Some results on controllability for linear and nonlinear heat equations

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tem (1) when a ∈ L∞(Q) and B ∈ L∞(Q)N . Observe that the approach in [3] and ... Secondly, we will analyze the controllability properties of system (2) when ..... apply Lemma 1.1 of [12] and deduce the existence of a function ˜η0 such that.
Some results on controllability for linear and nonlinear heat equations in unbounded domains∗ ´lez-Burgos† and Luz de Teresa‡ Manuel Gonza August 14, 2007

Abstract In this paper we present some results concerning the null controllability for a heat equation in unbounded domains. We characterize the conditions that must satisfy the auxiliary function that drives to a global Carleman inequality for the adjoint problem and then to get a null controllability result. We give some examples of unbounded domains (Ω, ω) that satisfy these sufficient conditions. Finally, when Ω \ ω is bounded, we prove the null controllability of the semilinear heat equation when the nonlinearity f (y, ∇y) grows slower than |y| log3/2 (1 + |y| + |∇y|) + |∇y| log1/2 (1 + |y| + |∇y|) at infinity (generally in this case in the absence of control, blow up occurs). In this aim we also prove the linear null controllability problem with L∞ -controls.

1

Introduction and main result

Let Ω ⊂ RN be an unbounded connected open set with boundary ∂Ω at least of class C 0,1 uniformly. Let ω ⊂ Ω be a nonempty open subset and assume T > 0. We will consider the parabolic linear problem ( yt − ∆y + B · ∇y + ay = v1ω in Q = Ω × (0, T ), (1) y = 0 on Σ = ∂Ω × (0, T ), y(x, 0) = y0 (x) in Ω, where B ∈ L∞ (Q)N , a ∈ L∞ (Q), y 0 ∈ L2 (Ω) and v ∈ L2 (Q). In (1), 1ω denotes the characteristic function of the subset ω, y = y(x, t) is the state and v = v(x, t) is the control function (which acts on the system through the subset ω). ∗ AMS

subject classification: 93B05, 93B07, 35M20. E.D.A.N., Universidad de Sevilla, Aptdo. 1160, 41080 Sevilla, Spain. E-mail: [email protected]. Supported by D.G.E.S. (Spain), grant BFM2003–06446. ‡ Instituto de Matem´ aticas, Universidad Nacional Aut´ onoma de M´ exico, Circuito Exterior, C.U., 04510 D.F., M´ exico. E-mail: [email protected]. Supported by projects 37046-E of CONACyT and FENOMEC (Mexico). † Dpto,

1

We will also consider the nonlinear problem ( yt − ∆y + f (y, ∇y) = v1ω in Q, (2) y = 0 on Σ, y(x, 0) = y0 (x) in Ω, where f : R × RN → R is a locally Lipschitz-continuous function, y 0 is given in an appropriate Banach space and v is a control which also acts on the system through the open set ω. The main goal of this paper is to analyze the controllability properties of (1) and (2). It will be said that (1) (resp., (2)) is approximately controllable in L2 (Ω) at time T if, for any y0 , yd ∈ L2 (Ω) (resp., y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω) , yd ∈ L2 (Ω)) and any ε > 0, there exists a control function v ∈ L2 (Q) such that the solution y of (1) satisfies (3)

||y(·, T ) − yd ||L2 ≤ ε,

(resp., (2) possesses a solution y ∈ C 0 ([0, T ]; L2 (Ω)) satisfying (3)). On the other hand, it will be said that (1) (resp., (2)) is null controllable at time T if for every y0 ∈ L2 (Ω) (resp., y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω)) there exists a control v ∈ L2 (Q) such that the solution y of (1) satisfies (4)

y(·, T ) = 0

in Ω,

(resp., (2) possesses a solution y ∈ C 0 ([0, T ]; L2 (Ω)) satisfying (4)). The controllability of linear and semilinear parabolic problems has been analyzed when Ω is a bounded open set in several recent papers. Among them, let us mention [17], [12], [7], [1], [10] and [5] in what concerns null controllability and [6], [10] and [5] for approximate controllability. Nevertheless, there are few results when Ω is an unbounded set. On one hand, in [23] and [24] the authors extended to unbounded domains the results on approximate controllability obtained in [6]. On the other hand, the first result on null controllability of parabolic problems was negative: in [19] the authors considered the case Ω = (0, ∞) ⊂ R, a ≡ 0 and B ≡ 0, and proved that there is no initial data in any negative Sobolev space that may be driven to zero in finite time by a control function acting on the boundary {x = 0} (also see [20] for similar results when Ω = RN + ). The first positive result on null controllability of (1) and (2) was obtained in [3]. There, the authors considered the semilinear heat equation (2) with a globally-Lipschitz continuous function f and proved the null controllability of the problem when the distributed control acts along the open set ω satisfying that the uncontrolled region Ω \ ω is bounded. In [4] the authors studied the null controllability of system (1) when a ≡ 0, B ≡ 0, Ω = (0, ∞) and ω is an unbounded open set of the form [ ω= ωn . n≥0

Under some technical assumptions on ωn , they proved two null controllability results: one result with control functions in L2 (Q) for initial data in some 2

weighted L2 (Ω)-spaces and another one for arbitrary initial data in L2 (Ω) using control functions in a weighted L2 (Q)-space. In particular the authors show that (1) is null controllable in L2 (Ω) (with controls in L2 (Q)) for some open ω such that the uncontrolled region Ω \ ω is unbounded. Finally, in [21] the author also considers the heat equation (a ≡ 0 and B ≡ 0) with Dirichlet boundary condition in an unbounded domain Ω and proves several results about the null controllability of this system with distributed and boundary controls. To be precise, the author shows the lack of null controllability of the heat equation under a geometric condition on Ω and ω and also proves a positive result on null controllability for unbounded product domains. The results that we present in this work are related to the results in the previous papers and will be obtained when Ω is an unbounded domain: • Firstly, we will obtain a global Carleman inequality for a heat equation (and then, a positive null controllability result for system (1)) under general assumptions on Ω and ω. As a consequence, we will be able to prove the results of [3] and [21] about the null controllability in L2 (Ω) of system (1) when a ∈ L∞ (Q) and B ∈ L∞ (Q)N . Observe that the approach in [3] and [21] (and also in [4]) is different from ours: in these works the authors directly prove an observability inequality for the so-called adjoint problem which gives the corresponding null controllability result for system (1) (in [21], a ≡ 0, B ≡ 0). In the present paper, this observability inequality will be obtained as a consequence of a global Carleman inequality for the adjoint problem. • Secondly, we will analyze the controllability properties of system (2) when Ω is an unbounded domain and the function f has a superlinear growth at infinity. As in [3] we will consider the situation in which Ω\ω is a bounded set. In the study of the null controllability of semilinear parabolic equations with superlinear nonlinearities, additional technical difficulties arise. As in [27, 6, 9, 5, 3]..., the technique to deal with the null controllability problem for system (2) combines a fixed point reformulation of this one. Due to the superlinear growth of the nonlinearity f , we will solve the null controllability problem for (1) with controls in L∞ (Q) and with explicit bounds of the L∞ -norm of this control with respect to T , ||a||∞ and ||B||∞ . All along this paper we will assume that Ω ⊂ RN is an unbounded connected open set with boundary at least of class C 0,1 uniformly, such that (5)

DΩ (−∆) = H 2 (Ω) ∩ H01 (Ω).

This “non classical” assumption is needed to give examples in unbounded set that are the cartesian product of regular sets. Observe that even in the simple case in which Ω is the cartesian product of two (bounded or unbounded) intervals, Ω is only a C 0,1 set, however it is a “good one” in the sense that (5) holds. We will also suppose that ω ⊂ Ω is a nonempty open subset such that there 3

exist an open set ω0 and a function η 0 satisfying (6)

ω0 ⊂ ω ⊂ Ω

with d0 = dist (ω0 , Ω \ ω) > 0,

and, for two positive constants C0 and C1 ,  0 η ∈ C 2 (Ω), η 0 ≥ 0 in Ω,      |∇η 0 | ≥ C0 > 0 in Ω \ ω0 , (7) X ¯ ¯ ∂η 0   ¯Dβ η 0 ¯ ≤ C1 in Ω.  ≤ 0 on ∂Ω,   ∂n |β|≤2

As said above, we first prove a global Carleman inequality for the linear problem  N X  ∂Fi  −ϕ − ∆ϕ = F + in Q, t 0 ∂xi (8) i=1   ϕ = 0 on Σ, ϕ(x, T ) = ϕ0 (x) in Ω, where F0 , Fi ∈ L2 (Q) (1 ≤ i ≤ N ) and ϕ0 ∈ L2 (Ω). To this end, let us introduce (9)

α(x, t) =

eλm||η

0

||∞

− eλ(m||η t(T − t)

0

||∞ +η 0 (x))

eλ(m||η ||∞ +η t(T − t) 0

,

ξ(x, t) =

0

(x))

with s and λ positive real numbers and m > 1 is a fixed real number. The first important result of this paper is the following one: Theorem 1.1. Under assumptions (5), (6) and (7), there exist three positive constants σ1 , λ1 ≥ 1 and C1 only depending on C0 , C1 and d0 such that, for every ϕ0 ∈ L2 (Ω) and F0 , Fi ∈ L2 (Q) (1 ≤ i ≤ N ), the corresponding solution to (8) satisfies ZZ ZZ  2 −2sα 2 3 4   sλ e ξ|∇ϕ| + s λ e−2sα ξ 3 |ϕ|2 ≤    Q Q  Ã   ZZ ZZ   3 4 −2sα 3 2 C s λ e ξ |ϕ| + e−2sα |F0 |2 + 1 (10) ω×(0,T ) Q    !  N ZZ  X   2 2 −2sα 2 2  s λ e ξ |Fi | ,   i=1

Q

for all s ≥ s1 = σ1 (T + T 2 ) and λ ≥ λ1 . In (10), the functions α and ξ are given by (9). Remark 1.1. The inequality (10) was proved in [15] when Ω is a regular bounded domain and it is based on a similar Carleman inequality for the heat equation with a right-hand side in L2 (Q) (see [12] for the proof in the bounded case and [9] for the precise way s1 depends on T ). On the other hand, if Ω is a bounded open set of class C 2 and ω ⊂ Ω then conditions (5)–(7) are readily fulfilled. Indeed, it is enough to consider ω0 ⊂⊂ ω and apply Lemma 1.1 of [12].

4

Our second main result is the following one: Theorem 1.2. Under assumptions of Theorem 1.1, given y0 ∈ L2 (Ω), there exists a control vb ∈ L2 (Q), with Supp vb ⊂ ω × [0, T ], such that the solution yb of system (1) satisfies yb(x, T ) = 0 in Ω. Moreover, the control vb can be chosen in such a way that (11)

||b v ||2L2 (Q) ≤ eCH(T,||a||∞ ,||B||∞ ) ||y0 ||2L2

where C is a positive constant only depending on C0 , C1 and d0 , and H is given by H (T, ||a||∞ , ||B||∞ ) ≡ 1 +

1 2 + ||a||2/3 ∞ + T ||a||∞ + (1 + T )||B||∞ . T

Theorem 1.2 provides a positive result on null controllability of system (1) for open sets Ω and ω under the technical assumptions (5), (6) and (7). The key point in the proof of this result is to obtain an appropriate observability inequality for the corresponding adjoint problem ( −ϕt − ∆ϕ − ∇ · (Bϕ) + aϕ = 0 in Q, (12) ϕ = 0 on Σ, ϕ(x, T ) = ϕ0 (x) in Ω, with ϕ0 ∈ L2 (Ω). It is by now a classical result that inequality (10) implies an observability inequality for the solutions to the adjoint problem (12), i.e.: ZZ 2 CH(T,||a||∞ ,||B||∞ ) (13) ||ϕ(·, 0)||L2 ≤ e |ϕ|2 , ω×(0,T )

where C is a positive constant only depending on C0 , C1 and d0 and H is as in Theorem 1.2 (in fact, the positive constant C appearing in (11) is the same constant that appears in (13)) and this one proves the existence of a control vb ∈ L2 (Q) which satisfies Theorem 1.2 (see for instance [9]). As said above, we are interested in solving the null controllability problem of system (1) with control functions in L∞ (Q). Our next result establishes the existence of this L∞ -control: Theorem 1.3. Assume that hypotheses (5)–(7) hold. In addition, assume that ∂ω ∩ ∂Ω is of class C 2 uniformly when ∂ω ∩ ∂Ω 6= ∅. Then, given y0 ∈ L2 (Ω), there exists a control v ∈ L2 (Q) ∩ L∞ (Q) such that the corresponding solution y to (1) satisfies (4). Furthermore, the control v can be chosen such that (14)

||v||L2 (Q) + ||v||∞ ≤ eCK(T,||a||∞ ,||B||∞ ) ||y0 ||2L2

where C is a positive constant only depending on Ω and ω, and K is given by K (T, ||a||∞ , ||B||∞ ) ≡ 1 + T +

1 2 + ||a||2/3 ∞ + T ||a||∞ + (1 + T )||B||∞ . T 5

Remark 1.2. The proof of Theorem 1.3 is a consequence of the regularizing effect of the heat equation. This idea has been used in several previous papers, for instance, in [1], [7], [10], [5] and [2]. The proof given in [10] and [5] cannot be adapted to the case in which Ω and ω are unbounded sets such that |Ω| = |ω| = ∞ because they use the fact that Lr (Ω) ⊂ Ls (Ω) (1 ≤ s < r ≤ ∞) when |Ω| < ∞. To be precise, in [10] and [5] the refined observability inequality ÃZ Z !2 ||ϕ(·, 0)||2L2 ≤ C

|ϕ| ω×(0,T )

is proved for the solutions ϕ to the adjoint problem (12). When |ω| = ∞ this last inequality is not enough to ensure the existence of a control v in L∞ (Q) which leads to zero the solution y to system (1). In this paper we use the technique introduced in [2] for the study of the existence of insensitizing controls for the heat equation. This technique allows to build a L∞ -control v which gives the null controllability of the linear system (1) starting from a L2 -control vb which also drives to zero system (1). Furthermore, we can estimate the L∞ -norm of the new control v with respect to ||b v ||L2 (Q) . This paper is also devoted to the study of the null controllability of the semilinear heat equation (4) when Ω is an unbounded open set and ω is a nonempty subset of Ω such that Ω \ ω is a bounded set. We shall assume that f : R × RN → R is a locally Lipschitz-continuous function such that f (0, 0) = 0 and, therefore, we can write f (s, p) = g(s, p)s + G(s, p) · p

∀(s, p) ∈ R × RN ,

for some L∞ loc functions g and G. Our last result is the following one: Theorem 1.4. Let Ω be an unbounded domain of class C 2 uniformly, and let ω be a nonempty open subset of Ω such that Ω \ ω is bounded. Assume that f is a locally Lipschitz-continuous function, f (0, 0) = 0 and (15)

lim

|(s,p)|→∞

|g(s, p)| log

3/2

(1 + |s| + |p|)

= 0,

lim

|(s,p)|→∞

|G(s, p)| log

1/2

(1 + |s| + |p|)

= 0.

