Carleman estimates and observability inequalities for parabolic equations with interior degeneracy Genni Fragnelli∗ Dipartimento di Matematica Universit`a di Bari ”Aldo Moro” Via E. Orabona 4 70125 Bari - Italy email:
[email protected] Dimitri Mugnai Dipartimento di Matematica e Informatica Universit`a di Perugia Via Vanvitelli 1, 06123 Perugia - Italy email:
[email protected]
Abstract We consider a parabolic problem with degeneracy in the interior of the spatial domain, and we focus on Carleman estimates for the associated adjoint problem. The novelty of interior degeneracy does not let us adapt previous Carleman estimate to our situation. As an application, observability inequalities are established.
Keywords: degenerate equation, interior degeneracy, Carleman estimates, observability inequalities MSC 2010: 35K65, 93B07
1
Introduction
In this paper we focus on two subjects that in the last years have been object of a large number of papers, i.e. degenerate problems and Carleman estimates. Indeed, as pointed out by several authors, many problems coming from physics (boundary layer models in [9], models of Kolmogorov type in [6], models of Grushin type in [5], . . . ), biology (Wright-Fisher models in [32] and FlemingViot models in [20]), and economics (Black-Merton-Scholes equations in [16]) are described by degenerate parabolic equations. ∗ Research supported by the GNAMPA project Equazioni di evoluzione degeneri e singolari: controllo e applicazioni
1
On the other hand, the fields of applications of Carleman estimates are so wide that it is not surprising that also several papers are concerned with such a topic. A first example is their application to global null controllability (see [1], [2], [3], [4], [11], [12], [13], [18], [21], [22], [27], [28], [26], [29], [34] and the references therein): for all T > 0 and for all initial data u0 ∈ L2 ((0, T ) × (0, 1)) there is a suitable control h ∈ L2 ((0, T ) × (0, 1)) such that the solution of ( ut − (aux )x = h(t, x)χω (x), (t, x) ∈ (0, T ) × (0, 1), (1.1) u(0, x) = u0 (x) with some boundary conditions, satisfies u(T, x) = 0 for all x ∈ [0, 1]. Here χω denotes, as usual, the characteristic function of the set ω, i.e. χ(x) = 1 if x ∈ ω and χ(x) = 0 if x 6∈ ω. Moreover, Carleman estimates may be a fundamental tool in inverse problems, in parabolic, hyperbolic and fractional settings, e.g. see [15], [25], [30], [31], [35], [36] and their references. However, all the previous papers deal with problems like (1.1) which are non degenerate or admit that the the function a degenerates at the boundary of the domain, for example a(x) = xk (1 − x)α ,
x ∈ [0, 1],
where k and α are positive constants. To our best knowledge, [33] is the first paper treating a problem with a degeneracy which may occur in the interior of the spatial domain. In particular, Stahel considers a parabolic problem in a bounded smooth domain Ω of RN with Dirichlet, Neumann or mixed boundary conditions, associated to a N ×N matrix a, which is positive definite and symmetric, but whose smallest eigenvalue might converge to 0 as the space variable approaches a singular set contained in the closure of the spatial domain. In this case, he proves that the corresponding abstract Cauchy problem has a solution, provided that a−1 ∈ Lq (Ω, R) for some q > 1, where a(x) := min{a(x)ξ · ξ : kξk = 1}. Moreover, while in [33] only the existence of a solution for the parabolic problem is considered, in [23] the authors analyze in detail the degenerate operator Au := (aux )x in the space L2 (0, 1), with or without weight, proving that in some cases it is nonpositive and selfadjoint, hence it generates a cosine family and, as a consequence, an analytic semigroup. In [23] the well-posedness of (1.1) with Dirichlet boundary conditions is also treated, but nothing is said about controllability properties. This paper is then concerned with several inequalities (Carleman estimates, observability inequalities, Hardy–Poincar´e inequalities) related to the parabolic
2
equation with interior degeneracy ut − (aux )x = h(t, x)χω (x), u(t, 0) = u(t, 1) = 0, u(0, x) = u0 (x),
(t, x) ∈ (0, T ) × (0, 1), (1.2)
