Ji(i= 1,2). In other words, once fhas been fixed, we look for a couple (v1, v2) with vi∈L2(Oi×(0, T)) such that J1(f;v1, v2) = min ˆv1J1(f; ˆv1, v2), J2f;v1, v2= min ˆv2J2(f;v1,ˆv2).(2) Note that, if the functionals Ji(i= 1,2) are C1and convex, then (v1, v2) is a Nash equilibrium if and only if J0 1(f;v1, v2)(ˆv1,0) = 0,∀ˆv1∈L2(O1×(0, T)) , vi∈L2(Oi×(0, T)) and J0 2(f;v1, v2)(0,ˆv2) = 0,∀ˆv2∈L2(O2×(0, T)) , vi∈L2(Oi×(0, T)). (In fact, this is also true if Jiis C1and convex in the i-th variable.) 2. Let us fix an uncontrolled trajectory of (1), that is, a sufficiently regular solution to the system    yt−∆y+a(x, t)¯y= 0 in Q, y= 0 on Σ, y(·,0) = y0in Ω. (3) Once the Nash equilibrium has been identified and fixed for each f, we look for an optimal control ˆ f∈L2(O × (0, T)) such that J(ˆ f) = min fJ(f), subject to the exact controllability restriction y(·, T) = y(·, T) in Ω.(4) In [1] it is proved that, if µis large enough, for every f∈L2(O × (0, T )) there exists a unique Nash equilibrium (v1, v2) for (J1, J2), given by vi=−1 µφi1Oi, i = 1,2, where (y, φ1, φ2) is the unique solution to the optimality system              yt−∆y+a(x, t)y=f1O− 2 X i=1 1 µφi1Oiin Q, −φi t−∆φi+a(x, t)φi=αi(y−yi,d)1Oi,d in Q, y= 0, φi= 0 on Σ, y(·,0) = y0, φi(·, T) = 0 in Ω. (5) The main result of this paper concerns the exact controllability to the trajectories of (1)–(2). It is the following: 2 Theorem 1. Suppose that Oi,d ∩ O 6=∅, i = 1,2.(6) Also, assume that one of the following two conditions holds: O1,d =O2,d (7) or O1,d ∩ O 6=O2,d ∩ O.(8) Then, there exists µ0>0, only depending on Ω,O,T,Oi,Oi,d,αiand kakL∞(Q)and a positive function ˆρ= ˆρ(t)blowing up at t=Tsuch that, if µ≥µ0, the yi,d are such that ZZOi,d×(0,T ) ˆρ2|y−yi,d|2dx dt < +∞, i = 1,2 and ¯yis the unique solution to (3) associated to the initial state y0∈L2(Ω), there exist controls f∈ L2(O × (0, T)) and associated Nash equilibria (v1, v2)such that the corresponding solutions to (1) satisfy (4). Remark 2. It is worth mentioning that, in [1], the authors have proved this result in the particular case in which (6) and (7) are satisfied. Fig. 1–3 illustrate some situations where this fails and (6) and (8) hold simultaneously.  1,d 2,d Figure 1 1,d 2,d Figure 2 1,d 2,d Figure 3 Note that, if we introduce the new variable z=y−¯y, (5) can be rewritten in the form              zt−∆z+a(x, t)z=f1O− 2 X i=1 1 µφi1Oiin Q, −φi t−∆φi+a(x, t)φi=αi(z−zi,d)1Oi,d in Q, z= 0, φi= 0 on Σ, z(·,0) = z0, φi(·, T) = 0 in Ω, (9) where zi,d =yi,d −¯yand z0=y0−y0and (4) is equivalent to the null controllability property for z, that is, z(·, T) = 0 in Ω.