ESAIM: Control, Optimisation and Calculus of Variations June 2002, Vol. 8, 621{661URL: http://www.emath.fr/cocv/ DOI: 10.1051/cocv:2002047EXACT CONTROLLABILITY TO TRAJECTORIES FOR SEMILINEAR HEATEQUATIONS WITH DISCONTINUOUS DIFFUSION COEFFICIENTSAnna Doubova1, A. Osses2 and J.-P. Puel3Abstract. The results of this paper concern exact controllability to the trajectories for a coupledsystem of semilinear heat equations. We have transmission conditions on the interface and Dirichletboundary conditions at the external part of the boundary so that the system can be viewed as a singleequation with discontinuous coecients in the principal part. Exact controllability to the trajectoriesis proved when we consider distributed controls supported in the part of the domain where the diusioncoecient is the smaller and if the nonlinear term f(y) grows slower than |y| log3=2(1 + |y|) at innity.In the proof we use null controllability results for the associate linear system and global Carlemanestimates with explicit bounds or combinations of several of these estimates. In order to treat the termsappearing on the interface, we have to construct specic weight functions depending on geometry.Mathematics Subject Classication. 35B37.Received October 23, 2001. Revised February 7, 2002.1. Introduction and hypothesisLet Ω RN , N 1 be a bounded connected open set with boundary Γ of class C2. Let ! Ω be anonempty open subset and T > 0. We will use the following notation: Q = Ω (0; T ), = Γ (0; T ). For anyp 2 [1;+1], we will denote by jj jjp the usual norm in Lp(Q).There are two dierent situations that will be analyzed in this paper. More precisely, let Ω0 and Ω1 be apartition of Ω in two non empty open sets such thatCase 1: Ω0 Ω; Ω1 = ΩnΩ0 (see Fig. 1, left); (1)Case 2: Ω1 Ω; Ω0 = ΩnΩ1 (see Fig. 1, right): (2)Keywords and phrases: Carleman inequalities, controllability, transmission problems.1 Departamento E.D.A.N., Universidad de Sevilla, Tara s/n, 41012 Sevilla, Spain and Ecole Polytechnique, 91128 PalaiseauCedex, France; e-mail: dubova@numer.us.es, doubova@cmapx.polytechnique.frThis work has been partially supported by D.G.E.S., Spain, Grants PB98{1134.2 Departamento de Ingenera Matematica, Facultad de Ciencias de Fsicas y Matematicas, Universidad de Chile, Casilla 170/3 -Correo 3, Santiago, Chile and Centro de Modelamiento Matematico, UMR 2071 CNRS-Uchile; e-mail: axosses@dim.uchile.clThis work has been partially supported by FONDECYT grants No. 1000955 and 7000955.3 Laboratoire de Mathematiques Appliquees, Universite de Versailles Saint-Quentin, 45 avenue des Etats Unis, 78035 VersaillesCedex, France and Ecole Polytechnique, 91128 Palaiseau Cedex, France; e-mail: jppuel@cmapx.polytechnique.frc© EDP Sciences, SMAI 2002622 A. DOUBOVA, A. OSSES AND J.-P. PUELΩ 0Ω1S−S+Γnn S-S+Ω 0Ω1ΓnnFigure 1. Two geometrical cases covered in this paper depending on Ω0 Ω or Ω1 Ω.We denote by S = Ω0 \Ω1 the interface, which will be supposed of class C2 and by n the outward unit normalto Ω1 at the points of S and also the outward unit normal to Ω at the points of Γ. Let S+ (resp. S−) be thepart of S corresponding to the positive (resp. negative) direction of the normal n.Remark 1.1. The two cases mentioned above are not exhaustive, we do not treat other possible geometricalsituations in this paper.In both cases mentioned above, we will consider the following transmission problem for semilinear heatequation 8>>>>>>>><>>>>>>>:@ty − div(a0(x)ry) + f(y) = v1! + g0 in Ω0 (0; T );@ty − div(a1(x)ry) + f(y) = v1! + g1 in Ω1 (0; T );yjS+(0;T ) = yjS−(0;T );a0(x) @nyjS+(0;T ) = a1(x) @nyjS−(0;T ) ;y = 0; on y(x; 0) = y0 in Ω:(3)Here f : R ! R is a locally Lipschitz-continuous function, @ny denotes the outward normal derivative to Ω1,y0 2 L2(Ω) and v 2 Lr(0; T ;Lr(!)), gi 2 Lr(0; T ;Lr(Ωi)), i = 0; 1 with r such that8<:1r+N2r< 1 if N 2;r = 2 if N = 1:(4)Remark 1.2. We could in fact consider v 2 Lp(0; T ;Lq(!)), gi 2 Lp(0; T ;Lq(Ωi)), i = 0; 1 with 1=p+N=(2q)< 1 in order to have L1 solutions, but in the sake of simplicity we take p = q = r.Remark 1.3. Without loss of generality we can assume y0 2 L1(Ω). Otherwise, taking v = 0 for t 2 (0; ), > 0 and thanks to the regularizing eect of parabolic equations, y() 2 L1(Ω) for some > 0 [21, 22].In (3), y = y(x; t) is the state and v = v(x; t) is the control which acts on the system through ! since 1! isthe characteristic function of the set !.We will assume that the diusion coecient in (3) satises the following:ai 2 C2(Ωi) for i = 0; 1;a0jS+ 6= a1jS− :(5)System (3) represents the coupling between two parabolic semilinear equations whose diusion coecient hasa jump. At the interface S, we impose the continuity of the solution y and also of the fluxes.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 623Let us seta(x) =a0(x) if x 2 Ω0;a1(x) if x 2 Ω1: (6)We also setg(x) =g0(x) if x 2 Ω0;g1(x) if x 2 Ω1: (7)Taking into account notations (6) and (7), problem (3) can be written in the divergence form (with discontinuousdiusion coecients) as follows:8><>:@ty − div(a(x)ry) + f(y) = v1! + g in Q;y = 0 on ;y(x; 0) = y0 in Ω:(8)We will require a to satisfya(x) > 0 a.e. in Ω (9)and the following additional hypothesis:a0jS+ a1jS− : (10)We assume that for each > 0, there exists C > 0 such thatf(s)− f(s0)s− s02=3 C + log(1 + js− s0j) 8 s; s0 2 R: (11)Let us also consider an \ideal" trajectory y, solution of the problem (without control)8><>:@ty − div(a(x)ry) + f(y) = g in Q;y = 0 on ;y(x; 0) = y0 in Ω(12)where y0 2 L2(Ω) and g 2 Lr(0; T ;Lr(Ω)), with r as in (4). We know that under conditions (9) and (11),problem (12) possesses exactly one local solution in time (cf. [21] and [22]). Moreover, we can say that there existsa time T > 0, such that for T < T , the solution y of (12) satises y 2 C0([0; T ];L2(Ω)) \L1(; T ;L1(Ω)),for every > 0.The main goal of this paper is to analyze the controllability properties of (8).Denition 1.1. We say that (8) is exactly controllable to the trajectories if, for any trajectory y solutionof (12) and for any initial condition y0 2 L2(Ω), for every T < T , there exists a control v 2 Lr(0; T ;Lr(!))such that (8) has a solution y on (0; T ) satisfyingy(x; T ) = y(x; T ) in Ω: (13)624 A. DOUBOVA, A. OSSES AND J.-P. PUELDenition 1.2. System (8) is said null controllable at time T if, for each y0 2 L2(Ω), there exists v 2Lr(0; T ;Lr(!)) such that the corresponding initial boundary problem (8) admits a solution y 2 C0([0; T ];L2(Ω))satisfyingy(x; T ) = 0 in Ω: (14)For linear problems, it is easy to see that the notions of null controllability and exact controllability to thetrajectories are equivalent, but this is not true for nonlinear systems.Denition 1.3. It will be said that (8) is approximately controllable in L2(Ω) at time T if, for any y0 2 L2(Ω),any yd 2 L2(Ω) and any " > 0, there exists a control v 2 Lr(0; T ;Lr(!)) such that the corresponding initialboundary problem (8) possesses a solution y 2 C0([0; T ];L2(Ω)), withky(; T )− ydkL2(Ω) ": (15)In the case in which the diusion coecients are suciently regular, the controllability of linear and semilinearparabolic systems has been analyzed in several recent papers. Among them, let us mention [1, 5, 11, 13, 15{17],and [8] concerning null controllability [9,12,13,25] and [8] for approximate controllability [16] and [13] for exactcontrollability to the trajectories.2. Main result2.1. Geometric hypothesis and main resultIn order to state the main result of this work, we need the following geometrical conditions.Condition 2.1 (corresponding to case (1)). We assume that there exists a vector eld : Ω1 7! RN , 2C1(Ω1), such that(x) n(x) < 0 8x 2 Γ; (16)(x) n(x) > 0 8x 2 S; (17)(x) 6= 0 8x 2 Ω1 (18)and if we consider the characteristics associated to 8<:dx(t)dt= (x(t)); t > 0;x(0) = x0;(19)with x0 2 Γ, we also assume that for some time T1 > 0 and for every x0 2 Γ, there exists t1(x0) < T1 such thatthe solution x(t) of (19) veriesx(t) 2 Ω1 for 0 < t < t1(x0) (20)andx(t1(x0)) 2 S for x0 2 Γ. (21)Remark 2.1. Condition 2.1 implies that Γ and S are isotopic, but it is not clear whether isotopy is sucientto ensure this condition.Remark 2.2. Notice that Condition 2.1 is fullled for usual domains, see for example the cases of Figure 2.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 625(a) (b) (c) (d) (e)Figure 2. Condition 2.1 is fullled in situations (a, c, e) but not in (b) and (d). The boundaryΓ is represented by a solid line and the interface S by a dashed line, the dashed region representsΩ0 and the black dot the location of the control zone.