where (t, x) ∈ QT := (0, T ) × (0, 1), u0 ∈ L2 (0, 1), a degenerates at x0 ∈ (0, 1) and the control h ∈ L2 (QT ) acts on a nonempty subdomain ω of (0, 1) such that x0 ∈ ω. We shall admit two types of degeneracy for a, namely weak and strong degeneracy. In particular, we make the following assumptions: Hypothesis 1.1. Weakly degenerate case (WD): there exists x0 ∈ (0, 1) such that a(x0 ) = 0, a > 0 on [0, 1] \ {x0 }, a ∈ C 1 [0, 1] \ {x0 } and there exists K ∈ (0, 1) such that (x − x0 )a0 ≤ Ka in [0, 1] \ {x0 }. Hypothesis 1.2. Strongly degenerate case (SD): there exists x0 ∈ (0, 1) such that a(x0 ) = 0, a > 0 on [0, 1] \ {x0 }, a ∈ C 1 [0, 1] \ {x0 } ∩ W 1,∞ (0, 1) and there exists K ∈ [1, 2) such that (x − x0 )a0 ≤ Ka in [0, 1] \ {x0 }. Typical examples for weak and strong degeneracies are a(x) = |x−x0 |α , 0 < α < 1 and a(x) = |x − x0 |α , 1 ≤ α < 2), respectively. The starting point of the paper is actually the analysis of the adjoint problem to (1.2) ( vt + (a(x)vx )x = h, (t, x) ∈ (0, T ) × (0, 1), (1.3) v(t, 1) = v(t, 0) = 0, t ∈ (0, T ). In particular, for any (sufficiently regular) solution v of such a system we derive the new fundamental Carleman estimate having the form Z T Z 1 2 2 2 3 3 (x − x0 ) v e2sϕ dxdt sΘa(vx ) + s Θ a 0 0 ! (1.4) Z TZ 1 Z T x=1 2 2sϕ 2sϕ 2 h e dxdt + s aΘe (x − x0 )(vx ) dt x=0 , ≤C 0
0
0
for all s ≥ s0 , where s0 is a suitable constant. Here Θ(t) := [t(T − t)]−4 , and ϕ(t, x) := Θ(t)ψ(x), with ψ(x) < 0 given explicitly in terms of a, see (3.3). Of course, for the Carleman inequality, the location of x0 with respect to the control set ω is irrelevant, since ω plays no rˆole at all. For the proof of the previous Carleman estimate a fundamental rˆole is played by the second basic result of this paper, that is a general Hardy-Poincar´e type inequality proved in Proposition 2.3, of independent interest, that we establish for all functions w which are only locally absolutely continuous in [0, 1] \ {x0 } and such that Z 1 p(x)|w0 (x)|2 dx < +∞, w(0) = w(1) = 0, and 0
3
and which reads Z 0
1
p(x) w2 (x)dx ≤ C (x − x0 )2
Z
1
p(x)|w0 (x)|2 dx.
0
Here p is any continuous function in [0, 1], with p > 0 on [0, 1] \ {x0 }, p(x0 ) = 0 and there exists q ∈ (1, 2) such that the function p(x) is nonincreasing on the left of x = x0 |x − x0 |q and nondecreasing on the right of x = x0 . x −→
Applying estimate (1.4) to any solution v of the adjoint problem (1.3) with h = 0, we shall obtain the observability inequality Z 1 Z TZ v 2 (0, x)dx ≤ C v 2 (t, x)dxdt, 0
0
ω
where now we consider the fact that x0 ∈ ω. Such a result is then extended to the complete linear problem ut − (a(x)ux )x + c(t, x)u = h(t, x)χω (x), (t, x) ∈ (0, T ) × (0, 1), u(t, 1) = u(t, 0) = 0, t ∈ (0, T ), u(0, x) = u0 (x), x ∈ (0, 1),
(1.5)
where c is a bounded function, previously proving a Carleman estimate associated to this problem, see Corollary 5.1 and Proposition 5.1. Finally, observe that on a we require that there exists K ∈ (0, 2) such that (x − x0 )a0 ≤ Ka in [0, 1], and K ≥ 2 is excluded. This technical assumption, which is essential in all our results, is the same made, for example, in [3], where the degeneracy occurs at the boundary of the domain and the problem fails to be null controllable on the whole interval [0, 1]. But since the null controllability for the parabolic problem and the observability inequality for the adjoint problem are equivalent ([26]), it is not surprising that we require K < 2. The paper is organized as follows. In Section 2 we give the precise setting for the weak and the strong degenerate cases and some general tools we shall use several times; in particular a general weighted Hardy–Poincar´e inequality is established. In Section 3 we provided the main result of this paper, i.e. a new Carleman estimate for degenerate operators with interior degeneracy. In Section 4 we apply the previous Carleman estimates together with a Caccioppoli type inequality to prove an observability inequality. Finally, in Section 5 we extend the previous results to complete linear problems. We conclude this introduction with the following Remark 1. At a first glance, one may think that our results can be obtained just by a “translation” of the ones obtained in [3], but this is not the case. Indeed, in [3] the degeneracy point was the origin, where the authors put suitable 4
homogeneous boundary conditions (Dirichlet for the (WD) case and weighted Neumann in the (SD) case) which coincide exactly with the ones obtained by the characterizations of the domains of the operators given in Propositions 2.1 and 2.2 below. In this way they can control a priori the possible uncontrolled behaviour of the solution at the degeneracy point, while here we don’t impose any condition on the solution in the interior point x0 .