(10) The proof of Theorem 1 relies on some duality arguments which reduce the null controllability property of a linear system to an observability inequality for the solutions to the associated adjoint system. In our case, the adjoint of (9) is                −ψt−∆ψ+a(x, t)ψ= 2 X i=1 αiγi1Oi,d in Q, γi t−∆γi+a(x, t)γi=−1 µψ1Oiin Q, ψ= 0, γi= 0 on Σ, ψ(·, T) = ψT, γi(·,0) = 0 in Ω. (11) 3 For the proof of the observability of (11), as usual, we must use Carleman estimates. Thus, the task is to estimate globally all the variables ψ,γ1and γ2by just one observation (ψin O × (0, T)). This is not trivial, especially when O1,d 6=O2,d. In fact, in [1], the assumption O1,d =O2,d was needed. The main idea in this paper is to use not one Imanuvilov function, but two. This is possible thanks to Lemma 5 (see below, in Section 2). Actually, we believe that this lemma, used together with the arguments in the proof of Proposition 11, can be a useful tool for the solution of other control problems. For instance, to control a 3 ×3 system with one scalar control. The observability estimate for (11) is given in the following result: Theorem 3. Under the assumptions of Theorem 1, there exists a constant C > 0, depending only on Ω,O, T,Oi,Oi,d,αiand kakL∞(Q)and a weight ˆρ= ˆρ(t)such that ZΩ |ψ(x, 0)|2dx + 2 X i=1 ZZQ ˆρ−2|γi|2dx dt ⩽CZZO×(0,T ) |ψ|2dx dt. (12) In order to prove (12), we will use some suitable Carleman estimates with a specific and unusual choice of the weight functions. This will be done precisely in the following section. Remark 4. The arguments in Section 3 of [1] concerning semilinear problems can be adapted to the present setting. This allows to prove a result similar to Theorem 1 for the system    yt−∆y+a(x, t)y=F(y) + f1O+v11O1+v21O2in Q, y= 0 on Σ, y(·,0) = y0in Ω, (13) where (for instance) F∈W1,∞(R). As there, we have to introduce the concept of Nash quasi-equilibrium and then an appropriate fixed-point mapping. We omit the details, since the process is rather standard and well known.  The paper is organized as follows. In Section 2, we introduce the weight functions needed to prove some Carleman estimates. In Section 3, we prove these Carleman estimates. In Section 4, we deduce the observability inequality (12) and, also, that (12) implies (10). 2 Some previous results In this section we define the weight functions needed in the proof of Theorem 1. We will also recall some known Carleman inequalities. Let us introduce a nonempty open set e O ⊂⊂ O such that Oi,d ∩e O 6=∅for i= 1,2 and the nonempty connected open sets ωiwith ωi⊂⊂ Oi,d ∩e O, i = 1,2, ω1∩ω2=∅.(14) Observe that this is possible thanks to assumptions (6) and (8). The next result will be crucial: Lemma 5. There exist functions ηi∈C2(Ω) (i= 1,2) such that    ηi>0in Ω,ηi= 0 on ∂Ω, |∇ηi|>0in Ω\ωi,η1=η2in Ω\e O. The proof of Lemma 5 can be found in an Appendix (Section 5), at the end of the paper. Remark 6. We know from [8, Lemma 1.1] that, for any open set ω0⊂Ω, there exists η0∈C2(Ω) satisfying    η0>0 in Ω, η0= 0 on ∂Ω, |∇η0|>0 in Ω \ω0. From the proof of this lemma (see [8, pp. 20–21]), it is also clear that the function η0can be chosen with a finite number of critical points.  4 Remark 7. Lemma 5 establishes the existence of functions η1and η2which coincide outside e Obut may be very different inside e O. Nevertheless, it will be seen in the proof that one can find η1and η2satisfying kη1k∞=kη2k∞. Remark 8. From (6), (8) and (14), we see that it can be assumed that either ω1∩ O2,d =∅and ω2∩ O1,d =∅(15) or ωi⊂ Oj,d and ωj∩ Oi,d =∅,with (i, j) = (1,2) or (i, j) = (2,1).