(a)(b)(c)(d)(e)Figure 3. Condition 2.2 is fullled in situations (a{d) but not in (e) with the same notationsas in the previous gure.Condition 2.2 (corresponding to case (2)). We assume that there exist two disjoint open sets O1 ;O2 Ω1(with always a unit outward normal n) and vectors elds i : Ω1 7! RN , i 2 C1(Ω1), i = 1; 2, such thati(x) n(x) > 0 8x 2 S;i(x) n(x) > 0 8x 2 @Oi ; i = 1; 2;i(x) 6= 0 8x 2 Ω1nOi(22)and for the characteristics associated to i8<:dxi(t)dt= −i(xi(t)); t > 0;xi(0) = xi0;(23)with xi0 2 S, we assume also that for some time T i2 > 0, and for all xi0 2 S, there exists ti2(xi0) < T i2 such thatthe solution xi(t) of (23) veriesxi(t) 2 Ω1nOi for 0 < t < ti2(xi0)andxi(ti2(xi0)) 2 @Oi for xi0 2 S, i = 1; 2.Remark 2.3. Notice that this hypothesis is essentially Condition 2.1 written for the case (2). It is also fullledin usual geometrical cases, see for example the cases in Figure 3.626 A. DOUBOVA, A. OSSES AND J.-P. PUELThe aim of this paper is to prove the following theorem:Theorem 2.1. Assume that in problem (8) the coecient a satises (5, 6, 9, 10), f is a locally Lipschitz-continuous function satisfying (11) or Condition 2:1 in case (1) or Condition 2.2 in case (2) are fullled. If! \ Ωi0 6= ;, for each connected component Ωi0 of Ω0, then for each case (1) or (2, 8) is exactly controllable tothe trajectories.The idea of the proof of Theorem 2.1 is the following. With a simple change of variables we reduce theproblem of exact controllability to the trajectories for (8) to null controllability for a still nonlinear similartransmission problem. For this null controllability result we use approximate controllability to the zero statefor an associated linear transmission problem with controls in Lr(0; T ;Lr(!)) for r as in (4) and then we applya xed point method. For this we need explicit estimates on the cost of approximate controllability whichis obtained from observability inequalities (see Props. 4.1 and 4.2). These estimates are deduced from globalCarleman inequalities. In case (1), we use one single global Carleman inequality (see Th. 3.3) with a suitableweight function, whose construction is presented in Lemma 3.1. Case (2) is more complicated and we have tocombine two dierent global Carleman inequalities (see Th. 3.4) with two appropriate weight functions whoseconstruction are given in Lemma 3.2. The growth condition of the non linear term f is analyzed using thearguments of [13].The idea of combining the controllability of a linearized system and a xed point argument in the proof israther general. It was introduced in [23] in the context of the boundary controllability of the semilinear waveequation. For other controllability results proved in a similar way, see for instance [9, 13, 15] and [8].In the proofs we will suppose that Ω0 and Ω1 are connected sets and we assume the simpler hypothesis! \ Ω0 6= ;. Otherwise the weight functions for Carleman inequalities are constructed analogously on eachconnected component of Ω0 and Ω1.The paper is organized as follows. In Section 3 we deduce global Carleman inequalities, that we use forproving the main result. Section 4 is devoted to obtain some observability estimates. In Section 5, we proveTheorem 2.1. Finally, in Section 6, we give an explicit construction of suitable weight functions, needed for theglobal Carleman inequalities.2.2. Some consequences and extensions1. Observe that, the controllability result holds if the control acts in the part of the domain where the diusioncoecient is smaller. To our knowledge, this result is the rst one in the literature related to exact controllabilityto the trajectories when the diusion coecients are discontinuous.2. In the case s0 = 0 and f(0) = 0, notice that assumption (11) can be simply read as follows:limjsj!+1f(s)jsj log3=2(1 + jsj) = 0: (24)The proof of Theorem 2.1 also gives the result of null controllability for (8) with a suitable hypothesis on gunder the hypothesis and the same geometrical cases considered in Theorem 2.1 by taking condition (24) insteadof (11).3. Notice that approximate controllability for a linear transmission problem is always true and it is independentof the choice of the part of the domain where the control acts as a consequence of unique continuation property.Nonlinear problem (8) with f growing as in (11) is still approximately controllable under the conditions ofTheorem 2.1. This is due to the fact that approximate controllability in this case can be proved as a consequenceof exact controllability to the trajectories. This idea is taken from [12], where approximate controllability forsemilinear heat equations is obtained in such a way.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 6274. We can also consider in (8) the more general case in which the diusion coecients are represented by a realsymmetric uniformly elliptic matrix A, i.e. there exists a constant > 0 such thatA(x; ; ) =NXi;j=1Aijij jj2 8 2 RN ; for a.e. x 2 Ω (25)and A is regular in each Ωi, i = 0; 1. In this case, condition (10) has to be replaced bydetAAn nS 0; (26)where [ ]S denotes the jump across S.Until now, null controllability for semilinear parabolic systems (in the divergence form) has been analyzedwhen the diusion coecients are suciently regular. More precisely, when A = (Aij), i; j = 1; : : : ; N withAij 2 C1;2(Q) (see [15]).2.3. Open problems related to Theorem 2.11. If ! Ω1 we do not know whether or not system (8) is exactly controllable to the trajectories. Is in thiscase null controllability also an open problem.2. In [13], it is proved that even in the case of regular diusion coecients, for each > 2, there exist functionsf = f(s) with f(0) = 0 andlimjsj!1jf(s)jjsj log (1 + jsj) = with > 0, (27)such that the corresponding system (for the semilinear heat equation) is not null-controllable for any T > 0. Inview of point 2 in Section 2.2, we see that, when f satises (27) with 3=2 2, null-controllability of (8) isan open question.3. On the other hand, it is proved in [13], that also in the case of regular diusion coecients, for each > 2,there exist functions f satisfying (27) such that the corresponding system (for the semilinear heat equation) isnot approximately controllable for all T > 0. Then, approximate controllability for the transmission problem (8)with 3=2 2, is also an open question.4. An abstract result due to Russell [20] shows that boundary exact controllability for the wave equation impliesboundary exact null controllability for the heat equation with the same type of control and geometry. Thisresult is proved in the case of smooth coecients. If we consider this principle still true in the case of nonsmooth coecients, the geometrical hypothesis that we consider here seems to be too restrictive in the caseN = 1 (cf. [7] for the controllability of the corresponding wave equation) but not for N 2.5. In [14], it is proved null controllability result for the one-dimensional linear heat equation like (x) −(a(x)zx)x +m(x)z = 0 with only BV coecients without any assumption on the control zone. However, theproof is denitely strictly one dimensional relying on the corresponding one for the wave equation and nullcontrollability result is true if the potential m depends only on space variable, but not on time. Then, even inthe one-dimensional case it is not clear how to treat with this method a similar nonlinear problem.628 A. DOUBOVA, A. OSSES AND J.-P. PUEL3. Global Carleman inequalitiesIn this section we will deduce two global Carleman inequalities, that we need for the proof of Theorem 2.1.For this purpose, we will introduce suitable weight functions. Let us rst consider the situation of case (1) (seeFig. 1, left). The rst weight function is given by the following result:Lemma 3.1. Assume that we have the geometrical situation of case (1) (see Fig. 1). Assume that the functiona dened in (5, 6) satises (9, 10) and that Condition 2.1 holds. If ! \ Ω0 6= ; then for every open set!0 ! \ Ω0 there exists a function e 2 C0(Ω), ei = ejΩi 2 C2(Ωi), i = 0; 1, e > 0 in Ω, such thate = 0 on Γ; (28)@n e < 0 on Γ; (29)e = 1 on S; (30)@ne0 > 0; @n e1 > 0 on S; (31)a0@n e0 = a1@n e1 on S (32)andjrej > 0 in Ωn!0 : (33)The proof of Lemma 3.1 will be given in Section 6.Now, we consider the geometrical case (2) (see Fig. 1, right). For the second Carleman inequality, which wewill use to treat the situation 2, we need two suitable weight functions.We have the following result:Lemma 3.2. Assume that we have the geometrical situation of case (2) (see Fig. 1). Assume that the functiona dened in (5, 6) satises (9, 10) and that there exist two open disjoint sets O1 ;O2 Ω1 verifying Condi-tion 2.2. Let Bi and eBi, i = 1; 2 be balls such that B1 eB1 O1 and B2 eB2 O2. If ! \Ω0 6= ; thenfor every open set !0 ! \ Ω0 there exist two functions e1 and e2 such thate1(x) = ( e0(x) if x 2 Ω0;e11(x) if x 2 Ω1; e2(x) =( e0(x) if x 2 Ω0;e21(x) if x 2 Ω1; (34)with the following properties: e0 2 C2(Ω0), e0 > 0 in Ω0,e0 = 0 on Γ; @n e0 < 0 on Γ; (35)@ne0 > 0 on S; e0 = 2 on S; (36)jre0j > 0 in Ω0n!0: (37)And for i = 1; 2, ei1 2 C2(Ω1), ei1 > 0 in Ω1,ei1 = e0 = 2 on S; (38)a0@ne0 = a1@n ei1 on S; i = 1; 2; (39)e11 2e21 in eB2; (40)e21 2e11 in eB1; (41)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 629andjrei1j > 0 in Ω1nBi ; i = 1; 2: (42)The proof of Lemma 3.2 will also be given in Section 6.Remark 3.1. Notice that in geometrical case (2) (Ω1 Ω) it is impossible to have a function e which isconstant on S and such that re 6= 0 in Ω1.Let us consider the functions = e +K; = 54maxΩ; (43)with K > 0 such that K 5 maxΩe; and e is given by Lemma 3.1.Let be a suciently large positive constant that only depends on Ω and !. It will be xed later on. Fort 2 (0; T ) and following [16] and [12], we introduce the following functions:’(x; t) =e(x)t(T − t) ; (x; t) =e − et(T − t) (44)Notice thatr = −’r; r’ = ’r: (45)Let us setZ0 = fq : q 2 C2(Ωi [0; T ]); i = 0; 1; qjS+(0;T ) = qjS−(0;T ) ;a0@nqjS+(0;T ) = a1@nqjS−(0;T ) ; q = 0 on gWe have the following Carleman estimate:Theorem 3.3. Assume that !