NULL CONTROLLABILITY OF DEGENERATE/SINGULAR PARABOLIC EQUATIONS

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1 Journal of Dynamical and Control Systems, Vol. 18, No., October 1, ( c 1) DOI: 1.17/s NULL CONTROLLABILITY OF DEGENERATE/SINGULAR PARABOLIC EQUATIONS M. FOTOUHI and L. SALIMI Abstract. The purpose of this paper is to provide a full analysis of the null controllability problem for the one dimensional degenerate/singular parabolic equation u t (a(x)u x) x λ x β u =, (t, x) (, T ) (, 1), where the diffusion coefficient a( ) is degenerate at x =. Also the boundary conditions are considered to be Dirichlet or Neumann type related to the degeneracy rate of a( ). Under some conditions on the function a( ) and parameters β, λ, we prove global Carleman estimates. The proof is based on an improved Hardy-type inequality. 1. Introduction In the recent years, the study of the controllability for parabolic equations has become an active research area. After the pioneering works [9, 1, 13, 17, 18], there has been substantial progress in understanding the controllability properties of parabolic equations with variable coefficients. In particular, the null controllability for the following class of nondegenerate and nonsingular parabolic operators is well-known: P u = u t (a(x)u x ) x, x (, 1), where the coefficient a(x) is a positive continuous function on [, 1]. On the contrary, when the coefficient a(x) is zero at some points, the equation will be degenerate and few results are known in this case, even though many problems that are relevant for applications are described by parabolic equations degenerating at the boundary of the space domain. For instance, in [,6,7,16], the reader will find a motivating example of a Croccotype equation coming from the study of the velocity field of a laminar flow on a flat plate. In [5, 15], where the degenerate/nonsingular model P u = u t (a(x)u x ) x, x (, 1), (1) Mathematics Subject Classification. 35K65, 93B5, 93B7. Key words and phrases. Degenerate parabolic equations, singular potential, null controllability, Carleman estimates, improved Hardy inequality /1/1-573/ c 1 Springer Science+Business Media, Inc.

2 57 M. FOTOUHI and L. SALIMI is considered, observability results for the adjoint problem are obtained using Carleman estimates. Then the null controllability of (1) on [, 1] is derived by some standard arguments. Also, in [] the null controllability on [, 1] of the semilinear degenerate parabolic equation P u = u t (a(x)u x ) x + f(t, x, u), x (, 1), is studied, where f is locally Lipschitz with respect to u. Now, consider N 3 and let Ω R N be a bounded open set with the boundary Γ, of class C such that Ω. The heat equation perturbed by a singular inverse-square potential, namely P u = u t u + λ u, x Ω, x arised in quantum mechanics and combustion theory. When λ (N ) /, it is proved in [] that the equation can be controlled to zero with a distributed control which surrounds the singularity. Next, it is shown in [1] that this geometric assumption is not necessary, and one can control the equation from any open subset as for the heat equation. But in the case λ > (N ) /, there exists a sequence of regularized potentials λ/( x +ɛ ) such that one cannot stabilize the corresponding systems uniformly with respect to ɛ >. Furthermore, Vancostenoble and Zuazua in [] showed that in dimension N = 1, the following singular operator is null controllable for λ 1/, m R and β <, P u = u t u xx + λ x u + m u, x (, 1). xβ Also, new Carleman estimates (and consequently null controllability property) were established in [19] for the one dimensional degenerate/singular operator P u = u t (x α u x ) x λ u, x (, 1), xβ with suitable boundary conditions and under the following assumptions α [, ), < β < α, λ R, α [, ) \ {1}, β = α, (1 α) λ. Also, it has been proved that the null controllability is false when α, [7]. In this paper, we study the following one dimensional operator that couples a degenerate diffusion coefficient with a singular potential: P u = u t (a(x)u x ) x λ x β u, x (, 1). () Under suitable condition on a, β and λ, we prove the null controllability for the operator () which completes the results of [] and [19].

3 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 575 The paper is organized as follows. Section is devoted to discussing the assumptions on the coefficient degeneracy a( ) and the parameters β, λ. Then we state functional setting and well-posedness in Section 3. In Section, we state main results, especially Carleman estimates and its applications to observability and controllability. In Section 5, we study the improved Hardy inequality which is necessary in the proofs of well-posedness, Carleman estimates and observability results. Finally Section 6 is devoted to the proofs.. Assumptions and statement of the problem.1. Hypothesis on the degeneracy coefficient. In order to study the controllability of operator (), we let some assumption on the degenerate diffusion coefficient a(x) in the following way. Hypothesis 1. We suppose that the degeneracy coefficient a( ) satisfies the following conditions: (i) a C([, 1]) C 1 ((, 1]), a(x) > in (, 1] and a() =. (ii) α (, ) such that xa (x) αa(x) for every x [, 1]. (iii) If α [1, ), there exist m > and δ > such that for every x [, δ ], we have a(x) m sup a(y). y x First of all, notice that every function which is nondecreasing near x =, satisfies in condition (iii) for m = 1. Thus Hypothesis 1 is weaker than the conditions of the coefficient a( ) stated in the paper [], If α (1, ), θ (1, α] such that x a(x) x θ is nondecreasing near x =, (3) If α = 1, θ (, 1) such that x a(x) x θ is nondecreasing near x =. In fact, the condition (3) will imply that the coefficient a(x) is nondecresing near x =. As stated in [], in the case α (1, ), a sufficient condition to prove the controllability of the equation u t (a(x)u x ) x = fχ ω is that the degeneracy coefficient a(x) satisfies conditions (i) and (ii) of Hypothesis 1 as well as the Hardy type inequality holds, i.e. there exists C > such that for any locally absolutely continuous function u on (, 1] which u(1) =, a(x) u (x) dx <,