Then, if y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω), there exists a control v ∈ L2 (Q) ∩ L∞ (Q) such that (2) admits a weak solution y ∈ C 0 ([0, T ]; L2 (Ω)) which satisfies (4). Theorem 1.4 is a consequence of Theorem 1.3 and will be proved using an appropriate fixed point reformulation of the null controllability problem. As in [3] and due to the fact that Ω is unbounded (and then, the compactness of the Sobolev imbedding theorems fails), technical difficulties arises which will be solved thanks to the assumption on Ω \ ω. Remark 1.3. Theorem 1.4 generalizes the result on null controllability of system (4) obtained in [3] when f is a globally Lipschitz-continuous function. On the other hand, in [5] the null controllability of system (4) was proved for nonlinearities f satisfying (15) but in regular bounded domains Ω. 6

Remark 1.4. In [10] it is proved that, if Ω is a bounded domain and ω 6= Ω, for each β > 2 there exist functions f = f (s) with f (0) = 0 and f (s) ∼ |s| logβ (1 + |s|)

as s → ∞

such that (4) is not null controllable in any time T > 0. This negative result also apply when Ω is an unbounded domain. The rest of the paper is organized as follows. In the next section we give examples of unbounded sets Ω and ω such that (5)–(7) hold. Section 3 is devoted to the proof of Theorem 1.1. In Section 4 we prove Theorems 1.2 and 1.3. Section 5 is devoted to prove the null controllability of system (2) when the nonlinearity has a slightly super linear growth (see Theorem 1.4) and when Ω \ ω is bounded. Finally in Section 6 we present some further results and open problems and conclude the paper with two appendices. The first one contains the proof of an auxiliary Carleman inequality used in the proof of Theorem 1.1. The second one is devoted to the proof of a regularity result used in the proof of Theorem 1.3.

2

Examples

This section is devoted to give some examples of unbounded sets Ω and ω which satisfy assumptions (5)–(7), and therefore the observability inequality (13). Some of them already existed in the literature, but the construction of the function η 0 was not explicit. In addition, except in [3], the observability inequality for the linear equation (12) was only obtained for a(x, t) ≡ 0 and B(x, t) ≡ 0. Example 3 is of particular importance because we exhibit sets Ω and ω where the observability inequality (13) holds and Ω \ ω is not of finite measure. Moreover, as far as we know, this particular example is the first time that appears in the literature of the field. Example 1. Here we present the construction of the η0 function in the case contemplated in [3], that is, let Ω ⊂ RN be an unbounded connected open set with boundary ∂Ω of class C 2 uniformly and ω ⊂ Ω satisfy Ω\ω

is bounded,

then, hypotheses (5), (6) and (7) are fulfilled. Indeed, we fix d0 > 0 and consider ω0 = {x ∈ ω : dist (x, Ω \ ω) > d0 }. Then, ω0 ⊂ ω ⊂ Ω, Ω \ ω 0 is also bounded and dist (ω0 , Ω \ ω) = d0 and therefore, condition (6) is fulfilled. In addition, there exist ω1 and ω2 open subsets and O a bounded open subset such that ω2 ⊂ ω1 ⊂ ω0 ,  Ω \ ω 0 ⊂ Ω \ ω 1 ⊂ Ω \ ω 2 are bounded open sets,     d /2 ≤ dist (ω , Ω \ ω ) ≤ d , d /2 ≤ dist (ω , Ω \ ω ) ≤ d , 0 1 0 0 2 0 0 1 2 e  Ω = Ω \ ω 2 ∈ C and O ⊂⊂ ω0 \ω 1 such that    e O has a non empty open subset in each connected component of Ω. 7

e is a regular bounded open set and O ⊂⊂ Ω, e we can Due to the fact that Ω apply Lemma 1.1 of [12] and deduce the existence of a function ηe0 such that ηe0 ∈ C 2 (RN ),

e ηe0 > 0 in Ω,

e and ηe0 = 0 on ∂ Ω

e \ O. |∇e η0 | ≥ C0 > 0 in Ω

On the other hand, since dist (ω2 , Ω \ ω 1 ) > 0, we infer that there exists ρe ∈ C02 (RN ) satisfying 0 ≤ ρe ≤ 1 in RN ,

Supp ρe ∩ ω 2 = ∅

and

ρe ≡ 1 in Ω \ ω 1 .

Let us consider the function η 0 defined as η 0 ≡ ηe0 ρe. As a consequence of the previous properties, it is readily checked that η 0 together with ω0 satisfy condition (7). Indeed, η 0 ∈ C02 (RN ), η 0 ≥ 0 in Ω, ∇η 0 |Ω\ω ≡ ∇e η 0 and then 0

|∇η0 | ≥ C0 > 0 in Ω \ ω 0 . Finally, η 0 = 0 on ∂Ω that together to property η 0 ≥ 0 in Ω gives ∂η 0 /∂n ≤ 0 on ∂Ω. Example 2. Here we recover the example presented in [4]. In this case, the obtention of the Carleman inequality allows to work with the linear equation (1) with potentials a, B 6≡ 0. In fact, conditions (5)–(7) are fulfilled when Ω = (0, ∞) ⊂ R (resp., Ω = R) and one considers [ [ ω= (an , bn ), (resp., ω = ((an , bn ) ∪ (a−n , b−n ))), n≥1

n≥1

where, for every n ≥ 1, 0 < an < bn < an+1 , lim an = lim bn = ∞ (resp., 0 > a−n > b−n > a−n−1 , lim a−n = lim b−n = −∞) and ( ( b−n − a−n−1 ≤ M < ∞ an+1 − bn ≤ M < ∞ ). (resp., a−n − b−n ≥ m > 0 bn − an ≥ m > 0 In order to prove it, we will consider the case Ω = (0, ∞). Let us set [  (an + m/8, bn − m/8), a0 = −m, b0 = 0,  ω0 = 

n≥1

An = an + m/2

and

Bn = bn − m/2,

∀n ≥ 0.

It can be easily checked that, for every n ≥ 0, an < An ≤ Bn < bn and condition (6) is satisfied for d0 = m/8. On the other hand, let us introduce the function β 0 given by β 0 (x) = 0 in (−∞, A0 ] and, for k ≥ 0,  if x ∈ [A2k , B2k ],   0    x − B 2k   if x ∈ [B2k , A2k+1 ],  A2k+1 − B2k 0 β (x) =  1 if x ∈ [A2k+1 , B2k+1 ],     A − x  2k+2   if x ∈ [B2k+1 , A2k+2 ], A2k+2 − B2k+1 8

which satisfies ¯ 0 ¯ ¯ dβ ¯ 1 1 ≤ ¯¯ (x)¯¯ ≤ , M +m dx m

∀x 6= An , Bn .

In order to verify condition (7), we define η 0 (x) = β 0 (x)ψ(x), x ∈ R, with ψ ∈ C 2 (R) a cut-off function satisfying 0 ≤ ψ(x) ≤ 1, ψ ≡ 0 in (−∞, B0 + m/8],    ψ ≡ 0 in [An − m/8, Bn + m/8] ∀n ≥ 0, ψ ≡ 1 in [bn − m/4, an+1 + m/4] ∀n ≥ 0 and   0 |ψ (x)| ≤ C/m, |ψ 00 (x)| ≤ C/m2 , ∀x ∈ R, where C is a positive constant. It is not difficult to see that η 0 ∈ C 2 (R), ∂η 0 /∂n(0) < 0, 0 ≤ η 0 ≤ 1 and, for a new positive constant C, |dη 0 /dx| ≤ C/m and |d2 η 0 /dx2 | ≤ C(1 + 1/m2 ) in R. Finally, let us point out that, if x ∈ Ω \ ω 0 , then ψ ≡ 1 in a neighborhood of x and therefore, ¯ 0 ¯ ¯ 0 ¯ ¯ dη ¯ ¯ dβ ¯ 1 ¯ ¯=¯ ¯≥ (x) (x) ¯ dx ¯ ¯ dx ¯ M + m , ∀x ∈ Ω \ ω 0 In conclusion, conditions (6) and (7) are fulfilled. Example 3. The example we are going to present is, as far as we know, authentically new. We give an example of unbounded sets Ω and ω such that Ω \ ω is an unbounded set with infinite measure and where the observability inequality holds. In fact, let Ω be an unbounded open set uniformly of class C 2 . The set ω is going to be constructed from Ω by eliminating an “infinite strip” around the boundary. That is, ω = {x ∈ Ω : dist (x, ∂Ω) > δ0 } for some δ0 > 0 that is going to be fixed later. For the choice of δ0 we will need to recall some geometric results and properties which are inherent to uniformly C 2 unbounded sets. To this aim, we will use results appearing in the paper of Fornaro et al. [11]. We recall that a set Ω satisfies a uniform interior sphere condition if for each point y0 ∈ ∂Ω there exists a ball By0 , depending on y0 , contained in Ω and such that B y0 ∩ ∂Ω = {y0 }, moreover the radii of these balls are bounded from below by a positive constant. The following results are extracted from [11]: Proposition 2.1. If ∂Ω is uniformly of class C 2 , then it satisfies a uniform interior sphere condition. Proposition 2.2. Assume that ∂Ω is uniformly of class C 2 and let δb be a positive constant such that at each point of ∂Ω there exists a ball which satisfies b Then: the interior sphere condition at y0 with radius greater or equal to δ. b there exists a unique 1. For every x ∈ Ωδb = {y ∈ Ω : dist (y, ∂Ω) < δ} ξ = ξ(x) ∈ ∂Ω such that |x − ξ| = dist (x, ∂Ω). 9