\Ω0 6= ;, a satises (5, 6, 9) and (10) and Condition 2.1 in case (1) is fullled.There exists 1(Ω; !; a) > 0 such that for each > 1 there exists a positive constant C that only depends on Ω,! and a, and s1() > 0 so that the following estimate holdss3ZZQe−2st−3(T − t)−3jqj2 dxdt+ sZZQe−2st−1(T − t)−1jrqj2 dxdt C s3ZZ!(0;T )e−2st−3(T − t)−3jqj2 dxdt+ZZQe−2sj@tq + div(a(x)rq)j2 dxdt (46)for all q 2 Z0 and s s1. Moreover, s1 is of the form s1 = 1(Ω; !; a; )(T 2 + T ), where 1 is a positiveconstant that only depends on Ω, !, a and .Proof of the Theorem 3.3. In the sequel, C will stand for a generic positive constant only depending on Ω, !and a, whose value can change from line to line. We will also use the usual convention of repeated indices.Let us assume q 2 Z0 and s > 0. We setf = @tq + div(a(x)rq)630 A. DOUBOVA, A. OSSES AND J.-P. PUELand = e−sq: (47)Notice that (0) = (T ) = 0: (48)We have the following equality:e−s(@t(es ) + div(a(x)r(es )) = e−sf: (49)Using (45), we can write (49) in the formM1 +M2 = e−sf + s’div(a(x)r) − s2’a(x)jrj2 ; (50)whereM1 = div(a(x)r ) + s22’2jrj2a(x) + s@t (51)andM2 = @t − 2s’a(x)rr − 2s2’a(x)jrj2 : (52)Let setfs = e−sf + s’div(a(x)r) − s2’a(x)jrj2 : (53)From (50), we obtainkM1 k22 + kM2 k22 + 2(M1 ;M2 ) = kfsk22 ; (54)where (; ) denotes the scalar product in L2(Q). Let us compute the scalar product in the left hand side of (54).We can write(M1 ;M2 ) = I110 + I120 + I130 + I210 + I220 + I230 + I310 + I320 + I330 : (55)In (55), all the integrals denote the respective scalar products for the terms of M1 and M2 . For simplicity,in the sequel, we will write a instead of a(x). We haveI110 =ZZQdiv(ar )@t dxdt = −ZZQar @t(r ) dxdt+Z T0ZSa1r n @t dxdt−Z T0ZSa0r n @t dxdt = −12ZZQa@t(jr j2) dxdt = 0:(56)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 631Here we have used (48) which says that (0) = (T ) = 0.I120 = −2s1Xl=0Z T0ZΩl’div(alr )alrr dxdt = 2s1Xl=0Z T0ZΩlal@xi @xi(’al@xj@xj ) dxdt−2sZ T0ZS−’a21(r1 r )(r n) d dt+ 2sZ T0ZS+’a20(r0 r )(r n) d dt−2sZ T0ZΓ’a21(r r )(r n) d dt:(57)Let us consider the rst term of (57). Also for simplicity, we will make the computation only for the integralsin Ω1. We setI1120 = 2sZ T0ZΩ1a1@xi @xi(’a1@xj@xj ) dxdt:We will use (45) andr = (r n)n+r; (58)r = (r n)n+r ; (59)where r and r denote the tangential gradients. Thanks to the choice of we know from (30) and (43)that is a constant on S, then r = 0 on S and we can writeI1120 = 2s2Z T0ZΩ1’(a1@xi @xi)(a1@xj@xj ) dxdt+ 2sZ T0ZΩ1’a1@xi @xi(a1@xj)@xj dxdt+sZ T0ZΩ1’a1(a1@xj)@xj (jr j2) dxdt:(60)Integrating now by parts in the third term of (60) we obtainI1120 = 2s2Z T0ZΩ1’(a1@xi @xi)(a1@xj@xj ) dxdt+ 2sZ T0ZΩ1’a1@xi @xi(a1@xj)@xj dxdt−s2Z T0ZΩ1’ja1rj2jr j2 dxdt− sZ T0ZΩ1’a1@xj (a1@xj)jr j2 dxdt−sZ T0ZΩ1’(@xja1)(a1@xj)jr j2 dxdt+ sZ T0ZS−’a21(@n1)jr j2 d dt+sZ T0ZΓ’a21(@n)jr j2 d dt:(61)For the integrals in Ω0 it is sucient to take into account that n is the outward unit normal to Ω1 and replacein (61), n by −n, S− by S+ and Ω1 by Ω0.632 A. DOUBOVA, A. OSSES AND J.-P. PUELConsequently, from (57) and (61), using again (59) we deduceI120 = −s2ZZQ’jarj2jr j2 dxdt+ 2s2ZZQ’(ar r)2 dxdt− sZ T0ZS−’(@n1)ja1@n j2 d dt+ sZ T0ZS+’(@n0)ja0@n j2 d dt+ sZ T0ZS−’a1(a1@n1)jr j2 d dt− sZ T0ZS+’a0(a0@n0)jr j2 d dt− sZ T0ZΓ’a21(@n)j@n j2 d dt+X1(62)withX1 = 2sZZQ’a@xi @xi(a@xj)@xj dxdt− sZZQ’a@xj (a@xj)jr j2 dxdt−sZZQ’(@xja)(a@xj)jr j2 dxdt:(63)Finally, we getI120 = −s2ZZQ’jarj2jr j2 dxdt+ 2s2ZZQ’(ar r)2 dxdt+ sZ T0ZS’ja@n j2 [@n]S d dt− sZ T0ZS’jr j2(a@n)[a]S d dt− sZ T0ZΓ’ja@n j2(@n) d dt+X1;(64)where X1 is given by (63), and [ ]S denote the jump on S. Notice that in (64), due to the choice of e theboundary integrals are nonnegative and this is essential. In fact, from (9) and (10) we have[a]S = a0 − a1 0 on S: (65)On the other hand, from (10, 31, 32) and (43), we deduce that[@n]S = @n0 − @n1 0 on S; (66)since n is the outward unit normal to Ω1. Moreover, thanks to (29) and (43), we have @n 0 on Γ. Thisjusties the above statement.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 633Let us compute the scalar product of the rst term of M1 and the third one of M2 .I130 = −2s21Xl=0Z T0ZΩldiv(alr )’aljrj2 dxdt = 2s2ZZQar r(’ajrj2 ) dxdt−2s2Z T0ZS−’a1(@n )a1jr1j2 d dt+ 2s2Z T0ZS+’a0(@n )a0jr0j2 d dt= 2s2ZZQ’jarj2jr j2 dxdt+ 2s21Xl=0Z T0ZΩl’alr(aljrj2)r dxdt+2s3ZZQ’a2rjrj2r dxdt−2s2Z T0ZS−’a1(@n )a1j@n1j2 d dt+ 2s2Z T0ZS+’a0(@n )a0j@n0j2 d dt:(67)In (67), again we have used (58) and the fact that r = 0 on S. Finally, we haveI130 = 2s2ZZQ’jarj2jr j2 dxdt+X2; (68)whereX2 = 2s2ZZQ’ar(ajrj2)r dxdt+ 2s3ZZQ’a2rjrj2r dxdt+2s2Z T0ZS’(a@n )(a@n) [@n]S d dt:(69)The scalar product of the second term of M1 with the rst one of M2 givesI210 = s22ZZQ’2jrj2a@t dxdt = −s22ZZQ’@t’jrj2aj j2 dxdt: (70)We now consider the scalar product between the second term ofM1 with the second one of M2 . The followingholds:I220 = −2s331Xl=0Z T0ZΩl’3a2l jrj2(r)r dxdt= 3s34ZZQ’3a2jrj4j j2 dxdt+ s331Xl=0Z T0ZΩl’3div(a2l jrj2r)j j2 dxdt−s33Z T0ZS−’3ja1@n1j2(@n1)j j2 d dt+ s33Z T0ZS+’3ja0@n0j2(@n0)j j2 d dt= 3s34ZZQ’3a2jrj4j j2 dxdt+ s33Z T0ZS’3ja@nj2 [@n]S j j2 d dt+X3;(71)634 A. DOUBOVA, A. OSSES AND J.-P. PUELwhereX3 = s331Xl=0Z T0ZΩl’3div(a2l jrj2r)j j2 dxdt: (72)Observe again that the last boundary integral in (71) is nonnegative because (66) holds.Now we consider the third and the second terms of M1 and M2 , respectively. Using (45) we can writeI320 = −2s21Xl=0Z T0ZΩl’@talr r dxdt = s2ZZQ’@t div(ar)j j2 dxdt+s22ZZQ’(@t)ajrj2j j2 dxdt− s22ZZQ’(@t’)ajrj2j j2 dxdt−s2Z T0ZS−’@t(a1@n1)j j2 d dt+ s2Z T0ZS+’@t(a0@n0)j j2 d:(73)Since (32) holds we know that a0@n0jS+ − a1@n1jS− = 0, then we getI320 = s2ZZQ’@t div(ar)j j2 dxdt+ s22ZZQ’(@t)ajrj2j j2 dxdt− s22ZZQ’(@t’)ajrj2j j2 dxdt:(74)The last integrals giveI230 = −2s34ZZQ’3a2jrj4j j2 dxdt; (75)I310 = sZZQ@t @t dxdt = −12sZZQ@2t j j2 dxdt (76)andI330 = −2s22ZZQ’a@tjrj2j j2 dxdt: (77)Finally from (55), taking into account (56, 64, 67, 70, 71, 74{76) and (77) we deduce that(M1 ;M2 ) = s2ZZQ’jarj2jr j2 dxdt+ 2s2ZZQ’(ar r)2 dxdt+s34ZZQ’3a2jrj4j j2 dxdt+ s33Z T0ZS’3ja@nj2 [@n]S j j2 d dt+sZ T0ZS’ja@n j2 [@n]S d dt− sZ T0ZS’jr j2(a@n)[a]S d dt−sZ T0ZΓ’ja@n j2(@n) d dt+X1 +X2 + I210 +X3 + I310 + I320 + I330 ;(78)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 635where X1, X2, I210 , X3, I310 , I320 , and I330 , are given by (63, 69, 70, 72, 74, 76) and (77) respectively. Thus, wehave from (54) the following identity:kM1 k22 + kM2 k22 + 2s2ZZQ’jarj2jr j2 dxdt+ 4s2ZZQ’(ar r)2 dxdt+2s34ZZQ’3a2jrj4j j2 dxdt+ 2s33Z T0ZS’3ja@nj2 [@n]S j j2 d dt+2sZ T0ZS’ja@n j2 [@n]S d dt− 2sZ T0ZS’jr j2(a@n)[a]S d dt−2sZ T0ZΓ’ja@n j2(@n) d dt = kfsk22 − 2(X1 +X2 + I210 +X3 + I310 + I320 + I330):(79)As we mentioned above, since @n 0 on Γ and thanks to (66) and (65), all the boundary integrals in the lefthand side of (79) are nonnegative. Moreover, we know that (9) and (33) hold. Then, for some 0(Ω; !; a) 1we have2’jarj2 C(Ω; !; a)2’ in (Ωn!0) (0; T ) 8 0(Ω; !; a);4’3a2jrj4 C(Ω; !; a)4’3 in (Ωn!0) (0; T ) 8 0(Ω; !; a):(80)On the other hand, we use thatja@xj(a@xj)j C(Ω; !; a); (81)ajrj2 C(Ω; !; a); (82)jr(aljrj2)j C(Ω; !; a); l = 0; 1; (83)div(a2l jrj2r) C(Ω; !; a); l = 0; 1; (84)div(ar) C(Ω; !; a): (85)Then, it is not dicult to check that for every " > 0, there exists C" > 0 such thatjX1j C(Ω; !; a)sZZQ’jr j2 dxdt; (86)jX2j C(Ω; !; a)sZZQ’jr j2 dxdt+ s3T 4ZZQ’3j j2 dxdt(87)+ "s2ZZQ’jr j2 dxdt+ C"s4T 4ZZQ’3j j2 dxdt+ "sZ T0ZS’ja@n j2 [@n]S d dt+ C"s3T 4Z T0ZS’3ja@nj2 [@n]S j j2 d dt;jX3j C(Ω; !; a)s33ZZQ’3j j2 dxdt: (88)In the estimate (87), we have used Young inequalities and the fact that’ C’3t2(T − t)2 CT 4’3636 A. DOUBOVA, A. OSSES AND J.