4 576 M. FOTOUHI and L. SALIMI we should have a(x) x u (x)dx C a(x) u (x) dx. () In fact, we replace the condition (iii) of Hypothesis 1 instead of Hardy inequality, however they are not equivalent. The following example shows us a degenerate coefficient a(x) which satisfies Hypothesis 1, but the Hardy inequality does not hold for that. We use the following proposition in this example which was proved in []. Proposition. Let a, b : [, 1] R be in C([, 1]) C 1 ((, 1]), a() = b() =, a > and b > on (, 1]. Moreover, assume that there exist two positive constants c 1 and c such that c 1 b a c b, in a neighborhood of zero. Then Hardy inequality () holds for the function a( ) if and only if it holds for b( ). Example 3. Consider a(x) = xe (α 1)x for some α (1, ), we will have xa (x) a(x) = 1 + (α 1)x 1 + (α 1) = α, for every x [, 1]. This type of function satisfies the conditions of Hypothesis 1, but the condition (3) or Hardy inequality does not hold. At first, we notice that there exists no θ (1, α] such that the function a (x) = a(x) is increasing near x =, because for every θ > 1, a (x) = e(α 1)x [1 θ + (α 1)x] x θ, which is negative near x =. Furthermore, if Hardy inequality holds for a( ), according to Proposition we conclude that Hardy inequality should hold for the function b(x) = x, because of equivalency c 1 b(x) a(x) c b(x) on the interval x [, 1] for suitable positive constants c 1 and c. But by substituting the functions u(x) = x r (1 x) in the Hardy inequality () for b(x) = x and let r +, we imply that the inequality can t hold. So, by the method used in [] one can not prove the controllability of the related equation to a(x) = xe (α 1)x. On the other hand a( ) satisfies the condition (iii) of Hypothesis 1 because of its increasing property. Now we state a lemma which is useful along the paper. This shows 1 a L1 (, 1) in the case of α (, 1), and 1 L 1 (, 1) for α [1, ). a 1 Note that if / L 1 (, 1), then the null controllability of () fails, see [7]. a x θ

5 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 577 Lemma. Let a( ) satisfies the Hypothesis 1, then the integral is finite for α (, 1) and dx a(x) is finite for α [1, ). dx a(x) 1 Remark 5. For α (1, ), the integral dx might be finite or a(x) infinite, for example consider the functions a(x) = x r e (α r)x where r α... Assumptions on the singular potential. First note that when α, the null controllability might be false. For example consider a(x) = x α for α. In this case the necessary condition for null controllability which 1 was proved in [7] does not hold, i.e. does not belong to L 1 (, 1). On a(x) the other hand, since α = 1 is a peculiar case in the following Hardy type inequality (which is proved in Lemma 18), a(x)u xdx λ u (a, α) dx, (5) x α this special case is considered separately. Also if α =, then the nonnegative function a( ) is decreasing on [, 1], therefore a(x) = for every x, which is impossible since a > in (, 1]. Now we assume that α (, ) \ {1}. For a given singular potential, Cabré and Martel in [3] proved that existence versus blow-up of positive solutions is connected to the existence of some Hardy inequality involving the considered potential, like as the following: a(x)u xdx λ u dx. (6) xβ Therefore, the inequality (5) implies that the critical exponent could be β = α, so we assume that β α. Here, we show well-posedness and controllability of () for { α (, ), < β < α, λ R, α (, ) \ {1}, β = α, λ < λ (7) (a, α), where λ (a, α) is the optimal constant in Hardy type inequality (5). When α = 1, the fact that a(x)u x L (, 1) does not imply that u/ x belongs to L (, 1), because λ may be equal to zero. So the case β = α = 1 is now forbidden. However, we have the following improved version of the preceding Hardy type inequality which is valid for all values n >, γ < α and some positive constant C = C (a, α, γ, n) >. (See Theorem 1) a(x)u xdx + C u u dx n x γ dx.

6 578 M. FOTOUHI and L. SALIMI Applying the above inequality for γ = β, from the fact a(x)u x L (, 1) u one can imply x β/ L (, 1) for all β < α. Thus, in this case we assume < β < α = 1, and no condition on λ is necessary..3. Statement of the problem. Consider the operator Au := (a(x)u x ) x + λ x β u (8) where the coefficient a( ) satisfies Hypothesis 1. As mentioned in the last subsection, we will study the operator A under one of the two assumptions (7). Now let ω be a nonempty subinterval of (, 1) and consider the following initial-boundary value problem in the domain Q T = (, T ) (, 1). u t Au = hχ ω, (t, x) Q T, u(t, ) = u(t, 1) =, in the case α (, 1), t (, T ), (a(x)u x )(t, ) = u(t, 1) =, in the case α [1, ), t (, T ), u(, x) = u (x), x (, 1), where the initial condition u is given in L (, 1) and h L (Q T ). Depending on the value of α, we choose some natural boundary conditions. The problem is considered in appropriate weighted spaces which will be described in the next section. We will see that in these spaces, the Dirichlet boundary condition makes sense when α (, 1). (See Lemma 7). But for α 1, the trace at x = does not make sense anymore, so we choose some suitable Neumann boundary condition in this case. (Lemma 1). We are interested in the null controllability of (9) in time T > with a distributed control supported in ω, i.e. for all u L (, 1), does there exist h L (Q T ) such that u(t, x) = for every x [, 1]? In this purpose, we drive Carleman estimates for operator A in Section. But before going any further, we describe in the following section the functional setting in which problem (9) is well-posed. 3. Functional setting and well-posedness In order to investigate the well-posedness of equation (9), the unbounded operator A : D(A) L (, 1) L (, 1) has to be studied in appropriate weighted spaces whose definitions depend on the values of α. Indeed a natural functional setting involves the space H 1 a(, 1) := {u L (, 1) H 1 loc((, 1]) : which is a Hilbert space for the scalar product u, v H 1 a(, 1), (u, v) H 1 a = a(x)u xdx < }, uv + a(x)u x v x dx. (9)