2. Let us define the function d(x) = dist (x, ∂Ω), then d ∈ Cb2 (Ωδb). 3. ∇d(x) = −n(ξ(x)) for every x ∈ Ωδb and where n(ξ) denotes the unit outward normal vector to ∂Ω at point ξ. With the previous results in mind, we are now able to define δ0 and prove that the sets Ω and ω satisfy (6) and (7) (obviously, (5) holds). In fact, let us b and take δ0 = δb − ε, where δb > 0 is as in Proposition 2.2 and ε ∈ (0, δ), ω0 = {x ∈ Ω : dist (x, Ω) > δb − ε/2}. It is easy to see that (6) holds with d0 ≡ ε/2. On the other hand, we define ( d(x) ≡ dist (x, ∂Ω) if x ∈ Ω \ ω0 , η˜0 (x) = b δ − ε/2 if x ∈ ω0 . Let us remark that Ω \ ω0 ⊂ Ωδb and dist (Ω \ ω0 , Ω \ Ωδb) = ε/2 > 0. We can then consider ρ ∈ C 2 (Ω) with 0 ≤ ρ(x) ≤ 1 in Ω, ρ(x) ≡ 1 in Ω \ ω0 , Supp ρ ⊂ Ωδb and |Dα ρ(x)| ≤ C/ε|α| for every x ∈ Ω and α ∈ NN with |α| ≤ 2. Finally, if we define η 0 (x) = η˜0 (x)ρ(x) and taking into account Propostion 2.2 and the previous properties we readily obtain that η 0 also satisfies condition (7). Example 4. Let us suppose that Ω ⊂ RN is an open set which satisfies (5), ω is an open subset of Ω and (Ω, ω) satisfies (6) and (7) with constants C0 , C1 and d0 . Then, if O ⊂ RM (M ≥ 1) is another open set with boundary of class C 0,1 e = Ω × O ⊂ RN +M satisfies (5), then uniformly, and such that the open set Ω e Ω with ω e = ω × O also satisfy assumption (6) and (7) with the same constants (and therefore, with constants which are independent of O). For the proof, let us assume that (Ω, ω) satisfies (6) and (7) for an open set ω0 ⊂ ω, a function η 0 ∈ C 2 (RN ) and positive constants d0 , C0 and C1 . Let O ⊂ RM be an open set of class C 0,1 uniformly. We define ω e0 = ω0 × O e \ω and ηe0 (x, y) = η 0 (x) for (x, y) ∈ RN +M . Then, one has dist (e ω0 , Ω e ) = d0 , 0 2 N +M 0 N +M ηe ∈ C (R ), ηe ≥ 0 in R and ¯ X ¯¯ βe X ¯ ¯ ¯ e ¯Dxβ η 0 (x)¯ ≤ C1 ∀(x, y) ∈ Ω. ¯D(x,y) ηe0 (x, y)¯ = e |β|≤2

|β|≤2

On the other hand, |∇(x,y) ηe0 (x, y)| = |∇x η 0 (x)| and, therefore |∇e η 0 | ≥ C0 > 0 e \ω in Ω e0 . Finally, ∂e η 0 /∂n(x, y) = ∂η 0 /∂nx (x) on ∂Ω × O and ∂e η 0 /∂n(x, y) = 0 0 e on Ω × ∂O and so, ∂e η /∂n(x, y) ≤ 0 on ∂ Ω. e In conclusion, the couple (Ω, ω e ) also satisfy assumption (6) and (7) with the same constants. Example 5. The previous example can be readily generalized as follows: Let us assume that Ω1 ⊂ RN , Ω2 ⊂ RM (N, M ≥ 1) are two open sets satisfying condition (5) and such that for Ω = Ω1 × Ω2 condition (5) holds. Suppose that ω1 ⊂ Ω1 , ω2 ⊂ Ω2 are two open subsets such that (Ω1 , ω1 ) and (Ω2 , ω2 ) satisfy (6) and (7). Then, conditions (6) and (7) come true for (Ω, ω) with ω = ω1 × ω2 . 10

Remark 2.1. The previous examples generalize some known results on null controllability of system (1) in unbounded domains that have been studied before. Firstly, we obtain again the null controllability result of the linear system (1) established in [3] when (Ω, ω) satisfies conditions of example 1. Secondly, by means of examples 4 and 5, we generalize the positive results on null controllability of the heat equation obtained in [21]. Observe that our approach permits to consider in system (1) general coefficients a ∈ L∞ (Q) and B ∈ L∞ (Q)N . Finally, we have considered the example 1 of [4] (the previous example 2) obtaining the same null controllability result as [4] but for general potential a and B. Let us remark that, apart from example 1, the null controllability result for system (1) proved in [4] (with a ≡ 0 and B ≡ 0) are of a different nature from ours. Observe that Cannarsa et al. show the null controllability of system (1) when either y0 or the control force v belong to appropriate weighted L2 -espaces. On the other hand, it is not difficult to see that example 2 and example 3 of [4] do not satisfy conditions (6) and (7). From our point of view, obtaining the kind of results proved in [4] by means of a Carleman inequality for the adjoint system seems to be complicated. Remark 2.2. In relation to example 4, let us observe that Theorem 1.1 and e = Ω × (0, ε) and Ω ⊂ RN is a Theorem 1.2 has been obtained in [25] when Ω regular bounded domain. To be precise, in [25] the authors consider problem (1) with B ≡ 0 and with Neumann boundary conditions at both ends and they prove a global Carleman inequality for the adjoint problem (and therefore the null controllability of the problem) with constants which are independent of ε. This uniform property allows them to pass to the limit and show that the limit of the control functions is a control function for the limit problem. The uniform Carleman estimate obtained in [25] has been the starting point which has permitted us to state Theorem 1.1 in a general framework. Remark 2.3. Theorems 1.1 and 1.2 provide, respectively, a global Carleman inequality for system (8) and a null controllability result for system (1), for some (bounded or unbounded) domains Ω only of class C 0,1 uniformly. This is the case QN when we consider a bounded hypercube Ω = i=1 (ai , bi ) (ai , bi ∈ R with ai < bi for every 1 ≤ i ≤ N ) and a nonempty subset ω ⊂ Ω. Indeed, condition (5) is fulfilled since Ω is a convex set. On the other hand, conditions (6)–(7) can be QN QN easily checked if we consider ω0 = i=1 (a0i , b0i ) ⊂⊂ ω and η 0 (x) = i=1 ηi0 (xi ), with η0i ∈ C 2 ([ai , bi ]) satisfying (6) and (7) for the open sets (ai , bi ) and (a0i , b0i ), 1 ≤ i ≤ N. By means of Example 5 we will give another example of domain Ω of class C 0,1 uniformly such that (Ω, ω) satisfies Theorems 1.1 and 1.2. Let us consider O ⊂ RN , an unbounded domain of class C 2 uniformly, and ϑ ⊂ O such that (O, ϑ) is as in Example 1 or 3. On the other hand, let us set Ω = O × (α, ∞) ⊂ RN +1 and ω = ϑ × (α0 , ∞), where α, α0 ∈ R and α < α0 . Thus, we know that (Ω, ω) satisfies conditions (6) and (7) (see Example 5), but also (5). Indeed, let us see that if u ∈ H01 (Ω) and f ≡ ∆u ∈ L2 (Ω), then u ∈ H 2 (Ω). To this end,

11

we define u e(x, s) =

½

u(x, s) if s > α, −u(x, 2α − s) if s ≤ α,

½ and fe(x, s) =

f (x, s) if s > α, −f (x, 2a − s) if s ≤ α.

It is not difficult to check that u e ∈ H01 (O × R), fe ∈ L2 (O × R) and ∆e u = fe −1 in H (O × R). Taking into account that O × R is an open set of class C 2 uniformly, we deduce that u e ∈ H 2 (O × R) and so, u ∈ H 2 (Ω). Evidently, other open subsets ω could be considered, for instance, ω = ϑ × ω0 with ω0 ⊂ (α, ∞) as in Example 2.

3

The global Carleman inequality. Proof of Theorem 1.1

This section is devoted to prove the global Carleman inequality stated in Theorem 1.1. To this end, firstly we will state a global Carleman inequality for the heat equation with a right-hand side in L2 (Q) (see Lemma 3.1). Then, as a consequence of Lemma 3.1 and following [15], we will show Theorem 1.1. The proof of Lemma 3.1 is by now standard. The proof of this lemma follows the steps of the corresponding proofs given in [12] and [9] (also see [8]). It is not difficult to see that, under assumptions (5), (6) and (7), the proofs given by the previous authors are still valid and the fact that Ω is unbounded does not play any part. In the sequel, unless otherwise specified, C will stand for a generic positive constant only depending on C0 , C1 and d0 (the positive constant appearing in (6)–(7)) whose value may change from line to line. Let us consider the linear heat equation ( −ϕt − ∆ϕ = F in Q, (16) ϕ = 0 on Σ, ϕ(x, T ) = ϕ0 (x) in Ω, where F ∈ L2 (Q) and ϕ0 ∈ L2 (Ω). One has the following result ([12], [9], [8]): Lemma 3.1. Under assumption (5), (6) and (7), there exist three positive constants σ0 , λ0 and C0 (only depending on C0 , C1 and d0 ) such that, for every ϕ0 ∈ L2 (Ω) and F ∈ L2 (Q), the corresponding weak solution to (16) satisfies ZZ ZZ  ¡ ¢ −1 −2sα −1 2 2 2   s e ξ |ϕt | + |∆ϕ| + sλ e−2sα ξ|∇ϕ|2    Q Q  Ã   ZZ ZZ  −2sα 3 2 3 4 3 4 e−2sα ξ 3 |ϕ|2 e ξ |ϕ| ≤ C s λ +s λ (17) 0  ω×(0,T ) Q   ¶  ZZ     + e−2sα |F |2  Q

for all s ≥ s0 = σ0 (T +T 2 ) and λ ≥ λ0 . The functions α and ξ are given by (9).

12

Remark 3.1. It is interesting to point out that the local terms appearing in the right-hand side of inequality (17) could be obtained in an open subset ω1 instead of ω, provided ω0 ⊂ ω1 ⊂ ω

and

d1 = dist (ω0 , Ω \ ω 1 ) > 0.

In this case, the positive constants σ0 , λ0 and C0 appearing in (17) only depend on C0 , C1 and d1 . Proof of Theorem 1.1: We will prove Theorem 1.1 following the technique developed in [15]. This strategy allows to obtain a global Carleman inequality for the heat equation with a right-hand side in L2 (0, T ; H −1 (Ω)) from a Carleman inequality for the heat equation with a second member in L2 (Q). Before begining the proof of Theorem 1.1, let us remark that problem (8) admits a unique weak solution ϕ ∈ L2 (0, T ; H01 (Ω)) ∩ C 0 ([0, T ]; L2 (Ω)) which satisfies ZZ Z T (18) ϕ(x, t)G(x, t) dx dt = − hF0 + ∇ · F , zi dt + (ϕ0 , z(T ))L2 (Ω) , Q

0

where h·, ·i denotes the duality product between H −1 (Ω) and H01 (Ω), G ∈ L2 (Ω) and z is the solution to the linear problem ( zt − ∆z = G in Q, z = 0 on Σ, z(0) = 0 in Ω. On the other hand, it is not difficult to see that, thanks to (7), we have  0   ∂i α = −∂i ξ ≡ −λ∂i η ξ ≤ C1 λξ, αt ≤ T ξ 2 , αtt ≤ CT 2 ξ 3 , ξt ≤ T ξ 2 , (19)   |∇αt | ≤ C1 T λξ 2 , and, if s ≥ CT 2 and l1 ≤ l2 , µ (20)

sl1 ξ(x, t)l1 ≤

1 4C

¶l1 −l2

sl2 ξ(x, t)l2 ≡ Csl2 ξ(x, t)l2

∀(x, t) ∈ Q.

We will do the proof in two steps: Step 1. An auxiliary null controllability problem Let s and λ be as in Lemma 3.1 and let us introduce the null controllability problem: Given f ∈ L2 (Q), find u ∈ L2 (Q) such that the solution z ∈ L2 (0, T ; H01 (Ω)) to ( zt − ∆z = s3 λ4 e−2sα ξ 3 f + u1ω in Q, z = 0 on Σ, z(0) = 0 in Ω.

13

satisfies z(x, T ) = 0 in Ω and ZZ ZZ ZZ −3 −4 2sα −3 2 2sα 2 3 4 s λ e ξ |u| + e |z| ≤ Cs λ e−2sα ξ 3 |f |2 . Q

Q

Q

(C is the positive constant appearing in (17) and then, only depending on C0 , C1 and d0 ). It is well know that inequality (17) implies the existence of such a control (see for instance [12], [15], [8], ...). In fact, if for each f ∈ L2 (Q) there exist a state z and a control u that solve the null controllability problem and satisfy the previous inequality, then inequality (17) for the solutions to (16) holds. Therefore, we can apply the previous considerations to the solution ϕ ∈ L2 (Q) to (8) and deduce the existence of a control u b ∈ L2 (Q) and a state 2 1 zb ∈ L (0, T ; H0 (Ω)) such that ( zbt − ∆b z = s3 λ4 e−2sα ξ 3 ϕ + u b1ω in Q, (21) zb = 0 on Σ, zb(0) = zb(T ) = 0 in Ω, and (22)

ZZ

ZZ

s−3 λ−4

ZZ

e2sα ξ −3 |b u|2 +

e2sα |b z |2 ≤ Cs3 λ4

Q

Q

e−2sα ξ 3 |ϕ|2 , Q

with s and λ such that s ≥ σ0 (T + T 2 ) and λ ≥ λ0 ≥ 1 (σ0 and λ0 furnished by Lemma 3.1). Now, if we multiply by s−2 λ−2 e2sα ξ −2 zb the equation satisfied by zb and we integrate by part, we infer ZZ ZZ   2sα −2 2 −2 −2 −2 −2  s λ e ξ |∇b z | = −s λ e2sα ξ −2 zb zbt    Q Q  ZZ ZZ   2sα −1 0 −1 −1 + 2s λ e ξ ∇η · ∇b z zb + 2s−2 λ−1 e2sα ξ −2 ∇η 0 · ∇b z zb  Q Q   ZZ ZZ     + sλ2 ξϕb z + s−2 λ−2 e2sα ξ −2 u b zb = H1 + H2 + H3 + H4 + H5 .  Q

Q

Due to (7), (19), (20) and the fact that s ≥ σ0 (T + T 2 ) and λ ≥ λ0 ≥ 1, we can estimate each term in the previous equality as follows: ZZ ZZ  ¡ 2sα −2 ¢ 1 −2 −2 2 −1 −2   H = − s λ e ξ |b z | ≤ Cs λ T e2sα |b z |2  1 t 2 Q Q ZZ  2sα 2   ≤C e |b z| , Q

ZZ 1 −2 −2 e2sα ξ −2 |∇b z |2 , H2 + H3 ≤ C e |b z| + s λ 2 Q Q ¶ µZ Z ZZ e2sα ξ 3 |ϕ|2 H4 ≤ C e2sα |b z | 2 + s 3 λ4 ZZ

2sα

2

Q

Q

14

and

µZ Z

¶ e2sα ξ −3 |b u|2 .