-P. PUELand we will chose " > 0 suciently small. Moreover, we havej@t’j = jT − 2tjt2(T − t)2 e C(Ω; !)T’2; (89)j@tj = jT − 2tj(e − e)t2(T − t)2 C(Ω; !)T et2(T − t)2 C(Ω; !)T e2t2(T − t)2 C(Ω; !)T’2; (90)j@2ttj =2jT 2 − 3T t+ 3t2j(e − e)t3(T − t)3 14T 2et3(T − t)3 C(Ω; !)T 2e2t3(T − t)3 C(Ω; !)T2’3: (91)In (90) and (91), we have used that e e2 . This is implied by the fact that =54maxΩ 2 minΩ; (92)which is a consequence of the choice of K in (43). Taking into account, equations (82, 85, 89, 90) and (91) wededucejI210 j C(Ω; !; a)Ts22ZZQ’3j j2 dxdt; (93)jI310 j C(Ω; !; a)T 2sZZQ’3j j2 dxdt; (94)jI320 j C(Ω; !; a)Ts22ZZQ’3j j2 dxdt; (95)jI330 j C(Ω; !; a)Ts22ZZQ’3j j2 dxdt: (96)On the other hand, from (53) we can write thatkfsk22 ke−sfk22 + C(Ω; !; a)s24ZZQ’2j j2 ke−sfk22 + C(Ω; !; a)s24T 2ZZQ’3j j2: (97)Using (80, 86{88, 93{96) and (97) in (79), we obtain:kM1 k22 + kM2 k22 +Cs34Z T0ZΩn!0’3j j2 dxdt+ Cs2Z T0ZΩn!0’jr j2 dxdt+2s33Z T0ZS’3ja@nj2 [@n]S j j2 d dt+ 2sZ T0ZS’ja@n j2 [@n]S d dt ke−sfk22 + Cs3T 4Z T0ZS’3ja@nj2 [@n]S j j2 d dt+ "sZ T0ZS’ja@n j2 [@n]S d dt+CsZZQ’jr j2 dxdt+ "s2ZZQ’jr j2 dxdt+ Cs33ZZQ’3j j2 dxdt+Cs2(4T 2 + 2T )ZZQ’3j j2 dxdt+ Cs(T 2 + 4T 4)ZZQ’3j j2 dxdt:CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 637From this, for 1(Ω; !; a) 0(Ω; !; a), with 1 not depending on T , and for " small enough, we can writekM1 k22 + kM2 k22 + s34ZZQ’3j j2 dxdt+ s2ZZQ’jr j2 dxdt+2s33Z T0ZS’3ja@nj2 [@n]S j j2 d dt Cke−sfk22+s34Z T0Z!0’3j j2 dxdt+ s2Z T0Z!0’jr j2 dxdt+ s(T 2 + 4T 4)ZZQ’3j j2 dxdt+s2(4T 2 + 2T )ZZQ’3j j2 dxdt+ s3T 4Z T0ZS’3ja@nj2 [@n]S j j2 d dt#:We take now s 0(Ω; !; a; )(T 2 + T ), then we also havekM1 k22 + kM2 k22 + s34ZZQ’3j j2 dxdt+ s2ZZQ’jr j2 dxdt C"ke−sfk22 + s34Z T0Z!0’3j j2 dxdt+ s2Z T0Z!0’jr j2 dxdt#: (98)Let us deduce from (98) that (46) holds for all s s1 where s1 = 1(Ω; !; a; )(T 2 + T ).Recall that = e−sq. Then,@xi = e−s(@xiq − s@xiq) = e−s(@xiq + s’@xi q):So we can write thate−s@xiq = @xi − se−s’@xi q:Consequently, we nd the following:s2ZZQe−2s’jrqj2 dxdt = s2ZZQ’jr − e−ss’rqj2 dxdt Cs2ZZQ’jr j2 dxdt+ C(Ω; !)s34ZZQe−2s’3jqj2 dxdt:Then, from (98) we haves34ZZQe−2s’3jqj2 dxdt+ s2ZZQe−2s’jrqj2 dxdt C(Ω; !; a)"ke−sfk22 + s34Z T0Z!e−2s’3jqj2 dxdt+ s2Z T0Z!0e−2s’jrqj2 dxdt#:(99)In order to conclude the proof of the Carleman inequality (46) it is sucient to prove thats2Z T0Z!0e−2s’jrqj2 dxdt C"ke−sfk22 + s34Z T0Z!e−2s’3jqj2 dxdt+ s2(4T 2 + 2T )Z T0Z!e−2s’3jqj2 dxdt+ s(3T 4 + 3T 2 + 2T 3)Z T0Z!e−2s’3jqj2 dxdt#:(100)638 A. DOUBOVA, A. OSSES AND J.-P. PUELIn fact, combining (99) and (100), we deduce that the global Carleman estimate (46) is true for s 1(Ω; !; a; )(T 2 + T 3=2 + T ). Notice that it is possible to drop the term in T 3=2 since T 3=2 1=2(T 2 + T ).In order to obtain (100), we consider a function 2 C10 (!) such that 1 in !0 and 0. We consider! Ω0 and the estimates obtained below remain true for larger !. Multiplying by se−2s’q the equation@tq + div(arq) = fand integrating in ! (0; T ), we obtains22Z T0Z!e−2s’@tjqj2 + s2Z T0Z!e−2s’div(arq)q dxdt = s2Z T0Z!e−2s’fq dxdt: (101)In (101), the second term, can be written after integration by parts as follows:s2Z T0Z!e−2s’@xi(a@xiq)q dxdt = −s2Z T0Z!e−2s’ajrqj2 dxdt− s22Z T0Z!@xi(e−2s’)a @xi jqj2 dxdt:Then, from (101) we deduces2Z T0Z!0e−2s’jrqj2 dxdt C(a)s22Z T0Z!e−2s’@tjqj2 dxdt+ C(a)s2Z T0Z!e−2s’fq dxdt+C(a)s2Z T0Z!e−2s@2xixi(e−2s’)jqj2 dxdt : (102)Let us consider the rst term of the right hand side of (102). We haveX4 =s22Z T0Z!e−2s’@tjqj2 dxdt = −s22Z T0Z!@t(e−2s’)jqj2 dxdt= s22Z T0Z!e−2s’@tjqj2 dxdt− s22Z T0Z!e−2s@t’jqj2 dxdt:(103)Using now (89) and (90) in (103) we obtainjX4j Cs22TZ T0Z!e−2s’3jqj2 dxdt+ Cs2T 3Z T0Z!e−2s’3jqj2 dxdt: (104)On the other hand, for the third term of the right hand side of (102), we can writes2Z T0Z!@2xixi(e−2s’)jqj2 dxdt Cs34Z T0Z!e−2s’3jqj2 dxdt+C(s24T 2 + s3T 4 + s3T 2)Z T0Z!e−2s’3jqj2 dxdt: (105)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 639This is a consequence of the particular form of and ’. Indeed, after the following calculation@xi(e−2s’) = e−2s [@xi’+ @xi’− 2s’@xi] ;@2xixj (e−2s’) = e−2s−2s’@xj@xi− 2s@xj@xi’+ 4s2’@xj@xi+ ’@2xixj+ @xi@xj’+ @xj@xi’+ @2xixj’− 2s’@xj@xi − 2s@xj’@xi − 2s’@2xixjiand using the following straightforward estimatesj’@xjj ’2jrj C’2 CT 2’3;j@xj@xi’j 2’2jrj2 C2’2 CT 22’3;j’@xj@xij 2’3jrj2 C2’3;j@xj’j ’jrj C’ CT 4’3j@2xixj’j ’jj + 2’2jrj2 C(’+ 2’2) C(T 4+ T 22)’3j’@2xixjj ’2jj+ 2’3jrj2 CT 2’3 + C2’3it is not dicult to see that (105) holds. On the other hand we haves2Z T0Z!e−2s’fq dxdt Cke−sfk22 + Cs24Z T0Z!e−2s’22jqj2 dxdt Cke−sfk22 + Cs24T 2Z T0Z!e−2s’3jqj2 dxdt:(106)Using (104, 105) and (106) in (102) we get (100). As we mentioned above, this ends the proof of Carlemaninequality (46) of Theorem 3.3.The situation of Case 2 is quite dierent. Let us consider the functionsi = ei +Ki; i = 54maxΩi; for i = 1; 2; (107)with Ki > 0 such that Ki 5 maxΩei ; and ei is given by the Lemma 3.2. We also introduce the followingweight functions:’i(x; t) =ei(x)t(T − t) ; i(x; t) =ei − eit(T − t) ; i = 1; 2: (108)Our second Carleman estimate is the following:Theorem 3.4. Assume that !\Ω0 6= ;, a satises (5, 6, 9) and (10) and Condition 2.2 in case (2) is fullled.There exists 2(Ω; !; a) > 0 so that for each > 2 there exists a positive constant C that only depends on Ω,640 A. DOUBOVA, A. OSSES AND J.-P. PUEL!, O1, O2 and a, and s6() > 0 so that the following estimate holdss3ZZQ(e−2s1+ e−2s2)t−3(T − t)−3jqj2 dxdt+ sZZQ(e−2s1+ e−2s2)t−1(T − t)−1jrqj2 dxdt Cs3ZZ!(0;T )(e−2s1+ e−2s2)t−3(T − t)−3jqj2 dxdt+ CZZQ(e−2s1+ e−2s2)j@tq + div(a(x)rq)j2 dxdt(109)for all q 2 Z0 and s s6. Moreover, s6 is of the form s6 = 6(Ω; !;O1 ;O2 ; a; )(T 2 + T ), where 6 is apositive constant that only depends on Ω, !, O1, O2, a and .Proof of Theorem 3.4. In order to obtain (109), we will apply the global Carleman inequality (46) fromTheorem 3.3 and the properties of weight functions (108).We observe, that from Lemma 3.2, we know that ri can vanish only in !0 and Bi for i = 1; 2, where theopen subsets B1 and B2 are xed balls dened in Lemma 3.2. Taking into account these statements, we canuse two weight functions given by (108) and write two Carleman estimates like (46). More precisely, there exista positive constant C and s1 that only depends on Ω ! and a, such thats3ZZQe−2sit−3(T − t)−3jqj2 dxdt+ sZZQe−2sit−1(T − t)−1jrqj2 dxdt C s3ZZ!(0;T )e−2sit−3(T − t)−3jqj2 dxdt+ s3ZZeBi(0;T )e−2sit−3(T − t)−3jqj2 dxdt+ZZQe−2si j@tq + div(a(x)rq)j2 dxdt (110)for i = 1; 2, for all q 2 Z0 and s s1. Moreover, s1 is of the form s1 = 1(Ω; !; a; )(T 2 + T ).Let us show that from (110), using the properties of the functions 1 and 2, we can deduce the Carlemanestimate (109). For this, it will be sucient to see that for each C > 0, there exists s4 such thate−2s2 2Ce−2s1 in eB1 ; (111)e−2s1 2Ce−2s2 in eB2 (112)for s s4 = 4(Ω; !;O1 ;O2; )T 2.Indeed, adding the two Carleman inequality (110), we deduces3ZZQ(e−2s1+ e−2s2)t−3(T − t)−3jqj2 dxdt+ sZZQ(e−2s1+ e−2s2)t−1(T − t)−1jrqj2 dxdt Cs3ZZ!(0;T )(e−2s1+ e−2s2)t−3(T − t)−3jqj2 dxdt+ C s3ZZeB1(0;T )e−2s1t−3(T − t)−3jqj2 dxdt+ s3ZZeB2(0;T )e−2s2t−3(T − t)−3jqj2 dxdt!+ CZZQ(e−2s1+ e−2s2)j@tq + div(a(x)rq)j2 dxdt(113)for s s5 = max(s1 ; s4).CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 641On the other hand, according to (111) and (112), we obtain thatCs3ZZeB1(0;T )e−2s1t−3(T − t)−3jqj2 dxdt+ Cs3ZZeB2(0;T )e−2s2t−3(T − t)−3jqj2 dxdt s32ZZeB1(0;T )e−2s2t−3(T − t)−3jqj2 dxdt+ s32ZZeB2(0;T )e−2s1t−3(T − t)−3jqj2 dxdt s32ZZQ(e−2s1+ e−2s2)t−3(T − t)−3jqj2 dxdt(114)for s s6 = 6(Ω; !;O1 ;O2 ; )T 2. Combining (113) and (114), we nd (109).To conclude the proof, let us justify (111) and (112). By construction, we have that21 211 in eB1 : (115)Using (115), we can deduce that for all 1 there exists a positive constant , which only depends on Ω, !,O1, O2 such that1 − 2 1 in eB1 : (116)Indeed, equation (116) is a consequence of the following:1 − 2 = e21 − e11t(T − t) e211 − e11t(T − t) e1 − e11t(T − t) = 1 in eB1 :Then, from (116) we obtain that, for each C > 0, there exists s2 such thate−2s2e−2s1 e2s1 e2smin 1 2C in eB1for s s2 = 2(Ω; !;O1 ;O2)T 2. This is exactly the inequality (111) for s s2.By similar arguments, using the fact that11 221 in O2 ; (117)it is easy to see that, for each C > 0, there exists s3 such thate−2s1e−2s2 2C in eB2for s s3 = 3(Ω; !;O1 ;O2)T 2. Then, equation (112) holds for s s3. Taking now s4 = max(s2; s3) weget (111) and (112) for s s4. This ends the proof of Theorem 3.4.4. Observability inequalities and technical resultsIn this section we will deduce some observability estimates as a consequence of appropriate global Carlemaninequalities and regularizing eect of the heat equation. This will be needed to prove the null controllabilityresult for a linear transmission problem with controls in Lr(0; T ;Lr(!)) with r suciently large, such that (4)holds.642 A. DOUBOVA, A. OSSES AND J.-P. PUELLet us consider the following linear (adjoint) transmission problem:8><>:−@tq − div(a(x)rq) + bq = 0 in Q;q = 0 on ;q(x; T ) = qT in Ω;(118)where a satises (5, 6, 9) and (10), b 2 L1(Q), and qT 2 L2(Ω). First of all, let us prove the observabilityestimate with L2(0; T ;L2(!))-norm in the right hand side. This can be used to deduce null controllability result(and estimates) for linear transmition problem with bounded potential, with controls in L2(0; T ;L2(!)). Wehave the following:Proposition 4.1. Assume that ! \ Ω0 6= ; and that Condition 2.1 (resp. Condition 2.2) in case (1) (resp. incase (2)) are fullled. Then for any a satisfying (5, 6, 9, 10), b 2 L1(Q) and qT 2 L2(Ω), there exists a positiveconstant C that only depends on Ω, ! and a (resp. Ω, ! O1, O2 and a), such thatjjq(; 0)jj2L2(Ω) expC1 +1T+ T jjbjj1 + jjbjj2=31ZZ!