(16) The cases (15) and (16) (with (i, j) = (2,1)) correspond, respectively, to Fig. 1 and Fig. 3.  Let us introduce the weight functions σi(x, t) := e4λkηik∞−eλ(2kηik∞+ηi(x)) t(T−t), ξi(x, t) := eλ(2kηik∞+ηi(x)) t(T−t) and the notation Ii m(ψ) := sm−4λm−3ZZQ e−2sσi(ξi)m−4(|ψt|2+|∆ψ|2)dx dt +Li m(ψ), where Li m(ψ) := sm−2λm−1ZZQ e−2sσi(ξi)m−2|∇ψ|2dx dt +smλm+1 ZZQ e−2sσi(ξi)m|ψ|2dx dt. We will need some known results concerning Carleman estimates for parabolic PDEs. Thus, let us consider the system                −ut−∆u+au =f+ n X k=1 ∂kfkin Q, u= 0 on Σ, u(·, T) = uTin Ω, (17) where uT∈L2(Ω) and f, f1, . . . , fn∈L2(Q). We have the following: Proposition 9. Assume that, in (17),fk= 0 for all k= 1, . . . , n. Then, there exists C(Ω,O)>0such that, for every s≥C(T+T2)and every λ≥C, the solution uto (17) associated to uT∈L2(Ω) and f∈L2(Q) satisfies Ij m(u)⩽C smλm+1 ZZωj×(0,T ) e−2sσj(ξj)m|u|2dx dt +sm−3λm−3ZZQ e−2sσj(ξj)m−3|f|2dx dt!, j = 1,2. If the functions fkare not necessarily zero, the following holds: Proposition 10. There exists C(Ω,O)>0such that, for every s≥C(T+T2)and every λ≥C, the solution uto (17) associated to uT∈L2(Ω),f∈L2(Q)and the fk∈L2(Q)satisfies Lj m(u)⩽C smλm+1 ZZωj×(0,T ) e−2sσj(ξj)m|u|2dx dt +sm−3λm−3ZZQ e−2sσj(ξj)m−3|f|2dx dt +sm−1λm−1 2 X k=1 ZZQ e−2sσj(ξj)m−1|fk|2dx dt!, j = 1,2. These results are nowadays well known. For instance, when m= 3, their proofs can be respectively found in [8] and [14]. 5 3 A new Carleman inequality In this section, we will prove a suitable Carleman inequality for the solutions to the adjoint system (11). In [1], the authors proved a similar estimate assuming that (6) and (7) hold. Proposition 11. Assume that (6) and (8) are satisfied. Then, there exists C(Ω,O)>0such that, for every s≥C(T+T2)and every λ≥C, the solution (ψ, γ1, γ2)to (11) associated to ψT∈L2(Ω) satisfies the following: (i) If (15) holds, then I1 0(γ1) + I2 0(γ2) + s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt ⩽Cs4λ5ZZO×(0,T ) (e−2sσ1(ξ1)4+e−2sσ2(ξ2)4)|ψ|2dx dt. (18) (ii) If (16) holds for (i, j)=(i0, j0), with (i0, j0) = (1,2) or (i0, j0) = (2,1), then Ij0 0(γj0) + Ii0 0(h) + s−3λ−2ZZQ e−2sσj0(ξj0)−3|ψ|2dx dt ⩽Cs4λ5ZZO×(0,T ) (e−2sσ1(ξ1)4+e−2sσ2(ξ2)4)|ψ|2dx dt, (19) where we have denoted h:= α1γ1+α2γ2. Remark 12. In order to prove the above Carleman inequalities, we will have to work with the equation satisfied by γi. However, the terms involving γiand hin the left-hand side of (18) and (19) will not be needed later, in the proof of the observability inequality (12).  Proof. The proofs of (i) and (ii) are slightly different and will be presented separately. •Proof of (i). We first apply Proposition 9 for m= 0 and j=ito the functions (x, t)7→ γi(x, T −t) for i= 1,2. This gives Ii 0(γi)⩽C λZZωi×(0,T ) e−2sσi|γi|2dx dt +s−3λ−3ZZOi×(0,T ) e−2sσi(ξi)−3|ψ|2dx dt!.