ZZ e2sα |b z |2 + s−3 λ−4

H5 ≤ C Q

Q

Thus, we obtain µ ZZ ZZ ZZ  2sα −2 2 −3 −4 2sα −3 2 −2 −2   s λ e ξ |∇b z | ≤ C s λ e ξ |b u | + e2sα |b z |2  Q Q Q ¶ ZZ  2sα 3 2  3 4  +s λ e ξ |ϕ| , Q

that together with inequality (22) gives ZZ ZZ  2sα −3 2 −3 −4   s λ e ξ |b u | + e2sα |b z |2  Q Q ZZ ZZ (23)  2sα −2 2 3 4 −2 −2   e−2sα ξ 3 |ϕ|2 e ξ |∇b z | ≤ Cs λ +s λ Q

Q

where s ≥ σ0 (T + T 2 ) and λ ≥ λ0 ≥ 1. Step 2. Proof of inequality (10) We begin applying inequality (18) for zb, the solution to (21), and the function G = s3 λ4 e−2sα ϕ + u b1ω , obtaining ZZ ZZ ZZ ZZ s3 λ4 e−2sα ξ 3 |ϕ|2 = − F0 zb + F · ∇b z+ ϕ1ω u b Q

µZ Z

e−2sα |F0 |2 + s2 λ2

≤ µZ Z

Q

ZZ

Q

e

ZZ 2sα

2

|b z| + s

Q

Q

Q

ZZ

e−2sα ξ 2 |F |2 + s3 λ4 Q

−2 −2

Q

e

ξ

2

|∇b z| + s

−3 −4

λ

Q

× ¶1/2

ZZ

2sα −2

λ

¶1/2

e−2sα ξ 3 1ω |ϕ|2 e

2sα −3

ξ

|b u|

2

.

Q

Now, taking into account (23), we deduce µZ Z  ZZ ZZ  −2sα 3 2 −2sα 2 2 2 3 4  s λ e ξ |ϕ| ≤ C e |F | + s λ e−2sα ξ 2 |F |2  0  Q Q Q ! ZZ (24)  −2sα 3 2 3 4  +s λ e ξ |ϕ| ,   ω×(0,T )

for s ≥ σ0 (T + T 2 ), λ ≥ λ0 ≥ 1 and C the constant appearing in (23). Finally, from this last inequality, it is not difficult to deduce (10). Indeed, if we multiply by sλ2 e−2sα ξϕ the equation satisfied by ϕ (see (8)), we infer ZZ ZZ  ¡ ¢¢ ¡¡ −2sα ¢ 1 2 −2sα 2 2   e ξ|∇ϕ| = − sλ e ξ t − ∆ e−2sα ξ |ϕ|2 sλ  2 Q Q ZZ ZZ ¡ ¢  −2sα 2   e−2sα F · ∇ e−2sα ξ ϕ. e (ξF0 ϕ − ξF · ∇ϕ) − sλ2 + sλ Q

Q

15

We use now the estimates  ¡ −2sα ¢ ¡ ¢ ξ t | ≤ T e−2sα 2sξ 3 + ξ 2 ≤ 3s2 λ2 e−2sα ξ 3 ,   | e¡ ¢ |∆ e−2sα ξ | ≤ (3C1 + 11C21 )s2 λ2 e−2sα ξ 3 ,  ¡ ¢  |∇ e−2sα ξ | ≤ 3C1 sλ e−2sα ξ 2 , which are valid for (x, t) ∈ Q, s ≥ T + T 2 /4 and λ ≥ 1 (for obtaining these inequalities we have taken into account (7), (19) and (20)). Thus, ZZ ZZ ZZ sλ2 e−2sα ξ|∇ϕ|2 ≤ Cs3 λ4 e−2sα ξ 3 |ϕ|2 + Cs−1 e−2sα ξ −1 |F0 |2 Q Q Q ZZ ZZ 1 2 −2sα 2 2 + sλ e ξ|∇ϕ| + Csλ e−2sα ξ|F |2 2 Q Q that, together with (24), gives (10) (we have again used (20)). The theorem is now proved.

4

Null controllability of system (1). Proof of Theorems 1.2 and 1.3

This Section is devoted to prove the null controllability property of system (1). Firstly, we will show the result on null controllability with distributed controls in L2 (Q) (Theorem 1.2). Secondly, we will see that, starting from the previous L2 control, it is possible to construct a new distributed control v in L2 (Q) ∩ L∞ (Q) that also gives the null controllability of system (1).

4.1

The linear null controllability problem with controls in L2 (Q). Proof of Theorem 1.2

As said above, the proof of Theorem 1.2 is a standard consequence of the observability inequality (13) for the solutions to the adjoint system (12) established in the following proposition: Proposition 4.1. Let us assume that a ∈ L∞ (Q) and B ∈ L∞ (Q)N . Then, under assumptions of Theorem 1.1, there exist a positive constant C (only depending on C0 , C1 and d0 ) such that, for every ϕ0 ∈ L2 (Ω), the corresponding solution to (12) satisfies (13) (H(·, ·, ·) is as in Theorem 1.2). Proof: Let ϕ be the weak solution to (12) associated to ϕ0 ∈ L2 (Ω). In particular, ϕ is the solution to (8) corresponding to F = Bϕ ∈ L2 (Q)N and

16

F0 = −aϕ ∈ L2 (Q). Thus, we can apply Theorem 1.1 to ϕ and deduce ZZ ZZ  2 −2sα 2 3 4   sλ e ξ|∇ϕ| + s λ e−2sα ξ 3 |ϕ|2 ≤    Q Q  Ã   ZZ ZZ  C1 s3 λ4 e−2sα ξ 3 |ϕ|2 + ||a||2∞ e−2sα |ϕ|2 +  ω×(0,T ) Q   ¶  ZZ    2 2 2 −2sα 2 2  s λ ||B||∞ e ξ |ϕ| ,  Q

for all s ≥ s1 and λ ≥ λ1 ≥ 1 (the functions α and ξ are given by (9)). Now, if 1/3 2/3 we take s ≥ 4−2/3 C1 T 2 ||a||∞ and s ≥ C1 T 2 ||B||2∞ , we obtain µ ¶ ZZ ZZ  2 −2sα 2 2 2 2 −2sα 2 2   C ||a|| e |ϕ| + s λ ||B|| e ξ |ϕ| ∞ ∞  1 Q Q ZZ 1 3 4  −2sα 3 2   e ξ |ϕ| , ≤ s λ 2 Q and therefore, we infer the existence of a new positive constant σ2 only depending on the geometric constants C0 , C1 and d0 , such that (25) Z Z ZZ ZZ sλ2

e−2sα ξ|∇ϕ|2 + s3 λ4 Q

e−2sα ξ 3 |ϕ|2 ≤ 2C1 s3 λ4 Q

e−2sα ξ 3 |ϕ|2 ω×(0,T )

´ ³ 2/3 for all s ≥ s2 = σ2 T + T 2 + T 2 ||a||∞ + T 2 ||B||2∞ and λ ≥ λ1 . From now on, we fix λ = λ1 and thus α(x, t) =

α0 (x) , t(T − t)

ξ(x, t) =

α1 (x) , t(T − t)

with (see (9)) α0 (x) = e2λ1 m||η

0

||∞

− eλ1 (m||η

0

||∞ +η 0 (x))

,

α1 (x) = eλ1 (m||η

0

||∞ +η 0 (x))

;

(we recall that m > 1 is a fixed real number). It can be easily verified that  0 0  inf α0 (x) ≥ e2λ1 m||η ||∞ − eλ1 (m+1)||η ||∞ ≡ m0 > 0,  x∈Q 2λ m||η   sup α0 (x) ≤ e 1

0

||∞

− eλ1 m||η

x∈Q

0

||∞

≡ M0 ,

and hence, (26) 0 2 ∀(x, t) ∈ Q,  e−2sα ξ 3 ≤ e3λ1 (m+1)||η ||∞ 26 T −6 e−8m0 s/T 12  e−2sα ξ 3 ≥ e3λ1 m||η0 ||∞ 2 T −6 e25 M0 s/3T 2 ∀(x, t) ∈ Ω × [T /4, 3T /4], 33 ¡ ¢ whenever s ≥ 3T 2 /8m0 . Therefore, if s ≥ s3 = max s2 , 3T 2 /8m0 we have (25) and (26). 17

Analyzing the constants s2 and s3 , we deduce that s3 ≤ s4 , where s4 is of the form ´ ³ 2 2 (27) s4 = σ4 T + T 2 + T 2 ||a||2/3 ∞ + T ||B||∞ and σ4 only depending on C0 , C1 and d0 . We will also set s = s4 and, so, coming back to (25) (written for λ = λ1 and s = s4 ), we deduce ZZ ZZ −2s4 α 3 2 e ξ |ϕ| ≤ 2C1 e−2s4 α ξ 3 |ϕ|2 . Q

ω×(0,T )

Finally, taking into account (26) (for s = s4 ) and (27), we readily obtain that for a positive constant C, the inequality ZZ ZZ 2/3 2 1 |ϕ|2 ≤ eC (1+ T +||a||∞ +||B||∞ ) (28) |ϕ|2 Ω×[T /4,3T /4]

ω×(0,T )

holds for any weak solution ϕ to (12). The observability inequality (13) is a consequence of (28) and the energy inequality for ϕ: ´ d ³ (2||a||∞ +||B||2∞ )t e ||ϕ(t)||2L2 (Ω) ≥ 0, dt

(29)

valid for all t ≥ 0. Indeed, if we integrate this inequality with respect to time variable in [T /4, t], with t ∈ [T /4, 3T /4], we obtain  (2||a||∞ +||B||2∞ )(T /4−t) ||ϕ(T /4)||2 2  ||ϕ(t)||2 2 L (Ω) ≥ e L (Ω) ≥ e−(2||a||∞ +||B||∞ ) 2 ||ϕ(T /4)||2L2 (Ω)



2

T

for every t ∈ [T /4, 3T /4]. Integrating this last inequality again with respect to t, we find ZZ 2 T T ||ϕ(T /4)||2L2 (Ω) ≤ e(2||a||∞ +||B||∞ ) 2 |ϕ(x, t)|2 . 2 Ω×(T /4,3T /4) Now, if we integrate (29) in the time interval [0, T /4] we deduce ||ϕ(0)||2L2 (Ω) ≤ e(2||a||∞ +||B||∞ ) 4 ||ϕ(T /4)||2L2 (Ω) , 2

T

and then, ||ϕ(0)||2L2 (Ω)

2 3T 2 ≤ e(2||a||∞ +||B||∞ ) 4 T

ZZ |ϕ(x, t)|2 . Ω×(T /4,3T /4)

Combining this last inequality with inequality (28), we obtain the observability inequality (13). This completes the proof of Proposition 4.1.

18

Proof of Theorem 1.2: We fix ε > 0 and consider the functional ZZ Z 1 1 Jε (v) = |v|2 dx dt + |yv (x, T )|2 dx, 2 2ε Q Ω with v ∈ L2 (Q) and yv the solution to (1) associated to v. It is easy to check that the functional Jε is well defined, strictly convex and coercive in L2 (Q). Then, it attains its minimum at a unique vε ∈ L2 (Q). Let yε be the solution to (1) corresponding to vε . Then the minimum vε is characterized by vε = ϕε 1ω , where ϕε is the solution to (12) associated to 1 ϕ0 = − yε (T ). ε If we multiply by ϕε the equation satisfied by yε and by yε the equation satisfied by ϕε , we integrate in Q and we add both inequalities, we readily obtain ZZ Z Z vε (x, t)ϕε (x, t) dx dt = yε (x, T )ϕε (x, T ) dx − yε (x, 0)ϕε (x, 0) dx, Q

i.e., ZZ



1 |ϕε (x, t)| dx dt + ε ω×(0,T )



Z

Z 2

2

|yε (x, T )| dx = − Ω

yε (x, 0)ϕε (x, 0) dx. Ω

In particular, the Cauchy-Schwartz inequality gives  Z ZZ 1  2  |yε (x, T )|2 dx |ϕ (x, t)| dx dt +  ε ε Ω ω×(0,T )  1 1  CH(T,||a|| ,||B|| ) 2 −CH(T,||a|| ∞ ∞ ∞ ,||B||∞ )  ≤ e ||y0 ||L2 (Ω) + e ||ϕε (0)||2L2 (Ω) 2 2 where C is the constant appearing in (13) and H(T, ||a||∞ , ||B||∞ ) is provided by Proposition 4.1. Taking into account the observability inequality (13) satisfied, in particular, by ϕε , we deduce: ZZ Z 2 2 |ϕε (x, t)| dx dt + |yε (x, T )|2 dx ≤ eCH(T,||a||∞ ,||B||∞ ) ||y0 ||2L2 (Ω) . ε Ω ω×(0,T ) Since vε = ϕε 1ω , we obtain 2 ||vε ||2L2 (Q) + ||yε (·, T )||2L2 (Ω) ≤ eCH(T,||a||∞ ,||B||∞ ) ||y0 ||2L2 (Ω) , ε whence we infer that the sequences {vε }ε>0 and {ε−1/2 yε (·, T )}ε>0 are bounded in L2 (Q) and L2 (Ω) respectively. We can then extract a subsequence weakly convergent in L2 (Q) to some vb ∈ L2 (Q) which satisfies Supp vb ⊂ ω × [0, T ], inequality (11) and such that the corresponding solution to (1), yb, also satisfies (4). This finishes the proof of Theorem 1.1. 19

Remark 4.1. Observe that the positive constant C appearing in formula (11) is the same constant which appears in (13). This constant depends on (Ω, ω) through C0 , C1 and d0 (see (6) and (7)). On the other hand, it is possible to consider another open subset ω1 satisfying ω0 ⊂ ω1 ⊂ ω

and

d1 = dist (ω0 , Ω \ ω 1 ) > 0.

and try to solve the null controllability problem for system (1) with 1ω1 instead of 1ω . But this is possible if we take into account Remark 3.1 and the proofs of Proposition 4.1 and Theorem 1.2: Thus, under the previous assumptions, it is not difficult to check that there exists a control ve ∈ L2 (Q) with Supp ve ⊂ ω 1 × [0, T ] such that the solution ye of system (1) (with 1ω1 instead of 1ω ) satisfies (4). In addition, we can choose ve such that ||e v ||2L2 (Q) ≤ eCH(T,||a||∞ ,||B||∞ ) ||y0 ||2L2 holds with C a new positive constant that only depends on C0 , C1 and d1 (H(·, ·, ·) is as in Theorem 1.2).