(0;T )jqj2 dxdt; (119)where q is the solution to the corresponding system (118).For simplicity, we only present the proof of Proposition 4.1 for the situation (1). We just note, that the proofcorresponding to the situation (2) is similar, it suces to take into account the dierent estimates (in space) forthe weight functions that appear in the global Carleman inequality (109). Thus we also obtain the constantsdepending on O1 and O2.Proof of Proposition 4.1. We will use global Carleman inequality (46) from Theorem 3.3 and some estimatesfor the weight functions. Let b and qT be given and let q be the solution to (118).Step 1: We will rst see thatZZΩ(T=4;3T=4)jqj2 dxdt expC1 +1T+ jjbjj2=31ZZ!(0;T )jqj2 dxdt: (120)By density, we can write (46) for q being the solution of (118). This givess3ZZQe−2st−3(T − t)−3jqj2 dxdt C s3ZZ!(0;T )e−2st−3(T − t)−3jqj2 dxdt+ZZQe−2sjbqj2 dxdt(121)for all s s1. We can estimate the second term in the right as follows:ZZQe−2sjbqj2 dxdt 2−6T 6jjbjj21ZZQe−2st−3(T − t)−3jqj2 dxdt:Thus, we deduce from (121) thatZZQe−2st−3(T − t)−3jqj2 dxdt CZZ!(0;T )e−2st−3(T − t)−3jqj2 dxdt (122)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 643provideds s7 = maxs1; CT2jjbjj2=31: (123)On the other hand, it can be easily veried thate−2st−3(t− T )−3 26T−6 exp (−CsT−2 8(x; t) 2 Q (124)ande−2st−3(t− T )−3 1633T−6 exp(−CsT−2 8(x; t) 2 Ω [T=4; 3T=4]; (125)whenevers s8 = max (s7; CT 2)(constants C in (124) and (125) may be dierent). If we analyze the structure of the constants s7 and s8, wesee that s8 s9, where s9 is of the forms9 = 9(Ω; !; a)T + T 2 + T 2jjbjj2=31: (126)Let us x the constant s = s9. We write (122) for s = s9 taking into account (124) and (125) and we deducethat (120) is satised for any solution q of (118).Step 2: Let us now prove thatjjq(x; 0)jj22 expC1T+ T jjbjj1ZZΩ(T=4;3T=4)jqj2 dxdt: (127)The estimate (127) together with (120) leads to the desired observability inequality (119).Let 0 2 C1[0; 1] be a function such that, 0 0 1, 0 = 1 in [0; 1=4], 0 = 0 in [3=4; 1]. Now, we considera function (t) = 0(t=T ) and we write (118) for (t)q. We obtain8><>:−@t(q)− div(a(x)r(q)) + b(q) = −q@t in Ω (0; 3T=4);q = 0 on @Ω (0; 3T=4);q(x; 3T=4) = 0 in Ω:(128)Multiplying (128) by q and integrating in Ω, we have−12ddtZΩjqj2 dx+ZΩajr(q)j2 dx = −ZΩbjqj2 dx−ZΩ(@t)jqj2 dx 8 t 0:Thus,− ddtZΩjqj2 dx+ 2ZΩajr(q)j2 dx 2jjbjj1ZΩjqj2 dx+ 2ZΩj@tjjqj2 dxand− ddtexp (2jjbjj1t)ZΩjqj2 dx 2 exp (2jjbjj1t)ZΩj@tjjqj2 dx (129)644 A. DOUBOVA, A. OSSES AND J.-P. PUELfor all t 0. Integrating this inequality with respect to time in [0; t] with t 2 [3T=4; T ], we obtainZΩjq(x; 0)j2 dx Z t0exp (2jjbjj1t) 2ZΩjqj2@t dx 2CTexp3T2jjbjj1ZZΩ(T=4 ;3T=4)jqj2 dxdt:(130)In (130), we have used the fact that 1 and j@tj = j@t0(t=T )j=T C=T . This justies the estimate (127)and ends the proof of Proposition 4.1.As we mentioned above, to analyze the controllability for nonlinear problem (8) we need the controls inLr(0; T ;Lr(!)) for r suciently large (1=r + N=(2r) < 1). For this we are going to prove a rened version ofthe observability inequality (119), i.e. with Lr0(0; T ;Lr0(!))-norm in the right hand side, where r0 is the dualexponent to r. We have:Proposition 4.2. Assume that ! \ Ω0 6= ; and that Condition 2.1 (resp. Condition 2.2) in case (1) (resp. incase (2)) are fullled. Then for any a satisfying (5, 6, 9) and (10), b 2 L1(Q), qT 2 L2(Ω) and any r0suciently small, there exist a positive constant C that only depends on Ω, !, a, r0 and N (resp. Ω, ! O1, O2,a, r0 and N) and a positive constant eC depending on Ω, !, a (resp. Ω, ! O1, O2 and a) such thatjjq(; 0)jj2L2(Ω) +ZZQe−2s eCT−1=(T−t)(T − t)−3jqj2 dxdt exp [CH(T; jjbjj1)] ZZ!(0;T )jqjr0dxdt!2=r0(131)for all s (Ω; !; a)(T 2 + T + T 2kbk2=31 ), where is a positive constant depending on Ω, ! and a (resp. Ω, !,a, O1, O2), H(T; jjbjj1) is given byH(T; jjbjj1) = 1 + 1T+ T + (T + T 1=2)jjbjj1 + jjbjj2=31 (132)and q is the solution to the corresponding system (118).As before, for simplicity, we only present the proof corresponding to case (1). We take into account theestimates for the weight functions in (109) for treatment of case (2). In the sequel, (Ω; !; a) will stand for ageneric positive constant only depending on Ω, ! and a, whose value can change from line to line. Let us rstprove the following technical lemma:Lemma 4.3. Let e! be a nonempty open set such that e! !. Then, for any a satisfying (5, 6) and (9),b 2 L1(Q), qT 2 L2(Ω) and any r0 suciently small, there exists C = C(Ω; !; a) > 0 such thatZZe!(0;T )e−2st−3(T − t)−3jqj2 dxdt CT−3TK(T; kbk1)γe−CsT−2 ZZ!(0;T )jqjr0 dxdt!2=r0(133)for all s (Ω; !; a)T 2, where is a positive constant depending on Ω, ! and a, , γ are positive numbers onlydepending of N , K(T; kbk1) is given byK(T; kbk1) = 1 + T 1=2(1 + kbk1) + T−5=2(s+ T 2) (134)and q is the solution to the corresponding system (118).CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 645Proof of Lemma 4.3. Let e! be a nonempty open set such that e! !. Notice that without loss of generalitywe can consider ! Ω0 or ! Ω1 with a smooth boundary and the estimates obtained below remain truefor a larger !. Let us consider a function 2 D(!), such that = 1 in e!. We setw(x; t) = (x)’(x; t)q(x; t);where q is the solution of (118) and ’ is given by’(x; t) =e−st3=2(T − t)3=2 (135)Notice that w(T ) = w(0) = 0. Taking into account (118), we deduce that w satisfy the following problem:8><>:−@tw − div(a(x)rw) = −bq’− q@t’− 2ar(’) rq − div(ar(’))q in Q;w = 0 on ;w(x; T ) = 0 in Ω:(136)For simplicity of the computations, we put ew(x; t) = ew(x; T − t) for (x; t) 2 Q. In a similar way, we introducethe functions ea, eb, e’ and eq. Then we have8><>:@t ew − div(ea(x)r ew) = −ebeq e’+ eq@t e’− 2ear(e’) req − div(ear(e’))eq in Q;ew = 0 on ;ew(x; 0) = 0 in Ω: (137)On the other hand, let z be the solution of the problem8><>:−@z − div(a(x)rz) = 0 in ! (0; t);z = 0 on @! (0; t);z(x; t) = in !;(138)where 2 L2(Ω) is given. Multiplying (137) by z and integrating in ! and in 2 (0; t), we obtain the followingfor t 2 (0; T ):( ew(t); z(t)) = Z t0Z!(−ebe’+ @t e’− div(ear(e’)))eqz dxd − 2 Z t0Z!ear(e’) req z dxd C(1 + kbk1)Z t0je’j keqkLr0(!)kzkLr(!) d + C Z t0je’j keqkLr0(!)krzkLr(!) d+Z t0j@t e’j keqkLr0(!)kzkLr(!) d;(139)where C is a positive constant depending on !, e! (i.e. on !) and a. In (139), we have used that jre’j Cje’jand that je’j Cje’j.Notice that, since the diusion coecients are suciently regular in ! and thanks to the regularizing eectof the heat equation (cf. [19] and [6]), we know that for all t > 0 and 1 p; q +1 the following holds:kS(t)ukLp(!) Ct−N2 ( 1q− 1p )kukLq(!) 8u 2 Lq(!); (140)kS(t)ukW 1;p(!) Ct−N2 ( 1q− 1p )− 12 kukLq(!) 8u 2 Lq(!); (141)where fS(t) : t 0g denotes the semigroup generated by the heat equation with Dirichlet boundary conditions.646 A. DOUBOVA, A. OSSES AND J.-P. PUELWe apply to z (solution of (138)) the estimates (140) and (141) with q = 2 and p = r, so we can writefrom (139) that( ew(t); z(t)) C(1 + kbk1)Z t0je’j (t− )−N2 ( 12− 1r )jjeq(; )jjLr0 (!)k kL2(!) d+CZ t0je’j (t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!)k kL2(!) d (142)+CZ t0j@t e’j (t− )−N2 ( 12− 1r )jjeq(; )jjLr0 (!)k kL2(!) dfor all t 2 [0; T ]. Thus, from (142) we havek ew(; t)kL2(e!) = supk kL2(!)=1( ew(; t); ) C(1 + kbk1)Z t0je’j (t− )−N2 ( 12− 1r )jjeq(; )jjLr0 (!) d+CZ t0je’j (t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!) d + C Z t0j@t e’j (t− )−N2 ( 12− 1r )jjeq(; )jjLr0 (!) dfor all t 2 [0; T ]. This gives,k ew(; t)kL2(e!) C(1 + T 1=2 + T 1=2kbk1)Z t0je’j (t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!) d+CT 1=2Z t0j@t e’j (t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!) d: (143)Using the denition of , it is not dicult to see thatje’j CT−3e−CsT−2 8 (x; t) 2 Q (144)ande−st7=2(T − t)7=2 CT−7e−CsT−2 8 (x; t) 2 Q (145)for s (Ω; !; a)T 2. Moreover, using (135) and (145), we can write thatj@t e’j = e−s(T − 2t)t7=2(T − t)7=2s+32tT − 32T 2 CT−6(s+ T 2)e−CsT−2 (146)for s (Ω; !; a)T 2. Thanks to (145) and (146), from (143) we havek ew(; t)kL2(e!) CT−3(1 + T 1=2 + T 1=2kbk1)e−CsT−2 Z t0(t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!) d+CT 1=2T−6(s+ T 2)e−CsT−2Z t0(t− )−N2 ( 12− 1r )− 12 jjeq(; )jjLr0 (!) d (147)for s (Ω; !; a)T 2. If r0 is such thatN212− 1r+12+1r0<32;CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 647that is to sayr0 >2(N + 2)N + 4; (148)then, we can apply Young’s inequality to (143) and estimate the L2(0; T ;L2(e!))-norm of ew as follows:jj ewjjL2(0;T ;L2(e!)) CT T−3(1 + T 1=2 + T 1=2jjbjj1)e−CsT−2 ZZ!(0;T )jqjr0 dxdt!1=r0+CT T 1=2T−6(s+ T 2)e−CsT−2 ZZ!(0;T )jqjr0 dxdt!1=r0;(149)where C is a new positive constant depending on Ω, !, e! r0 and N , depends on N and r and s (Ω; !; a)T 2.Notice that for N < 4 and r0 as in (4) the condition (148) is satised. For N 4 we apply again this process inorder to obtain the inequality (133) for r0 suciently small, in such a way that (4) holds for s (Ω; !; a)T 2and with and γ only depending on N (cf. [13]). This ends the proof of Lemma 4.3.Proof of Proposition 4.2. We will use a global Carleman inequality (46) from Theorem 3.3, some estimates forthe weight functions, classical parabolic estimates and technical result of Lemma 4.3.Step 1: Let b and qT be given and let q be the solution to (118). We will rst see thatZZΩ(T=4;3T=4)jqj2 dxdt T 6 expC1 +1T+ kbk2=31ZZ!