(20) Now, let θ3∈C2(Ω) be such that    θ3(x) = 0 for x∈e O, θ3(x) = 1 for x∈Ω\ O. From (11), we find that                −(θ3ψ)t−∆(θ3ψ) + a(θ3ψ) = 2 X j=1 αjθ3γj1Oj,d −2∇ · (ψ∇θ3)+∆θ3ψin Q, θ3ψ= 0 on Σ, (θ3ψ)(·, T) = θ3ψTin Ω. 6 Then, we apply Proposition 10 for m=−3 and j= 1 to θ3ψand we get L1 −3(θ3ψ)⩽C s−3λ−2ZZω1×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt +s−6λ−6ZZO1,d×(0,T ) e−2sσ1(ξ1)−6|γ1|2dx dt +s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ2|2dx dt +s−4λ−4ZZO×(0,T ) e−2sσ1(ξ1)−4|ψ|2dx dt!. (21) Using the fact that θ3equals 1 in Ω \O, we deduce that L1 −3(θ3ψ)≥s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt −s−3λ−2ZZO×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt. Putting together (21) and (20), we obtain I1 0(γ1) + I2 0(γ2) + s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt ⩽C λZZω1×(0,T ) e−2sσ1|γ1|2dx dt +s−6λ−6ZZO1,d×(0,T ) e−2sσ1(ξ1)−6|γ1|2dx dt +λZZω2×(0,T ) e−2sσ2|γ2|2dx dt +s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ2|2dx dt +s−3λ−3ZZO1×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt +s−3λ−3ZZO2×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt +s−3λ−2ZZO×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt!. (22) Observe that the second and the fifth terms in the right hand side of (22) can be absorbed by the left hand side by taking s≥CT2and λ≥C: s−6λ−6ZZO1,d×(0,T ) e−2sσ1(ξ1)−6|γ1|2dx dt +s−3λ−3ZZO1×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt ⩽εI1 0(γ1) + s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt. (23) The same happens to the fourth term in the right hand side of (22), since η1=η2(so σ1=σ2and ξ1=ξ2) in Supp (θ3)∩ O2,d: s−6λ−6 ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ2|2dx dt=s−6λ−6 ZZO2,d×(0,T ) e−2sσ2(ξ2)−6|θ3γ2|2dx dt ⩽εI2 0(γ2).(24) Using again that η1=η2outside O, we can estimate the sixth term in the right hand side of (22) for λ≥C as follows: s−3λ−3ZZO2×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt =s−3λ−3ZZ(O∩O2)×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt +s−3λ−3ZZ(O2\O)×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt ⩽s−3λ−3ZZO×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt +εs−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt. (25) 7 Plugging the estimates (23)–(25) in (22), we obtain: I1 0(γ1) + I2 0(γ2) + s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt ⩽C λZZω1×(0,T ) e−2sσ1|γ1|2dx dt +λZZω2×(0,T ) e−2sσ2|γ2|2dx dt +s−3λ−2ZZO×(0,T ) (e−2sσ1(ξ1)−3+e−2sσ2(ξ2)−3)|ψ|2dx dt!. (26) Now, we are going to estimate the first and the second term on the right hand side of (26). For the first one, let eω1be an open set satisfying ω1⊂⊂ eω1⊂⊂ O1,d ∩e O,eω1∩ O2,d =∅and let θ1∈C2 c(eω1) be such that θ1(x) = 1 for x∈ω1and 0 ⩽θ1⩽1. Using the PDE satisfied by ψ, integrating by parts in time and in space and using the PDE satisfied by γ1(see (11)), we deduce λZZω1×(0,T ) e−2sσ1|γ1|2dx dt ⩽λ α1ZZeω1×(0,T ) θ1e−2sσ1γ1(−ψt−∆ψ+aψ)dx dt =λ α1ZZeω1×(0,T )(∂t(θ1e−2sσ1)−∆(θ1e−2sσ1))γ1−2∇(θ1e−2sσ1)· ∇γ1ψ dx dt −λ α1µZZeω1×(0,T ) θ1e−2sσ1|ψ|21O1dx dt. (27) Using that |∂t(e−2sσ1)|+|∆(θ1e−2sσ1)|⩽Cs2λ2(ξ1)2e−2sσ1and |∇(θ1e−2sσ1)|⩽Csλξ1e−2sσ1, for s≥C(T+T2) and λ≥C, we obtain λZZω1×(0,T ) e−2sσ1|γ1|2dx dt ⩽Csλ2ZZeω1×(0,T ) ξ1e−2sσ1(sλξ1|γ1|+|∇γ1|)|ψ|dx dt ⩽εI1 0(γ1) + Cs4λ5ZZeω1×(0,T ) (ξ1)4e−2sσ1|ψ|2dx dt. (28) The same computations can be performed in order to estimate the second term in the right hand side of (26). Thus λZZω2×(0,T ) e−2sσ2|γ2|2dx dt ⩽εI2 0(γ2) + Cs4λ5ZZeω2×(0,T ) (ξ2)4e−2sσ2|ψ|2dx dt, (29) where eω2is an open set satisfying ω2⊂⊂ eω2⊂⊂ O2,d ∩e Oand eω2∩ O1,d =∅. In view of (28) and (29) and coming back to (26), we deduce the desired inequality (18). •Proof of (ii). In the proof of (i), in order to absorb the first and second terms on the right hand side of (26), the assumption (15) was crucial. However, it is clear that (15) does not cover all the cases where (8) holds (see, for instance, Fig. 3 in Remark 2). Thus, in order to treat this situation, we will proceed in a slightly different manner. We will prove (19) when (i0, j0) = (2,1), that is to say, in the case of Figure 3. Recall that h=α1γ1+α2γ2. Applying Proposition 9 (with m= 0 and j= 2) to the function (x, t)7→ h(x, T −t), one gets I2 0(h)⩽C λZZω2×(0,T ) e−2sσ2|h|2dx dt +s−3λ−3 2 X k=1 ZZOk×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt!.(30) 8 Summing up (20) for i= 1, (21) and (30), we obtain I1 0(γ1) + I2 0(h) + s−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt ⩽C λZZω1×(0,T ) e−2sσ1|γ1|2dx dt +s−6λ−6ZZO1,d×(0,T ) e−2sσ1(ξ1)−6|γ1|2dx dt +λZZω2×(0,T ) e−2sσ2|h|2dx dt +s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ2|2dx dt +s−3λ−3ZZO1×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt +s−3λ−3 2 X k=1 ZZOk×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt +s−3λ−2ZZO×(0,T ) e−2sσ1(ξ1)−3|ψ|2dx dt!. (31) Note that the first term in the right hand side of (31) can be estimated as in (27). For the second and fifth terms, we can proceed as in (23). Using the fact that η1=η2in Supp (θ3)∩ O2,d, we also see that the fourth term can be estimated as follows: s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ2|2dx dt ⩽s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3h|2dx dt +s−6λ−6ZZO2,d×(0,T ) e−2sσ1(ξ1)−6|θ3γ1|2dx dt (32) ⩽I2 0(h) + I1 0(γ1). For the sixth term, proceeding as in (25), one obtains s−3λ−3 2 X k=1 ZZOk×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt ⩽s−3λ−3ZZO×(0,T ) e−2sσ2(ξ2)−3|ψ|2dx dt +εs−3λ−2ZZQ e−2sσ1(ξ1)−3|ψ|2dx dt. (33) For the third term, let eω2be an open set satisfying ω2⊂⊂ eω2⊂⊂ O2,d ∩e Oand let θ2∈C2 c(eω2) be such that θ2(x) = 1 for x∈ω2and 0 ⩽θ2⩽1. Using that −ψt−∆ψ+aψ =hin eω2×(0, T ), one has λZZω2×(0,T ) e−2sσ2|h|2dx dt ⩽λZZeω2×(0,T ) θ2e−2sσ2h(−ψt−∆ψ+aψ)dx dt =λZZeω2×(0,T )(∂t(θ2e−2sσ2)−∆(θ2e−2sσ2))h−2∇(θ2e−2sσ2)· ∇hψ dx dt −λ 2 X k=1 1 µZZeω2×(0,T ) θ2e−2sσ2|ψ|21Okdx dt. (34) Using that |∂t(e−2sσ2)|+|∆(θ2e−2sσ2)|⩽Cs2λ2(ξ2)2e−2sσ2and |∇(θ2e−2sσ2)|⩽Csλξ2e−2sσ2, for s≥C(T+T2) and λ≥C, we obtain from (34) that λZZω2×(0,T ) e−2sσ2|h|2dx dt ⩽Csλ2ZZeω2×(0,T ) ξ2e−2sσ2(sλξ2|h|+|∇h|)|ψ|dx dt ⩽εI2 0(h) + Cs4λ5ZZeω2×(0,T ) (ξ2)4e−2sσ2|ψ|2dx dt. (35) 9