7 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 579 The elements of H 1 a(, 1) are characterized by the following property. Lemma 6. For α (, ) and every u H 1 a(, 1), we have lim x xu = and lim x xu =. Let us mention that for the well-posedness of boundary conditions in problem (9), we need to define the trace of u at boundary points x = and x = 1 for any u H 1 a(, 1). The trace at x = 1 obviously makes sense which allows to consider the Dirichlet boundary condition at this point. (Note that the function u H 1 a(, 1) belongs to the Sobolev space W 1, in a neighborhood of x = 1). On the other hand, if α < 1, the trace of u at x = is meaningful because of the following lemma. Lemma 7. If a( ) satisfies Hypothesis 1 and α (, 1), then for every u H 1 a(, 1) we have u W 1,1 (, 1) = {u L 1 (, 1) : u x L 1 (, 1)} and so u() is meaningful. Thus we could introduce the following space H 1 a,(, 1) depending on the values of α: Definition 8. (i) For α (, 1), we define (ii) For α [1, ), we let H 1 a,(, 1) := {u H 1 a(, 1) : u() = u(1) = }, H 1 a,(, 1) := {u H 1 a(, 1) : u(1) = }. There exists an important density result for the elements of H 1 a,. Proposition 9. (i) For α (, 1), the space Cc (, 1) is dense in Ha,(, 1 1). (ii) In the case α [1, ), the subset of C ([, 1]) which vanishes at x = 1 is dense in Ha,(, 1 1). Before going to define D(A) we state the following lemma which results that the boundary condition makes sense in the case α [1, ). Lemma 1. Assume that α [1, ) and the condition (7) holds. Then for all u H 1 a(, 1) such that (au x ) x + λ x β u L (, 1), we have au x W 1,1 (, 1). Definition 11. (i) For α (, 1), we define D(A) := {u H 1 a,(, 1) H loc((, 1]) : (au x ) x + λ x β u L (, 1)}, (ii) For α [1, ), we change the definition of D(A) in the following way D(A) := {u H 1 a,(, 1) H loc((, 1]) : (au x ) x + λ x β u L (, 1), (au x )() = }.

8 58 M. FOTOUHI and L. SALIMI Thus, in the case α (, 1), if u D(A), then u satisfies the Dirichlet boundary conditions u() = u(1) = and in the case α [1, ), every u D(A) satisfies the Neumann boundary condition (au x )() = and the Dirichlet boundary condition u(1) =. For the well-posedness of (9), it suffices to show that for suitable k the operator (A ki) is self-adjoint and positive. Proposition 1. Assume that the condition (7) holds, then there exists a constant k such that the operator ( (A ki), D(A)) is a self-adjoint and positive operator. Remark 13. As one can see in the proof of Proposition 1, the bilinear form associated to (A ki) is coercive in H 1 a,(, 1). Consequently, we have the following well-posedness result (see e.g. [8]). Theorem 1. Assume that the condition (7) holds and consider the problem (9) with h. Then, for all initial condition u L (, 1), the problem (9) has a unique solution u C ([, T ], L (, 1)) C ((, T ], D(A)) C 1 ((, T ], L (, 1)). (1) Moreover, if u D(A), then u C ([, T ], D(A)) C 1 ([, T ], L (, 1)). (11) In addition, the inhomogeneous problem (9) with h L (Q T ), has a unique solution u C ([, T ], L (, 1)) for all initial condition u L (, 1).. Carleman estimates and Applications to controllability As it is well-known, in order to get controllability results, we need to derive some observability inequalities for the adjoint problem v t + (a(x)v x ) x + λ x β v =, (t, x) Q T, v(t, 1) =, t (, T ), v(t, ) =, in the case α (, 1), t (, T ), (av x )(t, ) =, in the case α [1, ), t (, T ), v(t, x) = v T (x), x (, 1). More precisely, we need to prove the following inequality. (1) Proposition 15. Assume that the coefficient a( ) satisfies Hypothesis 1, let T > be given and ω be a nonempty subinterval of (, 1). Then there exists a positive constant C = C(T, a, α, λ) such that the following observability inequality is valid for every solution v of (1), v (, x)dx C v (t, x)dxdt. (13) Now, by standard arguments, a null controllability result follows. ω

9 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 581 Theorem 16. Assume that (7) holds. Let T > be given, and let ω be a nonempty subinterval of (, 1). Then, for all u L (, 1), there exists h L ((, T ) ω) such that the solution of (9) satisfies u(t ) in (, 1). Furthermore, we have the estimate for some C = C (T, a, α, λ) >. h L ((,T ) ω) C u L (,1), Here, we summarisely explain how one can deduce null controllability from observability. For the complete proof, one can see [11] and [1]. Let us fix T > and u L (, 1). We consider for any ɛ >, the functional v T J ɛ (v T ) on L (, 1): [ J ɛ (v T ) = 1 T v (t, x)dxdt] + ɛ v T L (,1) + v(, x)u (x)dx, ω where v is the corresponding solution of (1). One can see that v T J ɛ (v T ) is a continuous and strictly convex function on L (, 1). Moreover, J ɛ is coercive and achieves its minimum at a unique function ˆv ɛ. From ˆv ɛ, one can construct a control h ɛ such that u ɛ (T, ) L (,1) ɛ, for u ɛ be the corresponding solution of (9). Now, by observaility inequality (13), we obtain h ɛ L ((,T ) ω) C u L (,1). Therefore, by extracting an appropriate subsequence, we get h ɛ h weakly-* in L ((, T ) ω). Since it is for all ɛ >, we deduce that h is such that the correspondin solution u of (9) satisfies u(t ). For the proof of the observability inequality (13), we need Carleman estimates for the degenerate and singular problems v t + (a(x)v x ) x + λ x β v rv = h, (t, x) Q T, v(t, 1) =, t (, T ), v(t, ) =, in the case α (, 1), t (, T ), (av x )(t, ) =, in the case α [1, ), t (, T ), v(t, x) = v T (x), x (, 1), (1) where r is a nonnegative fixed constant. To define this estimates, consider < γ < α and σ(t, x) := θ(t)p(x) where θ(t) := 1 [t(t t)] k, k := 1 + γ > 1, (15)

10 58 M. FOTOUHI and L. SALIMI p(x) := c ( x ) 1 y α a(y) dy c. (16) Observe that there exists some c > such that for all t (, T ) θ t (t) cθ 1+ 1 k (t), θtt (t) cθ 1+ k (t). (17) 1 For the moment we assume c 1 > and c >, so that p(x) < a(1)( α) for all x [, 1]. As we shall see, a nice choice is c 1 = 3, and the following theorem is valid. Theorem 17. Assume that the function a( ) satisfies Hypothesis 1 and let T >. In the case α (, ), β < α and λ R, for every γ < α there exists R = R (a, α, γ, λ) > such that for all R R and all solutions v of (1), we have R 3 ( α) θ 3 + x a(x) v e Rσ(t,x) dxdt+r a(1)(1 α) + R + R 1 θ v x α erσ(t,x) dxdt θ v x γ erσ(t,x) dxdt h e Rσ(t,x) dxdt + 3Ra(1) α θa(x)v xe Rσ(t,x) dxdt θ(t)v x(t, 1)e Rσ(t,1) dt. (18) Also, in the case α (, ) \ {1}, β = α and λ < λ (a, α), for every γ < α there exists a constant R = R (a, α, γ, λ) > such that, for all R R and all solutions v of (1), we have R 3 ( α) θ 3 x a(x) v e Rσ(t,x) dxdt + RC λ v x α )erσ(t,x) dxdt + R 1 { } where C = min 1, λ λ. λ h e Rσ(t,x) dxdt + 3Ra(1) α θ(a(x)v x θ v x γ erσ(t,x) dxdt θ(t)v x(t, 1)e Rσ(t,1) dt, (19)