(0;T )e−2st−3(T − t)−3jqj2 dxdt: (150)Let us consider (122), that is true for s s7 with s7 given by (123). Then, using (125), we obtain thatZZΩ(T=4 ;3T=4)jqj2 dxdt CT 6 exp (CsT−2 ZZ!(0;T )e−2st−3(T − t)−3jqj2 dxdt (151)for all s s8. Thanks to the structure of the constant s9, given by (126), the inequality (151) written for s = s9implies that (150) is satised for any solution q of (118).Step 2: Let us now prove the rst part of desired observability estimate (131), i.e. the following inequality:kq(x; 0)k2L2(Ω) exp [CH(T; kbk1)] ZZ!(0;T )jqjr0 dxdt!2=r0(152)with H(T; kbk1) given by (132). Let e! be a nonempty open set such that e! ! \ Ω0. We consider (130),that is to saykq(x; 0)k2L2(Ω) 2CTexp3T2jjbjj1ZZΩ(T=4 ;3T=4)jqj2 dxdt:We apply to the right hand side of this inequality, the estimates (150) written for e! and we obtainkq(x; 0)k2L2(Ω) CT 5 exp1 +1T+ T jjbjj1 + kbk2=31ZZe!(0;T )e−2st−3(T − t)−3jqj2 dxdt: (153)648 A. DOUBOVA, A. OSSES AND J.-P. PUELOn the other hand, we notice that the constant (Ω; !; a)T 2 appearing in Lemma 4.3 is less then the constant s9given by (126). So, we can write technical estimate (133) for s = s9. This givesZZe!(0;T )e−2st−3(T − t)−3jqj2 dxdt CT−3K1(T; kbk1) ZZ!(0;T )jqjr0 dxdt!2=r0; (154)whereK1(T; kbk1) = exp1 +1T+ T jjbjj1 + kbk2=31T1 + T 1=21 + kbk1 + 1T 2+1Tkbk2=31γ: (155)Combining (153) and (154), we deduce thatkq(x; 0)k2L2(Ω) CT 2K1(T; kbk1) ZZ!(0;T )jqjr0 dxdt!2=r0: (156)Consequently, we also have (152).Step 3: Let us nally deduce thatZZΩ(0;T )e−2s eCT−1=(T−t)(T − t)−3jqj2 dxdt exp [CH(T; jjbjj1)] ZZ!(0;T )jqjr0 dxdt!2=r0; (157)where H(T; jjbjj1) is done by (132). Integrating (129) with respect to time and using that = 1 in [0; T=4], weobtain for all t 2 [0; 3T=4]exp (2jjbjj1t)ZΩjq(x; t)j2 dx Z 3T=4T=42 exp (2jjbjj1)ZΩj@tjjqj2 dxd: (158)Since 1, j@tj = j@t0(t=T )=T j C=T , from (158) we obtainZΩjq(x; t)j2 dx 2Texp3T2jjbjj1ZZΩ(T=4 ;3T=4)jqj2 dxdt (159)for all t 2 [0; T=4]. Then, we haveZZΩ(0;T=4)jq(x; t)j2 dx exp3T2jjbjj1ZZΩ(T=4 ;3T=4)jqj2 dxdt: (160)On the other hand, using the denition (44) of the function , we can say that there exist a positive constants Cand eC, depending on Ω, ! and a, such thate−2st−3(T − t)−3 CT−3(T − t)−3e−2s eCT−1=(T−t) 8x 2 Ω; 8 t 2 [T=4; T ] (161)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 649whenever s (Ω; !; a)T 2. We can writeZZΩ(0;T )e−2s eCT−1=(T−t)(T − t)−3jqj2 dxdtZZΩ(0;T=4)e−2s eCT−1=(T−t)(T − t)−3jqj2 dxdt+ CT 3ZZΩ(T=4;T )e−2st−3(T − t)−3jqj2 dxdt T−3e−2CsT−2ZZΩ(0;T=4)jqj2 dxdt+ CT 3ZZΩ(T=4;T )e−2st−3(T − t)−3jqj2 dxdt T−3eC(−sT−2+Tkbk1)ZZΩ(T=4;3T=4)jqj2 dxdt+CT 3ZZΩ(0;T )e−2st−3(T − t)−3jqj2 dxdt: (162)Here, we have used (161) and (160). Applying (151), written for e! to the rst term of the right hand sideof (161) and Carleman inequatily (122) (written also for e!) to the second one, we deduce that for s = s9 thefollowing holds:ZZQe−2s eCT−1=(T−t)(T − t)−3jqj2 dxdt CT 3 (1 + exp C (sT−2 + T kbk1 ZZe!(0;T )e−2st−3(T − t)−3jqj2 dxdt CT 3 expC1 +1T+ T kbk1 + kbk2=31ZZe!(0;T )e−2st−3(T − t)−3jqj2 dxdt:(163)Using (154) in (163), we get (157). This ends the proof of Proposition 4.2.5. Proof of Theorem 2.1We will rst prove a null controllability result for a corresponding linear problem which is interesting foritself. For this proof we use a result of approximate controllability to the zero state and then apply an observ-ability inequality to obtain estimates which enable us to pass to the limit. We could then try to use this nullcontrollability result and apply the xed point method to treat the nonlinear problem. For technical reasons,we prefer here to apply the xed point method to the approximate controllability problem to the zero state andthen use the observability inequality to obtain the desired result of exact controllability to the trajectories forthe nonlinear system.5.1. A null controllability result for a linear problemWe will consider the linear system8><>:@ty − div(a(x)ry) + by = v1! + k in Q;y = 0 on ;y(x; 0) = y0 in Ω;(164)650 A. DOUBOVA, A. OSSES AND J.-P. PUELwhere a satises (5, 6, 9), b 2 L1(Q), k 2 Lr(0; T ;Lr(Ω)), and y0 2 L2(Ω) are given. The following holds:Theorem 5.1. Let T > 0. Assume that ! \ Ω0 6= ;, Condition 2:1 (resp. Condition 2:2) in case (1) (resp. incase (2)) are fullled and a satises (5, 6, 9) and (10). We take data b 2 L1(Q), y0 2 L2(Ω), k 2 L2(Q) andsuch that ZZQe2s eCT−1=(T−t)(T − t)3jkj2 dxdt < +1 (165)for s (Ω; !; a)(T + T 2 + T 2kbk2=31 ). Then for each case (1) or (2), there exists a control bv 2 Lr(O (0; T ))such that the corresponding solution by of (164) veriesby(x; T ) = 0 in Ω: (166)Moreover, bv can be chosen satisfying the estimatejjbvjjLr(!(0;T )) bH(Ω; !; a; T; jjbjj1)jjy0jjL2(Ω) + jjes eCT−1=(T−t)(T − t)3=2kjjL2(Q) ; (167)withbH(Ω; !; a; T; jjbjj1) = exp C 1 + 1T+ T + (T + T 1=2)jjbjj1 + jjbjj2=31: (168)Remark 5.1. Notice that if k 2 Lr(0; T ;Lr(Ω)) such that (165) holds, then k vanishes exponentially at t = T .If k = 0 in a neighbourhood of t = T and k 2 Lr(0; T ;Lr(Ω)) then (165) is satised.Proof of Theorem 5.1. Let us x T > 0, b 2 L1(Q), y0 2 L2(Ω), k 2 Lr(0; T ;Lr(Ω)), with r verifying (4) andsuch that (165) holds. For every " > 0, let us consider the functional J" denite byJ"(qT ) =12 ZZ!(0;T )jqjr0 dxdt!2=r0+ "jjqT jjL2(Ω) +ZΩq(x; 0) y0(x) dx +ZZQkqdxdt 8qT 2 L2(Ω); (169)where q is the solution of (118) associated to qT 2 L2(Ω) and r0 is the dual exponent of r. It is easy to seethat J" is a continuous and strictly convex functional. Furthermore, from (122), it is immediate to deduce thefollowing unique continuation property for the adjoint problem (118):If q = 0 in ! (0; T ), then q 0.Then, arguing as in [9] and [10], we see thatlim infkqT kL2!1J"(qT )kqT kL2 "and, therefore, J" achieves its minimum at a unique point bqT;" 2 L2(Ω). Let bq" be the solution of (118) associatedto bqT;". Arguing again as in [9], we take in (164) v = bv", wherebv" = sgn(bq")jbq"jr0−1kbq"kLr0(!(0;T ))1! ; (170)then, we nd a solution by" satisfyingjjby"(; T )jjL2(Ω) ": (171)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 651Moreover, we can obtain the estimates of the Lr(! (0; T ))-norm of the control independent of ". We claimthat for a suitable C = C(Ω; !; a) > 0 we have:jjbv"jjLr(!(0;T )) bH(Ω; !; T; a; kbk1)jjy0jjL2(Ω) + jjes eCT−1=(T−t)(T − t)3=2kjjL2(Q) (172)for all " > 0 and with bH given by (168). Indeed, from (170) we can write thatjjbv"jjLr(!(0;T )) = ZZ!(0;T )jbq"jr0 dxdt!1=r0: (173)On the other hand, at the minimum bqT;", we haveJ"(bqT;") J"(0) = 0:Taking (169) into account, we see that12 ZZ!(0;T )jbq"jr0 dxdt!2=r0 −ZΩbq"(x; 0)y0(x) dx − ZZQkq dxdt kbq"(; 0)kL2(Ω)ky0(x)kL2(Ω)+ZZQe2s eCT−1=(T−t)(T − t)3jkj2 dxdt1=2ZZQe−2s eCT−1=(T−t)(T − t)−3jqj2 dxdt1=2: (174)In view of (131) and (165), the estimate (172) holds.Since bv" is uniformly bounded in Lr(! (0; T )), for an appropriate subsequence, we deduce thatbv" ! bv weakly in Lr(! (0; T )); (175)where bv 2 Lr(! (0; T )) and satises (167). Accordingly,by"(T ) ! by(T ) in L2(Ω);where by is the solution of (164) associated to bv. Since we have (171) for all " > 0, we also have (166). This endsthe proof of Theorem 5.1.Remark 5.2. Notice that, using the argument of [13] and [9], it is also possible to obtain the controlsin L1(0; T ;L1(!)).5.2. The xed point method. ConclusionWe are now ready to complete the proof of Theorem 2.1. We will apply a xed point method to obtain"-approximate controllability to the zero state for the nonlinear transmission problem, using the ideas of [9]and [10]. Then the main ingredients of our proof are the explicit estimates of the controls needed for control-lability of a linear transmission problem, which have been obtained in the previous section and the choice ofthe control time depending on the size of the potential. This will allow us to pass to the limit in the nonlinearproblem when " tends to zero and obtain the complete results of Theorem 2.1.652 A. DOUBOVA, A. OSSES AND J.