2
Preliminary Results
In order to study the well-posedness of problem (1.2), we introduce the operator Au := (aux )x and we consider two different classes of weighted Hilbert spaces, which are suitable to study two different situations, namely the weakly degenerate (WD) and the strongly degenerate (SD) cases: CASE (WD): Ha1 (0, 1) := u is absolutely continuous in [0, 1], √ 0 au ∈ L2 (0, 1) and u(0) = u(1) = 0 , and Ha2 (0, 1) := u ∈ Ha1 (0, 1)| au0 ∈ H 1 (0, 1) ; CASE (SD): Ha1 (0, 1) := u ∈ L2 (0, 1) | u locally absolutely continuous in [0, x0 ) ∪ (x0 , 1], √ 0 au ∈ L2 (0, 1) and u(0) = u(1) = 0 and Ha2 (0, 1) := u ∈ Ha1 (0, 1)| au0 ∈ H 1 (0, 1) . In both cases we consider the norms √ kuk2Ha1 (0,1) := kuk2L2 (0,1) + k au0 k2L2 (0,1) , and kuk2Ha2 (0,1) := kuk2Ha1 (0,1) + k(au0 )0 k2L2 (0,1) and we set D(A) = Ha2 (0, 1). The function a playing a crucial rˆole, it is non surprising that the following lemma is crucial as well: Lemma 2.1. Assume that Hypothesis 1.1 or 1.2 is satisfied.
5
1. Then for all γ ≥ K the map |x − x0 |γ is nonincreasing on the left of x = x0 a and nondecreasing on the right of x = x0 , x 7→
so that lim
x→x0
2. If K < 1, then
|x − x0 |γ = 0 for all γ > K. a
1 ∈ L1 (0, 1). a
1 1 3. If K ∈ [1, 2), then √ ∈ L1 (0, 1) and 6∈ L1 (0, 1). a a Proof. The first point is an easy consequence of the assumption. Now, we prove the second point: by the first part, it follows that K x0 (1 − x0 )K |x − x0 |K ≤ max , . a(x) a(0) a(1) Thus 1 ≤ max a(x)
xK (1 − x0 )K 0 , a(0) a(1)
1 . |x − x0 |K
Since K < 1, the right-hand side of the last inequality is integrable, and then 1 ∈ L1 (0, 1). Analogously, one obtains the third point. a 1 On the contrary, the fact that a ∈ C 1 ([0, 1]), and √ ∈ L1 (0, 1) implies that a Z x 1 1 6∈ L (0, 1). Indeed, the assumptions on a imply that a(x) = a0 (s)ds ≤ a x0 1 1 ≥C 6∈ C|x − x0 | for a positive constant C. Thus for all x 6= x0 , a(x) |x − x0 | L1 (0, 1). We immediately start using the lemma above, giving the following characterizations for the (SD) case which are already given in [23, Propositions 2.3 and 2.4], but whose proofs we repeat here to make precise some calculations. Proposition 2.1. Let X := {u ∈ L2 (0, 1) | u locally absolutely continuous in [0, 1] \ {x0 }, √ 0 au ∈ L2 (0, 1), au ∈ H01 (0, 1) and (au)(x0 ) = u(0) = u(1) = 0}. Then, under Hypothesis 1.2 we have Ha1 (0, 1) = X. 6
Proof. Obviously, X ⊆ Ha1 . Now we take u ∈ Ha1 , and we prove that u ∈ X. First, observe that (au)(0) = (au)(1) = 0. Moreover, since a ∈ C 1 ([0, 1]), then (au)0 = a0 u + au0 ∈ L2 (0, 1). Thus, for x < x0 , one has Z x (au)0 (t)dt (au)(x) = 0
R x0
This implies that there exists limx→x− (au)(x) = (au)(x0 ) = 0 L ∈ R. If L 6= 0, then there exists C > 0 such that
0
(au)0 (t)dt =
|(au)(x)| ≥ C for all x in a neighborhood of x0 , x 6= x0 . Thus, setting C1 := it follows that |u2 (x)| ≥
C2 max[0,1] a(x)
> 0,
C1 C2 ≥ , a2 (x) a(x)
for all x in a left neighborhood of x0 , x 6= x0 . But, since the operator is strongly 1 degenerate, 6∈ L1 (0, 1) thus u 6∈ L2 (0, 1). Hence L = 0. Analogously, starting a from Z 1 (au)(x) = − (au)0 (t)dt for x > x0 , x
one can prove that limx→x+ (au)(x) = (au)(x0 ) = 0 and thus (au)(x0 ) = 0. 0 From this it also easily follows that (au)0 is the distributional derivative of au, and so au ∈ H01 (0, 1), i.e. u ∈ X. Using the previous result, one can prove the following additional characterization. Proposition 2.2. Let D := {u ∈ L2 (0, 1) u locally absolutely continuous in [0, 1] \ {x0 }, au ∈ H01 (0, 1), au0 ∈ H 1 (0, 1), au is continuous at x0 and (au)(x0 ) = (au0 )(x0 ) = u(0) = u(1) = 0}. Then, under Hypothesis 1.2 we have Ha2 (0, 1) = D(A) = D. √ Proof. D ⊆ D(A) : Let u ∈ D. It is sufficient to prove that au0 ∈ L2 (0, 1). Since au0 ∈ H 1 (0, 1) and u(1) = 0 (recall that a > 0 in [0, 1] \ {x0 }), for x ∈ (x0 , 1] we have Z
1
x
[(au0 )0 u](s)ds = [au0 u]1x −
Z
1
(a(u0 )2 )(s)ds = −(au0 u)(x) −
x
Z
1
x
7
(a(u0 )2 )(s)ds.