4.2

The linear null controllability problem with controls in L2 (Q) ∩ L∞ (Q). Proof of Theorem 1.3

We will devote this section to prove the null controllability result for system (1) with control functions in L2 (Q) ∩ L∞ (Q). To this end, we will assume that (Ω, ω) satisfy (5)–(7) and the additional regularity hypothesis ω ∩ ∂Ω ∈ C 2 uniformly. We will prove Theorem 1.3 by adapting the technique introduced in [2]. This technique allows to construct new L∞ -controls which solve the null controllability problem for system (1) starting from L2 -controls. This strategy uses the local parabolic regularity effect of system (1). By reason of this, we will begin stating a technical result on parabolic regularity of the heat equation. Let O ⊂ RN be an open set with boundary ∂O of class C 2 uniformly and let u ∈ W (0, T ; O) be a function satisfying ( ut − ∆u + B · ∇u + au = f in O × (0, T ), (30) u(x, 0) = u0 (x) in O, where a ∈ L∞ (O × (0, T )), B ∈ L∞ (O × (0, T ))N , u0 ∈ L2 (O) and f ∈ L2 (O × (0, T )). Here W (0, T ; O) is the Banach space W (0, T ; O) = {u : u ∈ L2 (0, T ; H 1 (O)), ut ∈ L2 (0, T ; H −1 (O))} with the norm ||u||2W (0,T ;O) = ||u||2L2 (0,T ;H 1 (O)) + ||ut ||2L2 (0,T ;H −1 (O)) . If V ⊂ O is an open subset and ε is a positive real number, we will denote Vε = {x ∈ V : dist (x, O \ V) > ε}.

20

On the other hand, for arbitrary p ∈ (1, ∞) and δ ∈ [0, T ), we introduce the Banach space X p (δ, T ; V) = {u : u ∈ Lp (δ, T ; W 2,p (V)), ut ∈ Lp (V × (δ, T ))} and its natural norm ||u||X p (δ,T ;V) = ||u||Lp (δ,T ;W 2,p (V)) + ||ut ||Lp (V×(δ,T )) . With this notation, the following holds: Proposition 4.2. Assume that u ∈ W (0, T ; O) satisfies (30) and let p ∈ [2, ∞) be given. One has: ¡ ¢ (a) If f ∈ Lp (V×(0, T )) and u = 0 on ∂O ∩ V ×(0, T ), then, for every ε > 0 and δ ∈ (0, T ), u ∈ X p (δ, T ; Vε ) and there exist a constant I = I(N, p) ∈ N and a positive function M1 (increasing with respect its last two arguments) such that  µ ¶I+1  1 1  ||u||X p (δ,T ;Vε ) ≤ M1 (O, N, p, T, ||a||∞ , ||B||∞ ) + 2 × (31) δ ε  ¡ ¢  × ||u||W (0,T ;O) + ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) . Moreover, if p > N + 2, then u ∈ L∞ (δ, T ; W 1,∞ (Vε )) and one has   ||u||L∞ (δ,T ;W 1,∞ (Vε )) ≤ CT N4+2 DN +2 ||u||W (0,T ;O) ³ N +2 ´ (32)  + C T 4 DN +1 + T 1/4 + T λ ||f ||L2 (O×(0,T ))∩Lp (V×(0,T )) with C = C(O, N, p) a new positive constant, D ≡ ||a||∞ + ||B||∞ + δ −1 + ε−2 and 1 N +2 λ≡ − . 2 2p (b) Suppose, in addition, that u0 ∈ W 2−2/p,p (O) ∩ H01 (O). Then, if ε > 0 is fixed, u ∈ X p (0, T ; Vε ) and, for a positive function M2 (increasing with respect its last two arguments) one has ( ¡ ||u||X p (0,T ;Vε ) ≤ M2 (O, N, p, T, ||a||∞ , ||B||∞ )ε−2(I+1) ||u||W (0,T ;O) (33) + ||f ||L2 (O×(0,T ))∩Lp (V×(0,T )) + ||u0 ||W 2−2/p,p ∩H01 ) with I as in the previous paragraph. Moreover, if p > N + 2, then, u ∈ L∞ (0, T ; W 1,∞ (Vε )) and one has  N +2 e N +2 ||u||W (0,T ;O)  ||u||L∞ (0,T ;W 1,∞ (Vε )) ≤ CT 4 D   ³ ´  N +2 N +1 1/4 λ e 4 D + C T + T + T ||f ||L2 (O×(0,T ))∩Lp (V×(0,T )) (34)  ´ i h³  N +2   e + 1 ||u0 ||W 2−2/p,p ∩H 1 , e N +1 + T 1/4 + T λ D +C T 4 D 0 e ≡ ||a||∞ + ||B||∞ + ε−2 and λ as for a new positive constant C = C(O, N, p), D above. (c) Finally, assume that f ∈ Lp (O), u0 ∈ W 2−2/p,p (O)∩H01 (Ω) and u = 0 on ∂O×(0, T ). Then the solution u to (30) belongs to X p (0, T ; Ω)∩X 2 (0, T ; Ω) and 21

for a new positive function M3 (increasing with respect its last two arguments), the estimate ³ ( ||u||X p (0,T ;O)∩X 2 (0,T ;O) ≤ M3 (O, N, p, T, ||a||∞ , ||B||∞ ) ||u0 ||W 2−2/p,p ∩H01 ¢ + ||f ||L2 (O×(0,T ))∩Lp (V×(0,T )) is satisfied. In addition, if p > N + 2, one has u ∈ L∞ (0, T ; W 1,∞ (O)) and b = ||a||∞ + ||B||∞ instead of D. e estimate (34) for D Proposition 4.2 is a consequence of the analyticity in Lp (O), p ∈ (1, ∞), of the semigroup S(t) generated by the operator −∆ with Dirichlet boundary conditions. For the sake of completeness, we will give the proof in an appendix, at the end of the paper. Proof of Theorem 1.3: Since ∂ω ∩ ∂Ω ∈ C 2 uniformly and dist (ω0 , Ω \ ω) = d0 > 0, there exists an open subset U ⊂ ω with boundary ∂U of class C 2 uniformly and such that ω0 ⊂ U ⊂ ω

and dist (ω0 , Ω \ U) ≥ 3d0 /4.

On the other hand, let us set ωi = (ω0 + B(0; id0 /8)) ∩ Ω, with 1 ≤ i ≤ 4, which satisfy ( ωi ⊂ ωi+1 ⊂ U ⊂ ω, dist (ωi , Ω \ ω i+1 ) = d0 /8, 0 ≤ i ≤ 3, dist (ω4 , Ω \ U) ≥ d0 /4. Taking into account Remark 4.1, it is possible to consider the null controllability problem for system (1) with 1ω1 instead of 1ω and, thanks to Theorem 1.2, obtain a control ve ∈ L2 (Q) with Supp ve ⊂ ω 1 × [0, T ] such the corresponding solution ye to system (1) satisfies (4) and (35)

||e v ||2L2 (Q) ≤ eCH(T,||a||∞ ,||B||∞ ) ||y0 ||2L2

with C a positive constant depending on C0 , C1 and d0 and H as in Theorem 1.2. We will construct a new null control v which satisfies the statement of Theorem 1.3: Firstly, we consider a function η ∈ C ∞ ([0, T ]) such that η ≡ 1 in [0, T /3], η ≡ 0 in [2T /3, T ] and 0 ≤ η ≤ 1 and |η 0 | ≤ C/T in [0, T ]. On the other hand, due to previous properties of ωi (0 ≤ i ≤ 4) and U, there exists a function θ ∈ C ∞ (RN ) with θ ≡ 1 in ω2 ,    0 ≤ θ(x) ≤ 1, |Dα θ(x)| ≤ Cα ∀x ∈ RN , ∀α ∈ NN |α| d0   Supp θ ∩ Ω \ ω = ∅. 3 Let Y ∈ L2 (0, T ; H01 (Ω))∩C([0, T ]; L2 (Ω)) be the weak solution to system (1) corresponding to v ≡ 0. Let us set y(x, t) = [1 − θ(x)]e y (x, t) + θ(x)η(t)Y (x, t). 22

It is not difficult to see that y ∈ L2 (0, T ; H01 (Ω)) satisfies (4) and is the solution to (1) corresponding to the control v ∈ L2 (Q) given by (36) v = 2∇θ · ∇[e y − η(t)Y ] + ∆θ [e y − η(t)Y ] − (B · ∇θ)[e y − η(t)Y ] + θη 0 (t)Y. Let us prove that v satisfies the statement of Theorem 1.3. If we do w = ye − η(t)Y , then w is the weak solution to ( wt − ∆w + B · ∇w + aw = −η 0 (t)Y + ve1ω1 in Q, w = 0 on Σ, w(x, 0) = 0 in Ω. As a consequence of Proposition 4.2, we will prove that Y ∈ L∞ (ω4 × (T /6, T )) and w ∈ L∞ (0, T ; W 1,∞ (ω3 \ ω 2 )), together with appropriate estimates of the corresponding norms. Firstly, we apply Proposition 4.2 (a) to the function Y (the solution to (1) corresponding to v = 0) with f ≡ 0, u0 ≡ y0 , O = U, V = Ud0 /8 , ε = d0 /8 and δ = T /6, to conclude that Y ∈ X p (T /6, T ; Vε ) for every p ∈ [2, ∞). In particular, taking for instance p = N + 3, we obtain Y ∈ L∞ (T /6, T ; W 1,∞ (Vε )) and (32) reads ||Y ||L∞ (T /6,T ;W 1,∞ (Vε )) ≤ CT

N +2 4

DN +2 ||Y ||W (0,T ;U) ,

with C = C(U, N ) a positive constant and D = ||a||∞ + ||B||∞ + T −1 + d−2 0 . Taking into account the properties satisfied by ω4 and U, we deduce that ω4 ⊂ Vε . Thus, Y ∈ L∞ (T /6, T ; W 1,∞ (ω4 )) and the following estimate holds: (37)

2

||Y ||L∞ (T /6,T ;W 1,∞ (ω4 )) ≤ CDN +2 eCT (1+||a||∞ +||B||∞ ) ||y0 ||L2

with C a new positive constant only depending on Ω and ω (and N ). Secondly, we can also apply Proposition 4.2 (b) to w, this time with f ≡ −η 0 (t)Y + ve1ω , u0 ≡ 0, O = U, V = ω4 \ ω 1 and ε = d0 /16. Hence, we deduce that, for every p ∈ [2, ∞), w ∈ X p (0, T ; Vε ) and, since ω 3 \ ω2 ⊂ Vε , w ∈ X p (0, T ; ω3 \ ω 2 ). In fact, if we do p = 2(N + 2), w ∈ L∞ (0, T ; W 1,∞ (ω3 \ ω 2 )) can also be obtained, and from (34), one has  e N +2 ||w||W (0,T ;U)  ||w||L∞ (0,T ;W 1,∞ (ω3 \ω2 )) ≤ CT N4+2 D ³ N +2 ´ e N +1 + T 1/4 ||Y ||L2 (ω ×(T /6,T ))∩L∞ (ω ×(T /6,T )) .  +C T 4 D 4 4 In the previous inequality C > 0 is a constant only depending on U and N , and e is given by D e = ||a||∞ + ||B||2∞ + d−2 . Combining the energy inequalities D 0 satisfied by w and Y together with (37), the following estimate is obtained (38)  ¡ ¢ e N +2 eCT (1+||a||∞ +||B||2∞ ) ||e  ||w||L∞ (0,T ;W 1,∞ (ω3 \ω2 )) ≤ C D v ||L2 (Q) + ||y0 ||L2 ´ ³ e N +1 DN +2 eCT (1+||a||∞ +||B||2∞ ) ||y0 ||L2 .  +C 1+D Finally, let us recall that Supp (∇θ, ∆θ) ⊂ ω3 \ ω 2 . So, thanks to (36), (37), (38) and (35), we deduce that v ∈ L2 (Q)∩ ∈ L∞ (Q) and (14) is fulfilled. This finalizes the proof of Theorem 1.3. 23

Remark 4.2. From the proof of Theorem 1.3 we deduce, in fact, that Y ∈ X p (T /6, T ; ω4 ) and w = ye − η(t)Y ∈ X p (0, T ; ω3 \ ω 2 ) for every p ∈ [2, ∞). Taking into account the continuous imbedding X p (δ, T ; V) ,→ C 1+α,

1+α 2

([δ, T ] × V),

valid for p > N + 2 and α = 1 − (N + 2)/p, (see Lemma 3.3, p. 80, in [16]), we can obtain that the control given by (36) also satisfies v ∈ C α,α/2 (Q) for arbitrary α ∈ [0, 1). Let us recall that η ∈ C ∞ ([0, T ]) is a fixed function such that η ≡ 1 in [0, T /3], η ≡ 0 in [2T /3, T ] and 0 ≤ η ≤ 1 and |η 0 | ≤ C/T in [0, T ]. Analyzing the proof of Theorem 1.3, we observe that if v is the control given by (36) then, the corresponding solution y to (1) is y = η(t)Y + (1 − θ(x))w. Thereby, as a consequence of the proof of Theorem 1.3, we can deduce the following result: Corollary 4.3. Under the assumption of Theorem 1.3, for each y0 ∈ L2 (Ω), there exists a control v ∈ L2 (Q) ∩ L∞ (Q) such that (14) holds and the corresponding solution y to (1) satisfies (4) and Supp (y(·, t) − η(t)Y (·, t)) ⊂ Ω \ ω 0 for all t ∈ [0, T ] (Y is the solution to (1) for v ≡ 0).