11 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS Hardy-Type inequalities For the proof of Proposition 1, we need to use some Hardy-type inequalities. One of them is the following. Lemma 18. Let α (, ). There exists an optimal constant λ (a, α) such that for every u Ha,(, 1 1), we have a(x)u xdx λ (a, α) u dx. () x α In fact, one has λ (a, α) a(1)(1 α) /. Therefore λ (a, α) > for every α (, ) \ {1}, but λ (a, 1) might be equal to zero. Remark 19. Observe that in the case α = 1, the inequality () doesn t u take any estimate for dx. However, in a similar way one can prove a x weaker Hardy inequality in this case: for every u H 1 a,(, 1). a(x)u xdx a(1) Remark. For α (, ), by () and (1) we have a(x)u xdx C u dx, (1) x(ln x) u dx, u H 1 a,(, 1). Then we can consider the new equivalent norm u H 1 a, (,1) := ( ) 1 1 a(x)u xdx on Ha,(, 1 1). Another useful inequality to achieve the desired result is the following improved Hardy inequality. Theorem 1. Suppose that the function a( ) satisfies Hypothesis 1 and let λ < λ (a, α) be given. Then for all n > and γ < α, there exists some positive constant C = C (a, α, λ, γ, n) > such that, for all u Ha,(, 1 1), the following inequality holds: a(x)u xdx + C u dx λ u dx + n x α u dx. () xγ Proof. For the simplest example of the function a( ), namely x α, Vancostenoble in [19] proved that if α [, ), then for all n > and γ < α there exists some positive constant C = C (α, γ, n) > such that, for all u C c (, 1), the following inequality holds: x α u xdx + C u dx (1 α) u u dx + n dx. (3) x α xγ

12 58 M. FOTOUHI and L. SALIMI Also C (α, γ, n) is explicitly given by C (α, γ, n) = (n + 1) α+γ α γ α γ α + γ ( ) γ α γ γ. ( α) γ Now, considering Proposition 9, we imply that the inequality (3) is true for every u Ha,(, 1 1) in the case α (, 1). For α [1, ), we can prove the inequality (3) for every u C ([, 1]) such that u(1) =, by the similar method used in [19]. Thus the inequality will be true for every u Ha,(, 1 1) and α (, ). On the other hand for every x [, 1], we have a(x) a(1)x α, so for all n > and γ < α, there exists some positive constant C = C (a, α, γ, n) > such that for all u Ha,(, 1 1), the following inequality holds. a(x)u xdx+c u a(1)(1 1 α) u u dx dx+n dx. () x α xγ In fact, C = a(1)c (α, n γ, ). Now, suppose that improved Hardy () is a(1) true for some µ λ = λ (a, α), i.e. for all n > and γ < α there exists some positive constant C = C (a, α, γ, n) such that, for all u Ha,(, 1 1), a(x)u xdx + C u dx µ By summation with (), we obtain a(x)u xdx + C u dx λ + µ u dx + n x α u x γ dx. u x α dx + n u x γ dx. Therefore, if improved Hardy () is true for µ, then it is also true for λ + µ. Now, if we define the sequence {µ k } with a(1)(1 α) µ =, µ k+1 = λ + µ k, then the improved Hardy inequality is true for every λ < λ = lim k µ k. Remark. Clearly, if λ = a(1)(1 α) /, Improved Hardy () is true for λ. But if λ > a(1)(1 α) /, the above proof doesn t work for λ in the general case. In fact the constant coefficient C in the inequality tends to infinity in the limit case µ k λ. If we let C k = C k (a, α, γ, n) the constant value in the Improved Hardy () for µ k, we have C (a, α, γ, n) = a(1)c ( α, γ, n a(1) ), C k+1 (a, α, γ, n) = 1 Ck (a, α, γ, n).

13 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 585 So, for γ >, we obtain C k (a, α, γ, n) = 1 ( k C a, α, γ, k n ). Therefore one can not deduce Improved Hardy for λ in such cases with the same method used for λ < λ. 6. Proofs 6.1. Proof of Lemmas, 6, 7 and 1. Proof of Lemma. Since xa (x) αa(x) for every x, the function x a(x) x α is decreasing on (, 1), so a(x) a(1)xα and for α < 1 dx a(x) dx a(1)x α < +. Proof of Lemma 6. At first, we show that xu W 1,1. It is obvious that xu L 1 (, 1) for every u H 1 a(, 1). On the other hand (xu ) x = u + xuu x, ( ) x ( ) xuu x = u a(x)ux, a(x) and by Hypothesis 1 one can easily see that the function x increasing, so x a(x) 1 a(1) x a(x) u L (, 1) xuu x L 1 (, 1). x a(x) is Hence xu W 1,1 (, 1) and it follows that xu L as x. If L >, then one could have L u x + x / L (, 1), thus L =. Similarly one can see that lim x xu =. Proof of Lemma 7. For any u Ha(, 1 1) we have u L (, 1), so u L 1 (, 1). We show that u x L 1 (, 1). u x dx = ( ) 1 a(x)u x dx a(x) a(x) dx a(x)u xdx, 1 But by Lemma the integral a(x) dx is finite, so u x L 1 (, 1). Now, consider a sequence {u n } of smooth functions which converge to u in W 1,1 (, 1) and let χ be a smooth cut-off function such that χ [, 1 ] 1