-P. PUELStep 1: Let us consider a trajectory y, solution of the problem (12) without control. We introduce the changeof variable p = y − y, where y is a solution of (8). Then, we obtain that8><>:@tp− div(a(x)rp) + f(y + p)− f(y) = v1! in Q;p = 0 on ;p(x; 0) = p0 in Ω;(176)where p0 = y0 − y0 . Theorem 2.1 will be proved if we show that, for each p0 2 L2(Ω), there exists v 2Lr(! (0; T )) such thatp(x; T ) = 0 in Ω: (177)We will rst consider the case in which p0 2 L1 and f 2 C1 in R. We denote by h the following function:h(a; s) =8<:f(a+ s)− f(a)sif s 6= 0;f 0(a) if s = 0:Then h is continuous. Thanks to hypothesis (11) we know that for each > 0, there exists C > 0 (dependingonly of and the function f) such thatjh(y(x; t); s)j2=3 C + log(1 + jsj) 8 s 2 R; 8 (x; t) 2 Q: (178)Step 2: Let > 0 and R > 0 be given positive constants whose values will be xed later on.Let us x a timeTR = minnT; jjhjj−2=3L1(−R;R); jjhjj−1=3L1(−R;R)o (179)For simplicity, in the sequel, we will refer only to the case (1), but we also take into account the dependence ofthe constants corresponding to the case (2).Step 3: (a) We consider the truncation function TR : R 7! R; which is given as follows:TR(s) =s if jsj R;R sgn (s) otherwise.For each z 2 L2(Q), we consider the linear system8><>:@tp− div(a(x)rp) + h(y(x; t);TR(z))p = v1! in Ω (0; TR);p = 0 on Γ (0; TR);p(x; 0) = p0 in Ω:(180)Notice that (180) is of the form (164), with b = h(y;TR(z)) 2 L1(Q). Then we can apply the arguments ofthe proof of Theorem 5.1 to (180). In fact, we will apply this result in a time interval (0; TR), where TR is givenby (179). This is a key point in this proof that will drive to appropriate estimates (the idea is taken from [13]and it has been applied later in [8]).(b) More precisely, for every " > 0, let us consider the functional J" of the form (169). Arguing as in the proofof Theorem 5.1, we obtain the existence of a control v"z 2 Lr(! (0; TR)), minimizing the Lr(! (0; TR))-norm of the form v"z = sgn(q"z)jq"z jr0−1kq"zkLr0(!(0;T ))1!, (r0 > 1) with q"z the solution of the correspondingCONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 653problem (118), such that the solution p"z of (180) with v = v"z satiseskp"z(; TR)kL2(Ω) ": (181)Then, from (172) we havejjv"z jjLr(!(0;TR)) C1(Ω; !; a; TR; jjhjjL1(−R;R))jjp0jjL2(Ω) (182)whereC1(Ω; !; a; TR; jjhjjL1(−R;R)= eC(Ω;!;a)1+ 1TR+TR+(TR+TR1=2)jjhjjL1(−R;R)+jjhjj2=3L1(−R;R): (183)Using now the denition of TR, we deducejjv"z jjLr(!(0;TR)) eC2(Ω;!;a;T )1+jjhjj2=3L1(−R;R)jjp0jjL2(Ω): (184)Moreover, thanks to (178), from (184) for each z 2 L2(Q) we obtain the following:jjv"z jjLr(!(0;TR)) eC2(Ω;!;a;T )(1+C+ log(1+R))jjp0jjL2(Ω)= C3(Ω; !; a; T; ; p0)(1 +R)C2(Ω;!;a;T ): (185)Here we have used the fact that from (178) we can easily write thatkhk2=3L1(−R;R) C + log(1 + jRj): (186)(c) For each " > 0 and z 2 L2(Q) we introduce the mapping : L2(Q) 7! L2(Q) dened as follows: for eachz 2 L2(Q), (z) = p"z, where p"z is the the solution of (180) satisfying (181) with v = v"z constructed in the point(b) of this step. In fact, is of the following formz 2 L2(Q) 7! TR(z) 2 L1(Q) 7! h(y;TR(z)) 2 L1(Q) 7! v"z 2 Lr(! (0; TR)) 7! p"z 2 L2(Q):Arguing as in [9], we apply Schauder’s theorem and we deduce for each " > 0 the existence of a xed point p"(associated to v") of which verieskpz(; TR)kL2(Ω) ": (187)Notice that we have used that the solution p"z of (180) is bounded (uniformly in z) in L2(0; TR;H10 (Ω)) and itstime derivative @tp"z is bounded in L2(0; TR;H−1(Ω)).(d) Let p" be a xed point of associated to the control v" constructed as above. Since (185) holds for v",then v" is bounded in Lr(! (0; TR)) uniformly in ", p" is bounded in L2(0; TR;H10 (Ω)) and @tp"z is boundedin L2(0; TR;H−1(Ω)). For an appropriate subsequence, we deduce that as "! 0v" ! vR weakly in Lr(! (0; TR)); (188)where vR 2 Lr(! (0; TR)) also satises (185), andp" ! pR weakly in L2(0; TR;H10 (Ω));@tp" ! @tpR weakly in L2(0; TR;H−1(Ω));654 A. DOUBOVA, A. OSSES AND J.-P. PUELwhere pR is the solution of the following problem:8><>:@tpR − div(a(x)rpR) + h(y(x; t);TR(pR))pR = vR1! in Ω (0; TR);pR = 0 on Γ (0; TR);pR(x; 0) = p0 in Ω:(189)Thenp"(TR) ! pR(TR) in L2(Ω)and since we have (187) for all " > 0, we also havepR(TR) = 0: (190)On the other hand, since vR 2 Lr(0; TR;Lr(!)), with r as in (4), we can write (cf. for example [2] and [3]) thatjjpRjj1 eTRjjh(y;TR(pR))jj1 jjp0jj1 + TReTRjjh(y;TR(pR))jj1 kvRkLr(0;TR;Lr(!)) : (191)Using again the denition of TR and also taking into account (185) and (186), we deduce from (191) that pRveriesjjpRjj1 eC4(Ω;!;a;T )1+jjhjj2=3L1(−R;R) (jjp0jj1 + kvRkLr(0;TR;Lr(!)) C5(Ω; !; a; T; ; p0)(1 +R)C6(Ω;!;a;T ): (192)Notice that in (192) the constant C5 is independent of R and the constant C6 is independent of and R. Let usextend by zero pR and vR to the whole cylinder Q = Ω (0; T ) and for simplicity, we still call them pR and vR.It is clear that (192) holds and that vR is such thatpR(T ) = 0:(e) In order to conclude the proof of this theorem, it is sucient to check that for and R suitably chosen, pR(dened on Ω (0; T )) satiseskpRk1 R: (193)Then we can say that TR(pR) = pR. Of course, this implies the existence of a control v 2 Lr(0; T ;Lr(!)) suchthat the solution of (176) satises (177). Indeed, from (192) we can choose = 1=(2C6) and R > 0 such thatC5(Ω; !; a; T; p0)(1 +R)C6(Ω;!;a;T ) < R:Then we obtain (193). This proves Theorem 2.1 when p0 2 L1(Ω) and f 2 C1(R). We just mention that wetreat the case in which f is only locally Lipschitz continuous as for example in [9] and [13] using approximationsof f by C1 functions. Then, in this case we deduce the existence of a control v 2 Lr(0; T ;Lr(!)) such that thecorresponding solution to (176) veries (177).Finally, if p0 2 L2(Ω), for > 0 suciently small we set v 0 for t 2 (0; ). Using the regularizingeect of the heat equation (see, for example [21] and [22]), we deduce that the corresponding (local) solution pof (176) satises p(; ) 2 L1(Ω). Then, we argue as above for p in the interval [; T ] and we obtain a controlv 2 Lr(0; T ;Lr(!)) such that (177) holds. This ends the proof of Theorem 2.1.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 6556. Proofs of Lemma 3.1 and Lemma 3.2In this section we will present the construction of the weight functions we used for our global Carlemaninequalities.Proof of the Lemma 3.1. We will proceed in several steps.Step 1: Let and x be vector elds verifying Condition 2.1. First, we will construct a function e1 2 C1(Ω1),such thate1 > 0 in Ω1; (194)e1 = 0 on Γ; (195)@n e1 < 0 on Γ; (196)e1 = 1 on S; (197)@n e1 > 0 on S; (198)re1 6= 0 in Ω1: (199)For t 2 [0; t1(x0)], let us introduce the following change of variables: =tt1(x0); 2 [0; 1]; ex() = x(t):Observe that, if we take x 2 Ω1, then there exists x0 2 Γ such that x = x(t; x0) for t 2 [0; t1(x0)] or, in otherwords, there exists 2 [0; 1] such that ex(0) = x0 2 Γ, ex() = x(t) and ex(1) = x(t1(x0)) 2 S. Moreoverdexd=dxdtt1(x0) = (x(t))t1(x0): (200)Let us sete1(x(t)) = e1(ex()) = ; 2 [0; 1]: (201)This function veries the following properties:de1d(x) = 1 for all x 2 Ω1; (202)0 < e1(x) < 1 for all x 2 Ω1; (203)e1(x) = 0 for all x 2 Γ; (204)ande1(x) = e1(ex(1)) = 1 for all x 2 S: (205)On the other hand, from (200) and (202), for all x 2 Ω1, we havede1d(x()) = re1(ex())dexd () = re1(x)(x(t))t1(x0) = 1; (206)thereforere1(x) 6= 0 for all x 2 Ω1 :656 A. DOUBOVA, A. OSSES AND J.-P. PUELMoreover, using (200, 205) and (206), we can write thatre1(x)dexd = re1(x) ndexd n=re1(x) n ((x) n)t1(x0) = 1 for all x 2 S.Then, taking into account (17), we deduce thatre1(x) n > 0 for all x 2 S.This means (198). By the similar way, it is easy to check (196). Thus we also have (199).In order to obtain e1 2 C2(Ω1), we just notice that we can approximate the function of class C1, which wehave constructed above by an other function of class C2 (that we keep calling e1), such that it still satises theproperties (194{199).Let us now consider the diusion coecients ai, i = 0; 1 such that (5) holds and let !0 ! \ Ω0.Step 2: For " > 0 small enough, we setU"(S) = fx : x 2 Ω0 ; dist(x; S) < "gWe can construct a function 0 in U"(S), such that 0 2 C2(U"(S)) and0 = 1 on S; @n0 > 0 on S;0 > 0 in U"(S); r0 6= 0 in U"(S)(207)anda0@n0 = a1@n e1 on S. (208)Now, we extend this function in Ω0 to a function that we call again 0, with 0 2 C2(Ω0) and 0 > 0 in Ω0.Step 3: Thanks to the Morse theorem, we deduce that there exists a sequence of Morse functions k, k 1(functions with isolated critical points i.e. their gradient vanishes only in a nite number of points), such thatk ! 