5

Null controllability of system (2). Theorem 1.4

Proof of

This section is devoted to proving the null controllability property for the nonlinear system (2) when ω is such that Ω \ ω is a bounded set. In order to obtain the desired result, we will apply a fixed point argument. The structure of the proof (the controllability of a similar linear system together with a fixed point argument) is rather general and has been used in several previous works, for instance, in [27], [6], [10], [1], [3], etc. Let us recall that when Ω is an open set with boundary of class C 2 uniformly such that Ω \ ω is a bounded set then there exist ω0 and η 0 ∈ C 2 (RN ) such that Ω \ ω 0 is a bounded set and (6) and (7) hold. In fact, the regularity assumption on Ω implies that we can choose ω0 in such a way that Ω \ ω 0 is of class C 2 . We are now ready to prove Theorem 1.4. Firstly, observe that in the statement of this theorem we can assume that y0 ∈ W 2−2/p,p (Ω) ∩ H01 (Ω) with p > N + 2. Indeed, for v ≡ 0 and y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω), system (2) admits a unique local solution y ∈ L∞ (0, δ; W 1,∞ (Ω)) defined in [0, δ] (δ ≤ T ). Therefore, it suffices to set v ≡ 0 for t ∈ [0, δ/2] and to work in the time interval [δ/2, T ], looking at y(·, δ/2) as the initial state. As said above, a fixed point argument will be used. For convenience, it will be assumed in a first step that g and G are continuous.

5.1

The case in which g and G are continuous

Let y0 be given in W 2−2/p,p (Ω) ∩ H01 (Ω) with p > N + 2. We will assume that g ∈ C 0 (R × RN ), G ∈ C 0 (R × RN )N and (15) is satisfied. Then, for each ε > 0, 24

there exists Cε > 0 such that (39)

|g(s, p)|2/3 + |G(s, p)|2 ≤ Cε + ε log (1 + |s| + |p|)

∀(s, p) ∈ R × RN .

Let us set Z = L2 (0, T ; H01 (Ω)) and let R > 0 be a constant whose value will be determined below. Let us introduce the truncation functions TR : R 7→ R and T R : RN 7→ RN given by ½ s if |s| ≤ R, TR (s) = and T R (p) = (TR (pi ))1≤i≤N ∀p ∈ RN . R sgn (s) otherwise For each z ∈ Z, we consider the linear null controllability problem ( yt − ∆y + Bz · ∇y + az y = v1ω in Q, (40) y = 0 on Σ, y(x, 0) = y0 (x) in Ω, where az = g(TR (z), T R (∇z)) ∈ L∞ (Q) and Bz = G(TR (z), T R (∇z)) ∈ L∞ (Q)N . We are now going to solve the null controllability problem for system (40) in a convenient way. To this end, we reason as in [10] and [5], and we set  max |g(s, p)|, BR = max |G(s, p)|  aR = |s|≤R,|p|≤R



and

|s|≤R,|p|≤R

TR =

−1/3 min{T, aR }.

Consequently, we can apply Corollary 4.3 to problem (40) in the time interval (0, TR ) and deduce the existence of a control vbz ∈ L2 (Ω × (0, TR )) ∩ L∞ (Ω × (0, TR )) such that the solution ybz to (40) in Ω × (0, TR ) with v = vbz satisfies ybz (·, TR ) ≡ 0 in Ω,

Supp (b yz (·, t) − ηR (t)Yz (·, t)) ⊂ Ω \ ω 0 ∀t ∈ [0, TR ]

and, moreover, (41)

||b vz ||L2 (Ω×(0,TR ))∩L∞ (Ω×(0,TR )) ≤ eCK(TR ,||az ||∞ ,||Bz ||∞ ) ||y0 ||L2 .

In the previous expressions, Yz ∈ L2 (0, T ; H01 (Ω)) is the solution to (40) for v ≡ 0, C > 0 is a constant only depending on Ω and ω, K is as in Theorem 1.3 and ηR ∈ C ∞ ([0, TR ]) is a function such that ηR ≡ 1 in [0, TR /3], η ≡ 0 in 0 [2TR /3, TR ] and 0 ≤ ηR ≤ 1 and |ηR | ≤ C/TR in [0, TR ]. We will also denote ηR the extension by zero of ηR to the whole interval [0, T ]. If we now apply Proposition 4.2 (c) to ybz solution to (40) associated to vbz , it is not difficult to deduce ybz ∈ L∞ (0, TR ; W 1,∞ (Ω)) and the inequality (42)

||b yz ||L∞ (0,TR ;W 1,∞ (Ω)) ≤ eCK(TR ,||az ||∞ ,||Bz ||∞ ) ||y0 ||W 2−2/p,p ∩H01 ,

for a new positive constant depending on Ω and ω (and p). From the definitions of aR , BR , az , Bz and TR we can obtain K (TR , ||az ||∞ , ||Bz ||∞ ) ≤ 1 + T + 25

1 2/3 2 + (2 + T )aR + (1 + T )BR , T

which, together with (41) and (42), provides the inequalities ³ ´  2/3 2 C 1+aR +BR  ||b vz ||L2 (Ω×(0,TR ))∩L∞ (Ω×(0,TR )) ≤ e ||y0 ||L2 , ³ ´ 2/3 2  C 1+aR +BR ||b yz ||L∞ (0,TR ;W 1,∞ (Ω)) ≤ e ||y0 ||W 2−2/p,p ∩H01 , with C a positive constant which depends on Ω, ω and also on T . Let vez and yez be the extensions by zero of vbz and ybz to the whole cylinder Q = Ω ×(0, T ). Evidently, yez ∈ L∞ (0, T ; W 1,∞ (Q)) is the solution to (40) associated to vez ∈ L2 (Q) ∩ L∞ (Q), yez satisfies (4), Supp (e yz (·, t) − ηR (t)Yz (·, t)) ⊂ Ω \ ω 0 for each t ∈ [0, T ] and one has (43)

||e vz ||L2 (Q) + ||e vz ||∞ ≤ e

³ ´ 2/3 2 C 1+aR +BR

||y0 ||L2 ,

and (44)

||e yz ||L∞ (0,T ;W 1,∞ (Ω)) ≤ e

³ ´ 2/3 2 C 1+aR +BR

||y0 ||W 2−2/p,p ∩H01

with C = C(Ω, ω, T ) a positive constant. In the next step we will introduce a set-valued mapping on the Hilbert space Z = L2 (0, T ; H01 (Ω)) and we will check that it has a fixed point. To this end, for each z ∈ Z, let us define the set of admissible controls by UR (z) = {v ∈ L2 (Q) ∩ L∞ (Q) : v satisfies (43) and yv (·, T ) ≡ 0 in Ω}. In the previous expression, yv stands for the solution to (40) defined in Q with right-hand side v (in order to simplify the notation we have omitted the dependence on z). On the other hand, let us set ΛR (z) = {yv : v ∈ UR (z), yv ∈ L∞ (0, T ; W 1,∞ (Ω)), yv satisfies (44), and Supp (yv (·, t) − ηR (t)Yz (·, t)) ⊂ Ω \ ω 0 , ∀t ∈ [0, T ]}. We introduce the set-valued mapping ΛR : z ∈ Z = L2 (0, T ; H01 (Ω)) 7→ ΛR (z) ⊂ Z and we will prove that, for each R > 0, this mapping possesses at least one fixed point yR in Z. We will also prove the existence of some R > 0 for which every fixed point yR of ΛR satisfies yR ∈ L∞ (0, T ; W 1,∞ (Ω)) and (45)

||yR ||L∞ (0,T ;W 1,∞ (Ω)) ≤ R.

Evidently, this fact will finalize the proof of Theorem 1.3 when g and G are continuous functions. In order to prove the existence of a fixed point yR of ΛR , we will see that the Kakutani fixed point theorem can be applied to ΛR in Z (for the statement and proof of this result see, for instance, [26]). 26

Firstly, the considerations above prove that ΛR is a nonempty set. Also, due to the linearity of system (40), we readily obtain that ΛR (z) is, for every z ∈ Z, a closed convex set of Z. Compactness: Let us see that ΛR (Z) is a relatively compact set of Z. To this end, let us check that, if we fix sequences {zn }n≥1 ⊂ Z and {yn }n≥1 ⊂ Z such that yn ∈ ΛR (zn ) then, {yn }n≥1 is a relatively compact set of Z. Indeed, observe that yn is, for each n, the solution to problem (40) with second member vn ∈ UR (zn ) and coefficients an = azn = g(TR (zn ), T R (∇zn )) and Bn = Bzn = G(TR (zn ), T R (∇zn )). Moreover, from the definitions of an , Bn and the inequalities (43) and (44) we can infer that the sequences {an }n≥1 , {Bn }n≥1 , {yn }n≥1 and {vn }n≥1 are uniformly bounded in L∞ (Q), L∞ (Q)N , L∞ (0, T ; W 1,∞ (Ω)) and L2 (Q)∩L∞ (Q) respectively. On the other hand, if we apply Proposition 4.2 (c) to yn , we deduce that {yn }n≥1 is uniformly bounded in X p (0, T ; Ω) ∩ X 2 (0, T ; Ω). Now, if we denote by Yn the solution to (40) with v ≡ 0 and coefficients an and Bn , we can again apply Proposition 4.2 (c) and deduce that the sequence {Yn }n≥1 lies and is uniformly bounded in L∞ (0, T ; W 1,∞ (Ω)) ∩ X 2 (0, T ; Ω). From the definition of ΛR , Supp (yn − ηR (t)Yn ) ⊂ (Ω \ ω 0 ) × [0, T ], that is to say, yn = ηR (t)Yn + wn , Supp wn ⊂ (Ω \ ω 0 ) × [0, T ], with {wn }n≥1 ⊂ L∞ (0, T ; W 1,∞ (Ω \ ω 0 )) ∩ X 2 (0, T ; Ω \ ω 0 ) uniformly bounded. Let us remark that Ω\ω 0 is a regular bounded set and thus, H 2 (Ω\ω 0 )∩H01 (Ω\ ω 0 ) and H01 (Ω \ ω 0 ) are compactly embedded in H01 (Ω \ ω 0 ) and L2 (Ω \ ω 0 ), respectively. As a consequence, we also have that X 2 (0, T ; Ω \ ω 0 ) is compactly embedded in L2 (0, T ; H01 (Ω\ω 0 ))∩C 0 ([0, T ]; L2 (Ω \ ω 0 )) and thereby, we deduce that the sequence {wn }n≥1 is a relatively compact set of Z ∩ C 0 ([0, T ]; L2 (Ω)). This kind of compactness results are by now classical (see e.g. [22]). Next, we will prove that the sequence {Yn }n≥1 also is a relatively compact set of Z ∩ C 0 ([0, T ]; L2 (Ω)). Obviously, this fact will imply the compactness of the multiple-valued mapping ΛR . As said above, the sequences {an }n≥1 , {Bn }n≥1 and {Yn }n≥1 are uniformly bounded in L∞ (Q), L∞ (Q)N and L∞ (0, T ; W 1,∞ (Ω))∩X 2 (0, T ; Ω) respectively. Thus, we can deduce the existence of subsequences {an }n≥1 , {Bn }n≥1 and {Yn }n≥1 (which will be denoted with the same subindex n) such that ( an * a weakly-* in L∞ (Q) and a.e. in Q, Bn * B weakly-* in L∞ (Q)N and a.e. in Q, and (46)