14 586 M. FOTOUHI and L. SALIMI and χ vanishes in some neighbourhood of 1. Then one has χu n χu in W 1,1. On the other hand for every x, we have χu n (x) = x (χu n ) x (t)dt, which means that the lim x χu n (x) exists, thus lim x u n (x) exists and we define u() to be equal to this value. Proof of Lemma 1. Let us denote w = a(x)u x and choose M >, such that a(x) M for every x [, 1]. One has w dx = a(x)u x dx a(x) u xdx M a(x)u xdx < +, since u H 1 a(, 1). Next we write w x dx = (a(x)u x ) x dx (a(x)u x ) x + λ x β u dx + λ x β u dx (a(x)u x ) x + λ u x β u dx + λ x β dx. The first integral in the right hand side of the latter inequality is finite since (a(x)u x ) x + λu/x β L (, 1). Furthermore we have u 1 x β dx = 1 u u dx x β/ xβ/ x β x β dx. Note that β < 1 since 1 α <, thus the first integral in the right hand side is finite. Also the second one is finite by () using the fact that β α for α (1, ) and β < α for α = Proof of Proposition 9 and Lemma 18. We first state the following lemma which will be useful for the proof of Proposition 9. For the proof see []. Lemma 3. Suppose that α (, 1), then for every u Ha(, 1 1) such that u() =, we have a(x) x u (x)dx (1 α) a(x) u (x) dx. (5) Remark. As one can see in [], the inequality (5) is true for α (, ) \ {1} by the hypothesis on the function a( ) defined in []. But our hypothesis is weaker than one defined in [] for α (1, ) and the inequality fails in this case (See Example 3). We will prove a similar inequality for α (, ) in Lemma 18.

15 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 587 Proof of Proposition 9. (i): Since Cc (, 1) is dense in H 1 (, 1) and the embedding of H 1 (, 1) into Ha,(, 1 1) is continuous, it suffices to prove that H 1 (, 1) is dense in Ha,(, 1 1). Let v Ha,(, 1 1) be given and define the family {v δ } δ, with δ (, 1) in the following way x v(x), x δ, v δ (x) := δ v(x), δ < x 1. We want to show that: (1) v δ H 1 (, 1) for all δ (, 1). () v δ v in Ha,(, 1 1) as δ. One has v δ(x) dx = δ δ 1 δ v(x) + x δ v (x) dx + 1 δ v(x) + x δ v (x) dx + By Hypothesis 1 the function x a(x) x δ v (x) dx δ v (x) dx. (6) is decreasing on (, 1], so a(x) a(1)x for every x [, 1]. Therefore for every δ > we have δ x δ v (x) dx 1 a(1)δ δ a(x) v (x) dx. (7) On the other hand, since a(x) > in (, 1], there exists M δ > such that a(x) M δ in [δ, 1], So δ v (x) dx M 1 δ Combining (6), (7) and (8) we obtain C δ > such that δ a(x) v (x) dx. (8) v δ(x) dx C δ v(x) + a(x) v (x) dx The right part of the last inequality is finite since v Ha(, 1 1). v δ H 1 (, 1). Also v v δ H 1 a, = = δ δ v v δ + a(x) v v δ dx v x δ v δ dx + a(x) v v δ x δ v dx v dx + δ a(x) v dx + δ a(x) v δ dx. Then

16 588 M. FOTOUHI and L. SALIMI Now δ v dx + δ a(x) v dx as δ, since v Ha,(, 1 1). On the other hand by Lemma 3 we obtain δ a(x) v δ dx δ a(x) v x dx (1 α) δ a(x) v (x) dx. Indeed, we can rewrite the proof of Lemma 3 in the interval [, δ] instead of [, 1] and derive a new inequality with the same constant (1 α). Now the right part of the last inequality tends to zero as δ, because v Ha,(, 1 1) and the proof of (i) is complete. (ii): Similarly in this case, for v Ha,(, 1 1) it suffices to construct functions {v δ } δ such that (1) v δ H 1 (, 1) and v δ (1) =. () v δ v in Ha,(, 1 1) as δ. Define v(δ x), x δ, v δ (x) := v(x), δ < x 1. We have v (δ x), < x < δ, v δ(x) := v (x), δ < x < 1. Since a is strictly positive on (, 1] and in the computing of v δ dx, we are far from the boundary, it is easy to see that v δ H 1 (, 1). Also v δ v H 1 a (,1) = v δ v H 1 a (,δ) [ v H 1 a (,δ) + v δ H 1 a (,δ)]. Since v Ha(, 1 1), the term v Ha 1 (,δ) tends to zero as δ. Also if δ δ / where δ is the constant introduced in property (iii) of Hypothesis 1, then δ v δ(x) + a(x) v δ(x) dx = δ δ δ v (δ x) + a(x) v (δ x) dx v (x) + 1 m δ δ a(x) v (x) dx, which tends to zero as δ. Observe that v δ (1) =, so the subset of C ([, 1]) which vanishes at x = 1 is dense in H 1 a,(, 1). Before proving Lemma 18, we remark that for α (, 1) this lemma is a simple consequence of Lemma 3 and the fact that a(x) a(1)x α for x [, 1]. But our proof works for α (, ). Proof of Lemma 18. Note that by Proposition 9 it is enough to prove () for u C c (, 1) in the case α (, 1) and for u C ([, 1]) such that

17 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 589 u(1) = in the case α [1, ). In these two cases we write ( x α ux 1 α ) u dx. Thus we get x α u xdx + (1 α) x α u (1 α) 1 1 dx x α x 1 α (u ) x dx. Now, if we use integration by parts, in the case α (, 1] there exists no boundary term, also in the case α (1, ), since u C ([, 1]) the term u tends to zero as x. Therefore, we obtain x1 α x α u xdx Note that the function x a(x) x α for every x [, 1] and deducely (1 α) a(x)u (1 α) xdx a(1) u dx. x α is decreasing on (, 1], so a(x) a(1)xα u dx. x α Let k arbi- Proof of Proposition 1. (i) (A ki) is positive: trarily for the moment, then for all u D(A) we have (A ki)u, u = Now, we distinguish three different cases: a(x)u x λ x β u + ku. case 1: α 1, β α, λ < λ (a, α), case : α 1, β < α, λ λ (a, α), case 3: α = 1, β < α, λ R. For the first case, we can write ( ) a(x)u x λ u x β λ 1 1 λ (a, α) Therefore, by Lemma 18, we obtain + a(x)u x λ u x β λ λ (a, α) ( 1 a(x)u x (a(x)u x λ (a, α) u ) λ λ u H (a, α) a,. 1 x β )).