0 in C2(Ω0) if k ! +1: (209)If k is close enough to 0, the points where rk vanishes can not be in U"(S). Moreover, we can assume thatfor some > 0 we havejr0j > 0 in U"(S): (210)Let us construct a Morse function 2 C2(Ω0), such that = 1 on S; @n > 0 on S (211)a0@n = a1@n e1 on S (212)andr 6= 0 in U"(S): (213)For this, we consider ’ 2 D(U"(S)) and ’ = 1 in U"0(S), with 0 < "0 < ". We setk(x) = k(x) + ’(x)(0(x)− k(x)):CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 657It is clear thatk = 0 in U"0(S): (214)Then, the function k satises (211) and (212). Moreover, we haverk = rk in Ω0nU"(S)andrk = rk + ’(r0 −rk) +r’(0 − k) in U"(S): (215)Then, using (209) and (210) in (215), we deduce that there exist a positive number k0 = k0() such that, ifk k0 we havejrkj jrkj − 2k’kC1k0 − kkC1 2 in Ω0 \ U"(S):We choose k k0 and we set (x) = k(x). Then, is a Morse function whose gradient vanishes only in theset of points where the gradient of k vanishes. This, together with (214), implies that (211) and (213) hold.Step 4: On the other hand, arguing as in [15], we can deduce that there exists a mapping g : Ω0 7! Ω0 whichis a dieomorphism on Ω0, which leaves invariant U"(S) and transports the points where the gradient of vanishes in !0. We set e0(x) = (g(x)):Then, equation (33) holds. Thanks to the properties (211) of the function , we also have (30, 31) and (32).This ends the proof of Lemma 3.1.Proof of the Lemma 3.2. Assume that we are in the situation of Case (2). Let !0 ! \ Ω0 be an arbitraryxed open subset of Ω0.Step 1: We assume that there exist O1 ;O2 Ω1 two open disjoint subsets, such that Condition 2.2 holdsbetween Ω1 and each one of two sets O1 and O2. Then, as in the rst part of the proof of Lemma 3.1, weconstruct two functions i1 2 C1(Ω1nOi ), i1 > 0 in Ω1nOi, i = 1; 2, such thati1 = 2 on S, i = 1; 2;@ni1 > 0 on S, i = 1; 2;@ni1 > 0 on @Oi ; i = 1; 2;i1 = 1 on @Oi ; i = 1; 2;ri1 6= 0 in Ω1nOi ; i = 1; 2;(216)where n stands for the unit exterior normal to Ω1 and Oi, i = 1; 2.Step 2: Let eBi and Bi, i = 1; 2 be balls such that B1 eB1 O1 and B2 eB2 O2. We will presentonly the construction of e11 . The second function will be obtained by the same arguments. Let us setW" = fx : x 2 O1 ; dist(x; @O1) < "gFirst, we observe that since @n11 > 0 on @O1 and 11 = 1 on @O1, we construct a function 11 2 C1(W"), suchthat there exists > 0 such that11 1 in W" ; 0 < 11 1− 4 on @W"n@O1 and r11 6= 0 in W" :658 A. DOUBOVA, A. OSSES AND J.-P. PUELThen, we can extend this function by a function still called 11 2 C1(O1), such that0 < 11 1− 3 in O1nW" and r11 6= 0 in W" :Now, we approximate 11 by Morse functions in such a way that0 < 11 1− 2 in O1nW", (217)where we keep the name 11 for this approximation. The gradient of this function vanishes only in a nitenumber of points. As we already mentioned in the Step 3 of this proof, we can deduce the existence of adieomorphism on O1, which keeps invariant W" and transports the points where the gradient of 11 vanishesin B1. We obtain then a new function (that we keep on calling 11) such that 11 2 C1(Ω1), 11 > 0 in Ω1 andr11 6= 0 in Ω1nB1. (218)Moreover, from (217) we obtain that for > 0 we have11 1− in B1. (219)Analogously, we construct a function 21 2 C1(Ω1), 21 > 0 in Ω1, which veriesr21 6= 0 in Ω1nB2 (220)and21 1− in B2. (221)Step 3: Let us nally prove that the properties (40) and (41) are satised. For this, we will see that it ispossible to modify 11 (resp. 21) in B1 (resp. B2) in order to obtain the conditions (40) and (40). We will beable to do this without changing the values of these functions in O1nB1 and O2nB2. For simplicity, we willpresent the details of the construction of only one of such a function, because the same arguments will be validfor the other one.Let us dene a new function e11 as follows:e11(x) =(11(x) if x 2 Ω1nO1 ;(11(x)n(x) if x 2 O1 ; (222)withn(x) =111(x)p; (223)where p 2 N will be xed later on. We can write thate11 = (11n = en log 11 in O1. (224)Since 11 = 1 on @O1, we have that n = 1 on @O1 and then, from (224) we deduce thate11 = 11 = 1 on @O1. (225)CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 659Moreover, we have thatre11 = r11 on @O1 (226)and that the gradient of e11 vanishes only in B1, where the gradient of 11 is zero, i.e.re11 6= 0 in Ω1nB1. (227)Indeed, using (224) and (223), we havere11 = (11nrn log 11 + nr1111=(11nr11"−p log11(11)p+1 +1(11)p+1# (228)Taking into account (217), it is easy to deduce from (228), that (226) and (227) hold.In order to modify the values of e11 in B1, we rst use (219) and we obtain thatn 11− pNext, from (224) we deduce thate11 (1− )(1=(1−))p : (229)On the other hand, we know that the second function that we constructed in the Step 5 of this proof satises0 < 21 21 = maxB121 > 0 in B1:Choosing now p large enough, we can deduce from the estimate (229) the following:e11 12 21 in B1:This gives (41). The same arguments applied to the function 21 lead to the existence of a new functione21 2 C1(Ω1), e21 > 0 in Ω1, such that satises (38, 40) and (42) for i = 2.To conclude this step, we observe that we can approximate the functions of class C1 already constructed byfunctions of class C2, preserving the properties of the functions e11 and e21 .Step 4: For "; "0 > 0 small enough, we setV"(Γ) = fx : x 2 Ω0 ; dist(x;Γ) < "gandV"0(S) = fx : x 2 Ω0 ; dist(x; S) < "0gWe can locally construct a function 0 in V"(Γ), such that 0 2 C2(V"(Γ)) and0 = 0 on Γ; @n0 > 0 on Γ;0 > 0 in V"(Γ); r 6= 0 in V"(Γ):(230)660 A. DOUBOVA, A. OSSES AND J.-P. PUELOn the other hand, in V"0(S) we construct another function, which for simplicity, we also will denote by 0,such that 0 2 C2(V"0 (S)) and0 = 2 on S; @n0 > 0 on S;0 > 0 in V"0(S); r0 6= 0 in V"0 (S)(231)anda0@n0 = a1@ni1 on S; i = 1; 2: (232)Now, we extend both functions in Ω0, to a function which we keep on calling 0, with the following properties:0 2 C2(Ω0); 0 > 0 in Ω0;0 = 0 on Γ; @n0 > 0 on Γ;0 = 2 on S; @n0 > 0 on S:(233)Step 5: In the sequel, we will use same arguments as for the proof of Lemma 3.1.Thanks to the Morse theorem, we deduce that there exists a sequence of Morse functions k, k 1 (functionswith isolated critical points i.e. their gradient vanishes only in a nite number of points), such thatk ! 0 in C2(Ω0) if k ! +1: (234)If k is close enough to 0, the points where rk vanishes can not be in V"(Γ) [ V"0 (S). Moreover, we canassume that for some > 0 we havejr0j > 0 in V"(Γ) [ V"0(S): (235)We can construct a Morse function 2 C2(Ω0), such that = 0 on Γ; @n < 0 on Γ; (236) = 2 on S; @n > 0 on S (237)andr 6= 0 in V"(Γ) [ V"0 (S): (238)Indeed, it suces to consider ’ 2 D(V"(Γ) [ V"0(S)) and ’ = 1 in a neighborhood of Γ [ S and to denek(x) = k(x) + ’(x)(0(x)− k(x)):Arguing as in the proof of Lemma 3.1, we can choose k k0 and (x) = k(x) in such a way that is aMorse function with gradient vanishing only in the points contained in the set of points where the gradient of kvanishes and satisfying the previous properties.Step 6: Finally, we can deduce that there exists a mapping g : Ω 7! Ω which is a dieomorphism on Ω, whichleaves invariant V"(Γ) [ V"0(S) and transports the points where the gradient of vanishes in !0. We sete0(x) = (g(x)):This ends the proof of Lemma 3.2.CONTROLLABILITY FOR HEAT EQUATION WITH DISCONTINUOUS COEFFICIENTS 661We thank the referee for useful remarks and corrections concerning some technical points of this article.References[1] S. Anita and V. Barbu, Null controllability of nonlinear convective heat equation. ESAIM: COCV 5 (2000) 157-173.[2] D.G. Aronson and J. Serrin, Local behavior of solutions of quasilinear parabolic equations. Arch. Rational Mech. Anal. 25(1967) 81-122.[3] D.G. Aronson and J. Serrin, A maximum principle for nonlinear parabolic equations. Ann. Scuola Norm. Sup. Pisa 3(1967) 291-305[4] J.P. Aubin, L'analyse non lineaire et ses motivations economiques. Masson (1984).[5] V. Barbu, Exact controllability of the superlinear heat equation. Appl. Math. Optim. 42 (2000) 73-89.[6] T. Cazenave and A. Haraux, Introduction aux problemes d'evolution semi-lineaires. Ellipses, Paris, Mathematiques &Applications (1990).[7] S. Cox and E. Zuazua, The rate at which energy decays in a string damped at one end. Indiana Univ. Math. J. 44 (1995)545-573.[8] A. Doubova, E. Fernandez{Cara, M. Gonzalez{Burgos and E. Zuazua, On the controllability of parabolic system with anonlinear term involving the state and the gradient. SIAM: SICON (to appear).[9] C. Fabre, J.-P. Puel and E. Zuazua, (a) Approximate controllability for the semilinear heat equation. C. R. Acad. Sci. ParisSer. I Math. 315 (1992) 807-812; (b) Approximate controllability of the semilinear heat equation. Proc. Roy. Soc.Edinburgh Sect. A 125 (1995) 31-61.[10] C. Fabre, J.-P. Puel and E. Zuazua, Approximate controllability for the linear heat equation with controls of minimal L∞norm. C. R. Acad. Sci. Paris Ser. I Math. 316 (1993) 679-684.[11] E. Fernandez{Cara, Null controllability of the semilinear heat equation. ESAIM: COCV 2 (1997) 87-107.[12] E. Fernandez{Cara and E. Zuazua, The cost of approximate controllability for heat equations: The linear case. 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Pazy, Semigroups of linear operators and applications to partial dierential equations. Springer-Verlag, New York (1983).[20] D.L. Russell, A unied boundary controllability theory for hyperbolic and parabolic partial dierential equations. Stud.Appl. Math. 52 (1973) 189-211.[21] F.B. Weissler, Local existence and nonexistence for semilinear parabolic equations in Lp. Indiana Univ. Math. J. 29 (1980)79-102.[22] F.B. Weissler, Semilinear evolution equations in Banach spaces. J. Funct. Anal. 32 (1979) 277-296.[23] E. Zuazua, Exact boundary controllability for the semilinear wave equation, in Nonlinear Partial Dierential Equationsand their Applications, Vol. X, edited by H. Brezis and J.-L. Lions. Pitman (1991) 357-391.[24] E. Zuazua, Finite dimensional controllability for the semilinear heat equations. J. Math. Pures 76 (1997) 570-594.[25] E. Zuazua, Approximate controllability for semilinear heat equations with globally Lipschitz nonlinearities. Control andCybernetics 28 (1999) 665-683.

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