 ∞ 1,∞ (Ω)),   Yn * Y weakly-* in L (0, T ; W 2 Yn * Y weakly in X (0, T ; Ω),   Yn → Y and ∇Yn → ∇Y a.e. in Q, 27

with a ∈ L∞ (Q), B ∈ L∞ (Q)N and Y ∈ L∞ (0, T ; W 1,∞ (Q)) ∩ X 2 (0, T ; Ω). With the previous convergences in mind, it is not difficult to pass to the limit in the problem satisfied by Yn and check that Y is the weak solution to ( Yt − ∆Y + B · ∇Y + aY = 0 in Q, (47) Y = 0 on Σ, Y (x, 0) = y0 (x) in Ω. If we now set Wn = Yn − Y , then Wn is the weak solution to ( ut − ∆u + Bn · ∇u + an u = f in Q, (48) u = 0 on Σ, u(x, 0) = 0 in Ω, for f = Fn ≡ (B − Bn ) · ∇Y + (a − an )Y ∈ L2 (Q) ∩ L∞ (Q). As a consequence of the Lebesgue dominated convergence theorem, one has that Fn → 0 in L2 (Q). From the energy inequality satisfied by Wn we deduce ZZ 2 sup ||Wn (t)||2L2 + ||∇Wn ||2L2 (Q) ≤ eT (2||an ||∞ +||Bn ||∞ ) Fn Wn . t∈[0,T ]

Q

Observe that the integral in the right-hand side of the previous inequality converges to zero and, then, Yn → Y in the strong topology of Z ∩C 0 ([0, T ]; L2 (Ω)). We conclude that the sequence {Yn }n≥1 is a relatively compact set of Z ∩ C 0 ([0, T ]; L2 (Ω)). Finally, let us note that, if we consider K = co ΛR (Z), that is to say, the closed convex hull of ΛR (Z), then, K is a nonempty convex compact set of Z ∩ C 0 ([0, T ]; L2 (Ω)) and ΛR (K) ⊂ K. Continuity: Let us prove that the mapping ΛR is upper semi-continuous in Z. To this end, let {zn }n≥1 and {yn }n≥1 be two sequences satisfying zn → z in Z,

yn ∈ ΛR (zn ) and

yn → y in Z,

with z, y ∈ Z. We claim that y ∈ ΛR (z). Firstly, at least for a subsequence, we can write zn (x, t) → z(x, t) a.e. in Q. The continuity assumptions on g and G implies ( an = g(TR (zn ), T R (∇zn )) → a = g(TR (z), T R (∇z)) a.e. in Q, (49) Bn = G(TR (zn ), T R (∇zn )) → B = G(TR (z), T R (∇z)) a.e. in Q. Secondly, from the definition of ΛR (zn ) we deduce that there exists a control vn ∈ UR (zn ) such that yn is the solution to (40) with v = vn and coefficients an and Bn . In addition, vn ∈ L2 (Q) ∩ L∞ (Q) satisfies (43) and ∞ 1,∞ (Q)) satisfies (4), (44) and Supp (yn (·, ·) − ηR (·)Yn (·, ·)) ⊂ ¡yn ∈ L ¢ (0, T ; W Ω \ ω 0 × [0, T ] (as above, Yn is the solution to (40) corresponding to v ≡ 0 and coefficients an and Bn ). We can argue as in the proof of the compactness of ΛR and deduce that (46) holds for Y ∈ L∞ (0, T ; W 1,∞ (Ω)) solution to (47). On the other hand, the same 28

arguments allow us to check, that for a subsequence, one has:  2 ∞   vn * v weakly in L (Q), vn * v weakly-* in L (Q), yn * y weakly-* in L∞ (0, T ; W 1,∞ (Ω)), (50)   vn → v, yn → y, ∇yn → ∇y a.e. in Q, and y ∈ L∞ (0, T ; W 1,∞ (Ω)). From (46), (49) and (50), it is not difficult to show that y is the solution to (40), with coefficients a ≡ az and B ≡ Bz , corresponding to v ∈ L2 (Q) ∩ L∞ (Q), v and y satisfy (43) and (44), respectively, and ¢ ¡ Supp (y(·, ·) − ηR (·)Y (·, ·)) ⊂ Supp (yn (·, ·) − ηR (·)Yn (·, ·)) ⊂ Ω \ ω 0 × [0, T ] holds. Finally, owing to yn ∈ ΛR (zn ) ⊂ K, with K a compact set of the space Z ∩ C 0 ([0, T ]; L2 (Ω)), we deduce yn → y in C 0 ([0, T ]; L2 (Ω)) and therefore, y(·, T ) = 0 in Ω. We have then proved that v ∈ UR (z) and y ∈ ΛR (z). Thus, ΛR is an upper semi-continuous mapping in Z. Summarizing, we have showed that the mapping ΛR defined in the Hilbert space Z satisfies: (a) ΛR is upper semi-continuous in Z, (b) there exists a nonempty convex compact set K ⊂ Z such that ΛR (K) ⊂ K and (c) for every z ∈ Z, ΛR (z) is a nonempty closed convex set in Z. As a consequence, for any R > 0 the Kakutani theorem can be applied, ensuring the existence of a fixed point yR of ΛR . In order to finish the proof, we will show that we can choose R > 0 in such a way that any fixed point of ΛR satisfies (45). Thus, let yR be a fixed point of ΛR associated to a control vR ∈ UR (yR ). If we combine inequality (44), which is, in particular, satisfied by yR , inequality (39) and the definitions of aR and BR , then ( ||yR ||L∞ (0,T ;W 1,∞ (Ω)) ≤ eC(1+Cε +ε log(1+2R)) ||y0 ||W 2−2/p,p ∩H01 = eC(1+Cε ) (1 + 2R)Cε ||y0 ||W 2−2/p,p ∩H01 , is satisfied for a positive constant C = C(Ω, ω, T ). Taking ε = 1/2C, we find that ||yR ||L∞ (0,T ;W 1,∞ (Ω)) ≤ C(1 + 2R)1/2 ||y0 ||W 2−2/p,p ∩H01 , from which it can be deduced that there exists R large enough (depending on Ω, ω, T , g and G) such that (45) is satisfied. Therefore, TR (yR ) = yR and T R (∇yR ) = ∇yR , and yR together with vR solves the null controllability problem for system (2). We have then proved Theorem 1.4 when g and G are continuous.

5.2

The general case

In order to obtain the proof of the general case, we will reason as in [5]. To this end, we consider {ρn }n≥1 , a sequence of mollifiers, and the functions gn = ρn ∗ g and Gn = ρn ∗ G. Then, the following properties hold: 1. gn ∈ C 0 (R × RN ) and Gn ∈ C 0 (R × RN )N for all n ≥ 1. 29

2. If we put fn (s, p) = gn (s, p)s + Gn (s, p) · p for all (s, p) ∈ R × RN , then fn → f

uniformly in the compact sets of R × RN .

3. For any given M > 0, there exists C(M ) > 0 such that sup

(|gn (s, p)| + |Gn (s, p)|) ≤ C(M )

∀n ≥ 1.

|(s,p)|≤M

4. The functions gn and Gn satisfy (15) uniformly in n, that is to say, for any ε > 0, there exists M (ε) > 0 such that ( |gn (s, p)| ≤ ε log3/2 (1 + |s| + |p|), |Gn (s, p)| ≤ ε log1/2 (1 + |s| + |p|) whenever |(s, p)| ≥ M (ε) and for all n ≥ 1. We can apply the results of subsection 5.1 to fn to conclude the existence of a control function vn ∈ L2 (Q) ∩ L∞ (Q) such that the system ( yn,t − ∆yn + fn (yn , ∇yn ) = vn 1ω in Q, (51) yn = 0 on Σ, yn (x, 0) = y0 (x) in Ω possesses a solution yn ∈ L∞ (0, T ; W 1,∞ (Q)) satisfying (4). From the properties satisfied by gn and Gn and the reasoning done in subsection 5.1, we deduce that we can choose R > 0 and K, a compact set of Z ∩C 0 ([0, T ]; L2 (Ω)), independent of n, in such a way that, for every n ≥ 1, yn ∈ K and ||vn ||L2 (Q) + ||vn ||L∞ (Q) ≤ R,

||yn ||L∞ (0,T ;W 1,∞ (Ω)) ≤ R.

This uniform bound allow us to assure that, at least for a subsequence, (50) is satisfied for two functions v ∈ L2 (Q) ∩ L∞ (Q) and y ∈ L∞ (0, T ; W 1,∞ (Ω)) and yn → y in C 0 ([0, T ]; L2 (Ω)). On the other hand, property 1 together with the continuity of f gives fn (yn , ∇yn ) → f (y, ∇y) a.e. in Q. If we repeat the arguments given to show the upper semi-continuity of ΛR , we can pass to the limit in (51) and find a control v ∈ L2 (Q) ∩ L∞ (Q) such that (2) has a solution y which satisfies (4). This ends the proof of Theorem 1.4. Remark 5.1. As in the case when Ω is a bounded set (see [5]) and analyzing the proof of Theorem 1.4, we can deduce: 1. A null controllability result for system (2) when we change (15) by lim

|(s,p)|→∞

|g(s, p)| log

3/2

(1 + |s| + |p|)

≤ l1 ,

lim

|(s,p)|→∞

|G(s, p)| log

1/2

(1 + |s| + |p|)

≤ l2 ,

where l1 and l2 are small positive constants which depend on Ω, ω and T . 30

2. A local null controllability result for system (1) with a general nonlinear term f (s, p) satisfying f (0, 0) = 0. To be precise, it can be proved that if f is given, there exists a positive number δ = δ(Ω, ω, T, f ) such that for every y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω) with ||y0 ||W 1,∞ ∩H01 ≤ δ, there exists a control v ∈ L2 (Q) ∩ L∞ (Q) such that the corresponding solution to (2) satisfies (4). Remark 5.2. In Theorem 1.4 we can consider general nonlinearities of the form f = f (x, t; s, p), (x, t) ∈ Q and (s, p) ∈ R × RN , with f a Carath´eodory function satisfying: 1. f (x, t; 0, 0) = 0 for all (x, t) ∈ Q and f (·, ·; s, p) ∈ L∞ (Q), for all (s, p) ∈ R × RN , 2. f (x, t; ·, ·) is a locally Lipschitz-continuous function for (x, t) a.e. in Q, with Lipschitz constants independent of (x, t) in the bounded sets of R × RN , 3. f (·, ·; s, p) = g(·, ·; s, p)s + G(·, ·; s, p) · p for all (s, p) ∈ R × RN , with lim

|(s,p)|→∞

|g(x, t; s, p)| log

3/2

(1 + |s| + |p|)

= 0,

lim

|(s,p)|→∞

|G(x, t; s, p)| log

1/2

(1 + |s| + |p|)

0,

uniformly in (x, t) ∈ Q.

6

Further comments, results and open problems

1. As said above, conditions (5)–(7) are sufficient conditions which permit us to prove a null controllability result for system (1) (with initial conditions and control functions in L2 -spaces and for every T > 0) by means of a global Carleman inequality for the corresponding adjoint system. Observe that for general open sets Ω and ω these conditions are, in general, not easy to be checked. It would be then very interesting to give geometrical conditions on (Ω, ω) which assure the null controllability of (1) at time T > 0. 2. Conditions (5)–(7) also allow us to show a null controllability result for system (1) with control functions in L2 (Q) ∩ L∞ (Q) which, in addition, satisfy the estimate (14) (see Theorem 1.3). Let us observe that this kind of control functions plays an important part in the proof of null controllability results for nonlinear parabolic problems. The general strategy to prove a null controllability result for a nonlinear parabolic problem is to perform a fixed point argument. When Ω is an unbounded set some technical difficulties arise since the compactness of the Sobolev imbedding theorems fails. These difficulties have been overcome when Ω is a regular unbounded domain such that Ω \ ω is bounded. Nevertheless, proving a null controllability result for system (2) when f (·, ·) grows superlinearly at infinity and the couple (Ω, ω) satisfies the hypotheses in Theorem 1.3 is an open question. 3. Assume that (Ω, ω) satisfies the hypotheses of Theorem 1.4. Then, taking into account Remark 5.2 and reasoning as in [5], it is possible to show for 31

system (2) an exact controllability result to the trajectories of (2) which lies in L∞ (0, T ; W 1,∞ (Q)). To be precise, the following result holds: Theorem 6.1. Assume that f : R × RN → R is a locally Lipschitz-continuous function that satisfies ¯ ¯  ¯ ∂f ¯ 1  ¯ ¯  lim  ¯ ∂s (s0 + λs, p0 + λp) dλ¯ = 0,  |(s,p)|→∞ 3/2 log (1 + |s| + |p|) ¯ ¯ (52) ¯ ¯ ∂f  1  ¯ ¯  lim  |(s,p)|→∞ ¯ ∂pi (s0 + λs, p0 + λp) dλ¯ = 0 1/2 log (1 + |s| + |p|) uniformly in (s0 , p0 ) ∈ K for every compact set K ⊂ R × RN . Let y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω) be given and let y ∗ a solution to (2) in L∞ (0, T ; W 1,∞ (Ω)) corresponding to the data y0∗ ∈ W 1,∞ (Ω) ∩ H01 (Ω) and v ∗ ∈ L2 (Q) ∩ L∞ (Q). Then, there exists a control function v ∈ L2 (Q) ∩ L∞ (Q) such that system (2) has a weak solution y ∈ L∞ (0, T ; W 1,∞ (Ω)) such that y(x, T ) = y ∗ (x, T ) in Ω.