18 59 M. FOTOUHI and L. SALIMI Hence the result holds with k =. In the second and third case, since β < α, one can use () with γ = β and n = λ, so there exists C > such that ) (a(x)u x λ u + C u, x β then a(x)u x λ u x β + C u 1 a(x)u x = 1 u Ha,. 1 Therefore the result holds with k = C. (ii) (A ki) is self-adjoint: Clearly, it is sufficient to show that A is self-adjoint, i.e. that (A, D(A )) = (A, D(A)). We have D(A ) = { v L (, 1) : C >, u D(A), v, Au C u L (,1)}. We first show D(A) D(A ) and A D(A) = A. Suppose that v D(A), for every u D(A), we have v, Au = v, (a(x)u x ) x + λ x β u = a(x)u x v x + λ uvdx = Av, u, xβ so v D(A ) and A v = Av. Now we show that D(A ) D(A). If we define then by the part (i): u, u 1 = u, φ 1 := kuφ + a(x)u x φ x λ x β uφdx, ku + a(x)u x λ x β u dx C u H 1 a,, and by Lemma 18 there exists C > such that u, u 1 C u H 1 a,. Thus, 1 defines a scalar product on Ha, 1 which corresponding norm is equivalent with H 1 a,, therefore Ha, 1 with the inner product, 1 is a Hilbert space. So for every f L (, 1) there exists a unique u Ha, 1 such that Thus u, φ 1 = fφdx, φ H 1 a,. ku (a(x)u x ) x λ x β u = f, then u D(A), and satisfies in the equation (ki A)u = f. Now suppose that v D(A ), we want to show v D(A). Let w := kv A v L and consider the solution (ki A)u = w. So for every φ D(A), we will have v, (ki A)φ = (ki A )v, φ = w, φ = (ki A)u, φ = u, (ki A)φ, since u, φ D(A) and A D(A) = A. Now let φ be the solution of (ki A)φ = v u, we can imply that v = u. Therefore v D(A) and Av = A v.

19 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS Proof of the Carleman estimates. In order to prove Theorem 17 we define the function w(t, x) := e Rσ(t,x) v(t, x), where v is the solution of (1) and R > is a positive constant. Then w satisfies (e Rσ w) t + (a(x)(e Rσ w) x ) x + λ (e Rσ w) re Rσ w = h, x β (t, x) Q T, w(t, 1) =, t (, T ), w(t, ) =, for α (, 1), t (, T ), (aw x )(t, ) = R(σ x aw)(t, ), for α [1, ), t (, T ), w(t, x) = w(, x) =, x (, 1). (9) Thanks to the definitions of p and σ, we have (σ x aw)(t, x) = c 1 α xθ(t)w(t, x). Since, for t [, T ], the function w(t, ) is in H1 a(, 1), we deduce that xw(t, x) x= = for t [, T ] using Lemma 6. Thus (σ x aw)(t, x) x= = and the previous problem can be recast as follows. Set Lv := v t + (a(x)v x ) x + λ x β v rv, L Rw := e Rσ L(e Rσ w). Then (9) becomes L R w = he Rσ, (t, x) Q T, w(t, 1) =, t (, T ), w(t, ) =, in the case α (, 1), t (, T ), (aw x )(t, ) =, in the case α [1, ), t (, T ), w(t, x) = w(, x) =, x (, 1). We have the following proposition which implies the Carleman estimates. (3) Proposition 5. Let T > be given. (i) In the case α (, ), β < α, λ R, for every γ < α there exists a constant R = R (a, α, γ, λ) > such that, for all R R and all solutions w of (3), we have 19R 3 ( α) θ 3 x a(x) w dxdt + R a(1)(1 T α) + R 1 dxdt + R x α h e Rσ(t,x) dxdt + 3Ra(1) α θa(x)w xdxdt x γ dxdt θ(t)w x(t, 1)dt. (31) (ii) In the case α (, ) \ {1}, β = α, λ < λ (a, α), for every γ < α there exists a constant R = R (a, α, γ, λ) > such that, for all R R and all solutions w of (3), we have 1R 3 ( α) θ 3 x a(x) w dxdt + 3R θ(a(x)wx λ w )dxdt x α

20 59 M. FOTOUHI and L. SALIMI + R x γ dxdt 1 + 3Ra(1) α h e Rσ(t,x) dxdt θ(t)w x(t, 1)dt. (3) For the proof of this proposition, separate self-adjoint and skew-adjoint terms. Then one has L R w = L + R w + L R w, where Therefore, we have L + R w := (a(x)w x) x Rσ t w + R a(x)σ xw + λ x β w rw, L R w := w t Ra(x)σ x w x R(a(x)σ x ) x w. he Rσ = L + R w + L R w + L + R w, L R w L+ R w, L Rw, (33) where and, denote the usual norm and scalar product in L (Q T ), respectively. The proof of Proposition 5, is based on the computation of the scalar product L + R w, L Rw, which comes in the following lemma. The proof is similar to one stated in [, 19]. Lemma 6. The scalar product L + R w, L Rw may be written as a sum of a distributed term A and a boundary term B, L + R w, L Rw = A + B, where the distributed term A is given by A = R + R + R 3 θθ t p xa(x)w dxdt + R whereas the boundary term B is given by θ tt pw dxdt θ(a (x)p xx + a(x)a (x)p x )w xdxdt θ 3 (a (x)p xx + a(x)a (x)p x )p xw dxdt B = Ra(1)c 1 α where θ(t) and p(x) defined by (15) and (16). βλr θ(t)w x(t, 1)dt, θp x a(x) x β+1 w, Now we estimate the distributed term A in the following lemma. Lemma 7. (i) In the case α (, ), β < α, λ R for every γ < α, there exists a constant R = R (a, α, γ, λ) > such that, for all

21 R R we have A NULL CONTROLLABILITY OF PARABOLIC EQUATIONS R3 ( α) + θ 3 x a(1)(1 α) R a(x) w dxdt + R dxdt + R x α θa(x)w xdxdt x γ dxdt. (ii) In the case α (, )\{1}, β = α, λ < λ (a, α), for every γ < α, there exists a constant R = R (a, α, γ, λ) > such that, for all R R we have A 1R3 ( α) + R θ 3 x x γ dxdt. a(x) w dxdt + 3R θ(a(x)wx λ w )dxdt x α Proof. By Lemma 6, assumption xa (x) αa(x) and the fact that p x = c 1 x, one can estimate A in the following way: ( α)a(x) Let A R c 1 ( α) + Rc 1 βλrc 1 α A 1 := R c 1 ( α) A := R A 3 := Rc 1 A := βλrc 1 α x θθ t a(x) w dxdt + R θa(x)wxdxdt + R3 c 3 1 ( α) x β dxdt. θ tt pw dxdt, θθ t x a(x) w dxdt, θa(x)wxdxdt + R3 c 3 1 ( α) x β dxdt. θ 3 x θ 3 x θ tt pw dxdt a(x) w dxdt a(x) w dxdt, First, we estimate the term A 1. According to the relation (17), we know that θθ t cθ + 1 k cθ 3, and obtain Therefore A 1 R c 1 c ( α) θ 3 x a(x) w dxdt.