4. As a a consequence of the previous result and, again, following the ideas of [5], we can also obtain Theorem 6.2. Assume that (Ω, ω) and f are under the assumptions of Theorem 6.1. Then, for each y0 ∈ W 1,∞ (Ω) ∩ H01 (Ω), yd ∈ L2 (Ω) and ε > 0, there exists a control v ∈ L2 (Q) ∩ L∞ (Q) such that system (2) has a weak solution y ∈ L∞ (0, T ; W 1,∞ (Ω)) which satisfies (3). Observe that for a function f globally Lipschitz-continuous (independent of the gradient) the approximate controllability of (2) was proved in [23, 24] for unbounded domains and without restriction on the control set ω. Theorem 6.2 generalizes those results for the case of slightly superlinear nonlinearities, but imposes additional conditions on ω. It is an interesting open problem to answer if Theorem 6.2 is still valid when we take as control region ANY open nonempty subset of an unbounded set Ω.

A

Appendix. Proof of Proposition 4.2

As said above, Proposition 4.2 is a consequence of the regularizing effect of the heat equation. Let S(t) be the semigroup generated by the heat equation in O with homogenous Dirichlet boundary conditions. It is well know that S(t) is a bounded analytic semigroup in Lp (O) for every p ∈ (1, ∞) (see for instance [18]). We will consider the heat equation ( yt − ∆y = F in O × (0, T ), (53) y = 0 on ∂O × (0, T ), y(x, 0) = 0 in O, with F ∈ Lp0 (O × (0, T )) and p0 ∈ [2, ∞). Then, one has: 32

Lemma A.1. Assume F ∈ Lp0 (O × (0, T )) with p0 ∈ [2, ∞) and let y be the solution to (53). Then, 1. y ∈ X p0 (0, T ; O) = {u ∈ Lp0 (0, T ; W 2,p0 (O)) : ut ∈ Lp0 (O × (0, T ))}, and ||y||X p0 (0,T ;O) ≤ C||F ||Lp0 (O×(0,T )) , for a positive constant C = C(O, p0 ) which is independent of T . 2. If p0 ≤ N + 2 and pe ∈ (p0 , ∞] is given by p1 ∈ [p0 , pe), y ∈ Lp1 (0, T ; W01,p1 (O)) and (54)

1 1 1 = − , then, for each pe p0 N + 2

||y||Lp1 (0,T ;W 1,p1 (O)) ≤ CT µ(N,p0 ,p1 ) ||F ||Lp0 (O×(0,T )) ,

for a positive constant C = C(O, p0 , p1 , N ) and µ ¶ 1 N +2 1 1 µ(N, p0 , p1 ) = − − . 2 2 p0 p1 3. If p0 > N + 2, then y ∈ L∞ (0, T ; W 1,∞ (O)) and, for p1 = ∞ and a new positive constant C = C(O, p0 , N ), (54) is satisfied. Proof: The first point is a direct consequence of the results in [13]. On the other hand, we will prove the points 2 and 3 at the same time. Whether p0 ≤ N + 2 (and p1 ∈ [p0 , pe)) or p0 > N + 2 (and p1 = ∞), we can write 1/p1 = 1/p0 − 1/α with N + 2 < α ≤ ∞. Observe that the solution y to (53) can be written as Z t y(t) = S(t − s)F (s) ds, ∀t ∈ [0, T ], 0

and therefore Z t  ³ ´ 1 1 1 −N  2 p0 − p1 − 2  1,p ||F (s)||Lp0 ds ||y(t)|| ≤ C(O) (t − s)  W 1 0 Z t  N 1 1   = C(O) (t − s)− 2 α − 2 ||F (s)||Lp0 ds. 0 N 1

1

Since α > N + 2, if we now set g(t) = t− 2 α − 2 , then we have g ∈ Lσ (0, T ) with σ such that 1/σ = 1 − 1/α. If we apply the Young’s inequality to the previous inequality we deduce ||y(·)||W 1,p1 ∈ Lp1 (0, T ), ||y||Lp1 (0,T ;W 1,p1 (O)) ≤ C(O) ||g||Lσ (0,T ) ||F ||Lp0 (O×(0,T )) and inequality (54) for a positive constant C = C(O, α, N ). We have then proved the lemma. Proof of Proposition 4.2: For the proof of this result we will combine Lemma A.1 and a classical boot-strap argument. For simplicity, we will focus our attention on the proof of the two first points (a) and (b). The proof of point (c) can be easily obtained from the arguments below. 33

(a) Assume that f ∈ L2 (O × (0, T )) ∩ Lp (V × (0, T )) with V ⊂ O and let u ∈ W (0, T ; O) be a function which satisfies (30) and u = 0 on (∂O ∩ V) × (0, T ). If ε and δ are fixed, we take εi = iε/(I + 1) and δi = iδ/(I + 1) for 0 ≤ i ≤ I + 1 and I ∈ N given by1 I = min{N + 1, [(1 − 2/p) (N + 2)] + 1}. We also take the open sets Vi = Vεi which satisfy Vε ≡ VI+1 ⊂ VI ⊂ · · · ⊂ V1 ⊂ V0 ≡ V and ¡ ¢ dist Vi , O \ Vi−1 =

ε , I +1

1 ≤ i ≤ I + 1.

From these properties we can deduce that, for each i : 1 ≤ i ≤ I + 1, there exists a function θi ¡∈ C ∞ (RN¢) such that 0 ≤ θi (x) ≤ 1 for every x ∈ RN , θi ≡ 1 in Vi , Supp θi ∩ O \ Vi−1 = ∅ and |Dα θi (x)| ≤

Cα , (ε/(I + 1))α

∀x ∈ RN ,

∀α ∈ NN .

On the other hand, there exists ηi ∈ C ∞ ([0, T ]), 1 ≤ i ≤ I + 1, such that 0 ≤ ηi ≤ 1 in [0, T ], ηi ≡ 1 in [δi , T ], Supp ηi ⊂ (δi−1 , T ] and ¯ k ¯ ¯ d ηi ¯ Ck ¯ ¯≤ (t) ¯ dtk ¯ (δ/(I + 1))k , ∀t ∈ [0, T ], ∀k ≥ 1. We define y0 = u and yi = θi ηi yi−1 for each i, 1 ≤ i ≤ I + 1. Thus, Supp yi ⊂ Vi−1 × [δi−1 , T ],

yi = yi−1 in Vi × [δi , T ],

1 ≤ i ≤ I + 1.

Let us remark that yi = u in Vi × [δi , T ] for all 0 ≤ i ≤ I + 1. Therefore, our goal will be show that yI+1 satisfies the statement of the proposition. We readily check that yi is, for 1 ≤ i ≤ I + 1, the weak solution to (53) corresponding to a right hand side Fi given by Fi ≡ −θi ηi (B · ∇yi−1 + ayi−1 − f ) + ηi0 θi yi−1 − ηi [2∇θi · ∇yi−1 + (∆θi )yi−1 ] . Let us introduce the quantities 1 1 i = − qi 2 2(N + 2)

for i : 0 ≤ i ≤ I − 1 and

qI = p.

Thanks to the choice of I, we easily obtain qi ≤ N +2 and qi < p for every i, 0 ≤ i ≤ I − 1. Also, the hypotheses on f allow us to deduce that f ∈ Lq (Vi × (0, T )), for every q ∈ [2, p], and kf kLq (Vi ×(0,T )) ≤ ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) . 1 [x] stands for the function [x] = max{n ∈ Z : n < x}. Observe that [n] = n − 1 for every n ∈ Z.

34

In particular, F1 ∈ L2 (O × (0, T )) and we can apply Lemma A.1 2 to y1 with p0 = q0 = 2 and p1 = q1 obtaining y1 ∈ Lq1 (0, T ; W 1,q1 (O)) and, from (54),  µ(N,2,q1 ) ||F1 ||L2 (O×(0,T ))   ||y1 ||Lq1 (0,T ;W 1,q1 (O)) ≤ C(O, N )T 1/4 ≤ C(O, N )T D||y0 ||L2 (0,T ;H 1 (O))  ¡ ¢  + C(O, N )T 1/4 ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) , with D given by D ≡ ||a||∞ + ||B||∞ + δ −1 + ε−2 and µ(N, p0 , p1 ) as in Lemma A.1. For every i, 1 ≤ i ≤ I, we can continue applying Lemma A.1 2 to yi with F = Fi , p0 = qi−1 and p1 = qi obtaining again yi ∈ Lqi (0, T ; W 1,qi (O)) and ( ||yi ||Lqi (0,T ;W 1,qi (O)) ≤ C(O, p, N )T µ(N,qi−1 ,qi ) D||yi−1 ||Lqi−1 (0,T ;W 1,qi−1 (O)) ¡ ¢ + C(O, p, N )T µ(N,qi−1 ,qi ) ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) , (observe that µ(N, qi−1 , qi ) = 1/4 if i ≤ I − 1). Combining these two last inequalities, we infer that yI ∈ Lp (0, T ; W 1,p (O)) and there exists a new positive constant C = C(O, p, N ) such that (55)  µ(N,pI−1 ,p)+ I−1 4 D I ||y || 2  0 L (0,T ;H 1 (O))  ||yI ||Lp (0,T ;W 1,p (O)) ≤ C(O, p, N )T ¡ ¢ I−1 µ(N,pI−1 ,p)+ I−1 4 D ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) + C(O, p, N )T  ¡ ¢  + C(O, p, N )T µ(N,pI−1 ,p) ||f ||L2 (O×(0,T )) + ||f ||Lp (V×(0,T )) . Applying once again Lemma A.1 (part 1) to yI+1 , FI+1 and p, we obtain yI+1 ∈ X p (0, T ; O), i.e., u ∈ X p (δ, T ; Vε ) and the estimate (31). Finally, let us point out that if p > N + 2 then, I = N + 1 and we can apply once more Lemma A.1 (part 3) to yI+1 , FI+1 , p0 = p and p1 = ∞ and obtain that yI+1 belongs to L∞ (0, T ; W 1,∞ (O)) and, from (54), ||yI+1 ||L∞ (0,T ;W 1,∞ (O)) ≤ CT µ(N,p) ||FI+1 ||Lp (O×(0,T )) . This last inequality together with the expression of FI+1 and (55) shows that u ∈ L∞ (δ, T ; W 1,∞ (V)) and inequality (32). This concludes part (a) of Proposition 4.2. (b) Assume now that, in addition, u0 ∈ W 2−2/p,p (Ω)∩H01 (Ω) with p ∈ [2, ∞). Thus, the function u (a solution to (30) with u = 0 on (∂O ∩ V) × (0, T )) can be written as u = U + w with U (t) = S(t)u0 , t ≥ 0, and w a solution to ( wt − ∆w + B · ∇w + au = F in O × (0, T ), (56) w = 0 on (∂O ∩ V) × (0, T ), w(x, 0) = 0 in O, for F = f − B · ∇U − aU . Using again that S(t) is a bounded analytic semigroup in Lq (O), for every q ∈ (1, ∞), and the results in [13], we can infer U ∈ X p (0, T ; O) ∩ X 2 (0, T ; O), U ∈ C 0 ([0, T ]; W 2−2/p,p (Ω) ∩ H01 (Ω)), and for a positive constant C = C(O, p, N ), (57) ||U ||X 2 (0,T ;O)∩X p (0,T ;O) + max ||U (t)||W 2−2/p,p ∩H01 ≤ C||u0 ||W 2−2/p,p ∩H01 . t∈[0,T ]

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In particular, this last inequality implies F ∈ L2 (O × (0, T )) ∩ Lp (V × (0, T )). Accordingly, for a ε > 0 given, we can repeat for w and the problem (56) the arguments developed in the previous point (observe that it is not necessary to carry out the process of time localization) and conclude w ∈ X p (0, T ; Vε ) and ( ||w||X p (0,T ;Vε ) ≤ M1 (O, N, p, T, ||a||∞ , ||B||∞ )ε−2(I+1) ¡ ¢ × ||w||W (0,T ;O) + ||F ||L2 (O×(0,T )) + ||F ||Lp (V×(0,T )) , This inequality together with (57) proves u ∈ X p (0, T ; V² ) and estimate (33). Finally, when p > N + 2, we can deduce that w ∈ L∞ (0, T ; W 1,∞ (V² )) and  e N +2 ||w||W (0,T ;O)  ||w||L∞ (0,T ;W 1,∞ (Vε )) ≤ CT N4+2 D ³ N +2 ´ e N +1 + T 1/4 + T λ ||F ||L2 (O×(0,T ))∩Lp (V×(0,T ))  +C T 4 D e ≡ ||a||∞ + ||B||∞ + ε−2 and λ as with C = C(O, N, p) > 0 a new constant, D in Proposition 4.2. Taking into account this inequality, inequality (57) and the fact that, for p > N + 2, W 2−2/p,p (O) ,→ W 1,∞ (O) with continuous embedding, we obtain u ∈ L∞ (0, T ; W 1,∞ (Vε )) and estimate (34).

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