22 59 M. FOTOUHI and L. SALIMI A ( R 3 c 3 1 ( α) R c ) 1 c ( α) + Rc 1 θ 3 x a(x) w dxdt θa(x)w xdxdt + A + A. (3) In the following, we produce estimates of the last two terms A and A to complete the proof. We distinguish two cases: the case of a sub-critical exponent β < α and the case of a critical exponent β = α. First case: α (, ), < β < α and λ R. We want to prove the result for all γ satisfying < γ < α. However, if the result holds for any γ such that β γ < α, then it obviously also holds for all γ such that < γ < α. Therefore, we consider here γ, such that β γ < α and we study the term A. In the case λ >, we apply the improved Hardy inequality () with n = λc α λβc 1 α + 3 α, which gives: a(x)wxdx + C w a(1)(1 α) dx + w dx x α ( βλc1 α + 3 α ) w x γ dx, for a suitable C = C (a, α, n, γ) = C (a, α, λ, γ, c 1 ). Therefore we can write A = βλrc 1 α a(1)(1 α) R + R(3 α) For λ, we have A = βλrc 1 α x β dxdt βλrc 1 α x Applying () with n = 3 α, that is a(x)wx + C w α dxdt R x γ dxdt C R a(1)(1 α) dxdt. xβ x γ dxdt θa(x)w xdxdt θw dxdt. w w + (3 α) x α x γ, we obtain the same estimate as in the case λ >. It follows that ( R 3 c 3 1 A ( α) R c ) 1 c ( α) θ 3 x a(x) w dxdt + R(c 1 1) θa(x)w xdxdt

23 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 595 a(1)(1 T α) + R + R(3 α) + A C R x γ dxdt dxdt x α θw dxdt. (35) Finally, we need to estimate the two last terms in the above inequality. By (17), we have θ tt p Kθ 1+ k, for some K >. It follows that A C R θw dxdt RK θ 1+ k w dxdt, (36) for some K = K (a, α, λ, γ, c 1 ) >. At this stage, we use the special choice of k, that is and consider q = we have ɛ k = 1 + γ, k k 1 and q = k, so that 1 q + 1 q = 1. Then, for all ɛ >, θ 1+ k w dxdt = θ (1+ k 3 q )q a q q x q q w dxdt + C(ɛ) ) ) (θ 1+ k 3 q a 1 q x q w q (θ 3 q x q a 1 q w q dxdt θ 3 x a(x) w dxdt, (37) where C(ɛ) = (ɛq) q q q 1. Note that q (1 + k 3q ) = 1, q q = γ. Now if K > be such that a(x) q q K for every x [, 1], then we obtain R θ 1+ k w dxdt ɛk R x γ dxdt θ 3 x + C(ɛ)R Putting the estimate (38) in (36) and using (35), we obtain ( R 3 c 3 1 A ( α) R c ) 1 c ( α) C(ɛ)K R θ 3 x a(x) w dxdt a(x) w dxdt. (38)

24 596 M. FOTOUHI and L. SALIMI +R(c 1 1) +R(3 α ɛk K ) θa(x)w xdxdt + x γ dxdt. a(1)(1 T α) R dxdt x α Now take ɛ = ɛ(a, α, λ, γ, c 1 ) > such that 3 α ɛk K = 1, also let c 1 = 3. Thus there exists R = R (a, α, λ, γ) > such that for all R R A 19R3 ( α) + a(1)(1 α) R θ 3 x a(x) w dxdt + R dxdt + R x α θa(x)w xdxdt x γ dxdt. Second case: α (, ) \ {1}, β = α and λ < λ. Let us fix γ such that < γ < α. In the present case, we observe that A = λrc 1 x α. Therefore by (3) one has ( R 3 c 3 1 A ( α) R c ) 1 c ( α) θ 3 x a(x) w + Rc 1 θ (a(x)w x λ w Now, if we apply () with n =, we obtain a(x)w xdx λ w dx x α w x γ dx C for a constant C = C (a, α, γ). Therefore ( R 3 c 3 1 A ( α) R c ) 1 c ( α) θ 3 x a(x) w dxdt + Rc 1 λ w x α )dxdt + Rc 1 x γ dx Rc 1C x α w dx, ) + A. θ(a(x)w x θw dx + A. Next we need to estimate the term A and the last term in the above inequality. Proceeding as in the previous case, similar to the relations (36) and (38), we get A Rc 1C θw dxdt ɛk K R x γ dxdt + C(ɛ)K R θ 3 x a(x) w dxdt,

25 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 597 for suitable positive K and K. Then as in the previous case, we can set c 1 = 3 and there exists R = R (a, α, γ, λ) such that for every R R, one has A 1R3 ( α) + 3R θ 3 x a(x) w dxdt θ (a(x)w x λ w x α ) dxdt + R x γ dxdt. Now from Lemma 6, Lemma 7, the inequality (33) and the fact that c 1 = 3, we can easily imply Proposition 5. On the other hand, since v = e Rσ w and c 1 = 3, one has v x e Rσ ( w x + So, the left hand of (18) is smaller than ( ) R 3 ( α) + 18R3 ( α) + a(1)(1 α) R θ 3 x 9R x ) ( α) θ a(x) w. a(x) w dxdt+r dxdt + R x α θa(x)w xdxdt x γ dxdt. Also, note that w x (t, 1) = v x (t, 1)e Rσ(t,1), since v(t, 1) =. Now, by Proposition 5 part (i) of Theorem 17 follows immediately. For the proof of part (ii), note that the left hand of (19) is smaller than 1R 3 ( α) θ 3 x a(x) w dxdt + RC + λrc dxdt + R x α θ(a(x)wx λ w ) )dxdt x α x γ dxdt. Now, if λ, the above expression is smaller than the left hand of (3) and similar to (i) we can prove the desired result. But for λ >, we use Lemma 18 to get λrc dxdt λrc x α (λ λ) R ( θ θ (a(x)w x λ w a(x)wx λ a(x) w x α Therefore we can easily complete the proof of part (ii). x α ) dxdt ) dxdt. (39)

26 598 M. FOTOUHI and L. SALIMI 6.. Proof of the observability inequality. In this subsection we will prove Proposition 15 as a consequence of the Carleman estimates stated in Section. First, define ṽ(t, x) := e rt v(t, x), where v is the solution of (1) and r will be defined later. Then ṽ satisfies ṽ t + (a(x)ṽ x ) x + λ ṽ rṽ =, (t, x) Q x β T, ṽ(t, 1) =, t (, T ), ṽ(t, ) =, in the case α (, 1), t (, T ) (aṽ x )(t, ) = in the case α [1, ), t (, T ), ṽ(t, x) = e rt v T (x), x (, 1). () Furthermore, it is obvious that if the observability inequality (13) is true for ṽ, then it is also true for v. First, let us state the following standard inequality to be proved later on. Lemma 8. (Caccioppoli s inequality) Let ω ω, then there exist a positive constant C such that for every solution v of (1), ṽxe Rσ dxdt C ṽ dxdt. ω ω Proof of Proposition 15. We consider ω = (x, x 1) ω = (x, x 1 ) and a smooth cut-off function ψ 1 such that ψ = 1 in (, x ) and ψ = in (x 1, 1). Also define w := ψṽ where ṽ is the solution of (). Then w satisfies (1) with some right-hand side h explicitly given in term of ṽ and ṽ x and supported in ω. Applying Theorem 17 with γ = α and R = R (a, α, λ), we get (note that w x (t, 1) = ): R θw e Rσ dxdt R According to Lemma 8, we obtain R θw e Rσ dxdt C x γ erσ dxdt Ĉ (ṽx + ṽ )e Rσ dxdt. ω ω ṽ dxdt, next we use the definition of ψ to obtain a bound for v on (, x ) x θṽ e Rσ dxdt C ṽ dxdt. Using the properties of θ(t) and p(x), for a positive constant c(t ) we have 3T e c(t )R x ṽ dxdt C ṽ dxdt. T Clearly, using a similar cut-off argument, ṽ can be estimated on (x 1, 1) in the same way. In this case, we use the classical Carleman estimates, since R R ω ω

27 NULL CONTROLLABILITY OF PARABOLIC EQUATIONS 599 the operator A is neither degenerate nor singular on (x 1, 1). Therefore we get 3T 1 ṽ c(t )R Ce T (t, x)dxdt ṽ Ce c(t )R ṽ dxdt. T R Now, consider the function t ṽ (t, x)dx, its derivative is ṽṽ t dx = ṽ [ (a(x)ṽ x ) x λx ] β ṽ + rṽ dx = ω a(x)ṽ x λ x β ṽ + rṽ, and if let r = C where C be as defined in () when we set n = λ, then the last term is positive. (Note that this is true in the case β < α. For β = α, we can set r = and use Lemma 18). Therefore the function t ṽ (t, x)dx is nondecreasing in [, T ] and we obtain ṽ (, x)dx T 3T T ṽ c(t )R Ce (t, x)dxdt T ω ω ṽ dxdt. Proof of Lemma 8. Let us consider a smooth function ξ : R R such that ξ(x) 1, x R, ξ(x) = 1, x ω, ξ(x) =, x < x and x > x 1, and ξ > for x > x and x < x 1. Then = = d dt + = + + r = ξ e Rσ ṽ dxdt = (Rξ σ t e Rσ ṽ + ξ e Rσ ṽṽ t )dxdt Rξ σ t e Rσ ṽ dxdt (1) ( ξ e Rσ ṽ λ ) x β ṽ (aṽ x) x + rṽ dxdt Rξ σ t e Rσ ṽ dxdt λ x β ξ e Rσ ṽ dxdt (ξ e Rσ ṽ) x aṽ x dxdt () ξ e Rσ ṽ Rξ σ t e Rσ ṽ dxdt λ x β ξ e Rσ ṽ dxdt

28 6 M. FOTOUHI and L. SALIMI Hence, + + r a(ξ e Rσ ) x ṽṽ x dxdt + ξ e Rσ ṽ. ξ e Rσ aṽxdxdt = Case1: For λ, we obtain Thus, Therefore a(ξ e Rσ ) x ṽṽ x dxdt + ξ e Rσ aṽxdxdt + a(ξ e Rσ ) x ṽṽ x dxdt ( aξe Rσ ṽ x ) dxdt + ξ e Rσ aṽxdxdt + ( a (ξ e Rσ ) x ξe Rσ aξ e Rσ ṽ xdxdt (3) Rξ σ t e Rσ ṽ dxdt r λ x β ξ e Rσ ṽ Rξ σ t e Rσ ṽ dxdt Rξ σ t e Rσ ṽ dxdt ξ e Rσ ṽ dxdt. Rξ σ t e Rσ ṽ dxdt ( a (ξ e Rσ ) x ξe Rσ ṽ) dxdt. ) ṽ dxdt C ṽ dxdt. ω min a(x) e Rσ ṽxdxdt C ṽ dxdt. ω ω ω Case : For λ >, since the support of ξ is located in ω, then ξ e Rσ λ x β ṽ dxdt Ĉ ṽ dxdt, for some suitable Ĉ >. So, rewriting the relations of the case 1, we get min a(x) e Rσ ṽxdxdt ( C + Ĉ) ṽ dxdt. ω ω ω ω

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30 6 M. FOTOUHI and L. SALIMI 16. P. Martinez, J. P. Raymond, and J. Vancostenoble, Regional null controllability for a linearized Crocco-type equation. SIAM J. Control Optim. (3), D. L. Russell, A uniform boundary controllability theory for hyperbolic and parabolic partial differential equations. Stud. Appl. Math. 5 (1973), T. I. Seidman, Exact boundary control for some evolution equations. SIAM J. Control Optim. 16 (1978), J. Vancostenoble, Improved Hardy-Poincare inequalities and sharp Carleman estimates for degenerate/singular parabolic problems. Discrete Contin. Dyn. Syst. Ser.S (11), J. Vancostenoble and E. Zuazua, Null controllability for the heat equation with singular inverse-square potentials. Journal of Functional Analysis 5(7) (8), (Received September 3 11, received in revised form March 17 1) Authors addresses: M. Fotouhi Sharif University of Technology, Tehran, Iran fotouhi@sharif.edu L. Salimi Sharif University of Technology, Tehran, Iran salimi@mehr.sharif.ir