Thus (au0 u)(x) = −
Z
1
[(au0 )0 u](s)ds −
Z
1
(a(u0 )2 )(s)ds.
x
x
Since u ∈ D, (au0 )0 u ∈ L1 (0, 1). Hence, there exists lim (au0 u)(x) = L ∈ [−∞, +∞),
x→x+ 0
since no integrability is known about a(u0 )2 and such a limit could be −∞. If L 6= 0, there exists C > 0 such that |(au0 u)(x)| ≥ C for all x in a right neighborhood of x0 , x 6= x0 . Thus, by [23, Lemma 2.5], there exists C1 > 0 such that |u(x)| ≥
C C1 ≥p |(au0 )(x)| (x − x0 )
for all x in a right neighborhood of x0 , x 6= x0 . This implies that u 6∈ L2 (0, 1). Hence L = 0 and Z 1 Z 1 [(au0 )0 u](s)ds = − (a(u0 )2 )(s)ds. (2.1) x0
x0
If x ∈ [0, x0 ), proceeding as before and using the condition u(0) = 0, it follows that Z x0 Z x0 0 0 [(au ) u](s)ds = − (a(u0 )2 )(s)ds. (2.2) 0
0
By (2.1) and (2.2), it follows that Z 1 Z 1 [(au0 )0 u](s)ds = − (a(u0 )2 )(s)ds. 0
0
√
Since (au0 )0 u ∈ L1 (0, 1), then au0 ∈ L2 (0, 1). Hence, D ⊆ D(A). D(A) ⊆ D : Let u ∈ D(A). By Proposition 2.1, we know that au ∈ H01 (0, 1) and (au)(x0 ) = 0. Thus, it is sufficient to prove that (au0 )(x0 ) = 0. Toward this end, observe that, since au0 ∈ H 1 (0, 1), there exists L ∈ R such that lim (au0 )(x) = (au0 )(x0 ) = L. If L 6= 0, there exists C > 0 such that x→x0
|(au0 )(x)| ≥ C, for all x in a neighborhood of x0 . Thus |(a(u0 )2 )(x)| ≥
C2 , a(x)
for x in a neighborhood of x0 , x 6= x0 . By Lemma 2.1, this implies that √ all au0 6∈ L2 (0, 1). Hence L = 0, that is (au0 )(x0 ) = 0. 8
Now, let us go back to problem (1.2), recalling the following Definition 2.1. If u0 ∈ L2 (0, 1) and h ∈ L2 (QT ), a function u is said to be a (weak) solution of (1.2) if u ∈ C([0, T ]; L2 (0, 1)) ∩ L2 (0, T ; Ha1 (0, 1)) and Z
1
1
Z
Z u0 (x)ϕ(0, x) dx −
u(T, x)ϕ(T, x) dx − Z − aux ϕx dxdt +
0
0
uϕt dxdt = QT
Z
QT
hϕχω dxdt
QT
for all ϕ ∈ H 1 (0, T ; L2 (0, 1)) ∩ L2 (0, T ; Ha1 (0, 1)). As proved in [23] (see Theorems 2.2, 2.7 and 4.1), problem (1.2) is well-posed in the sense of the following theorem: Theorem 2.1. For all h ∈ L2 (QT ) and u0 ∈ L2 (0, 1), there exists a unique weak solution u ∈ C([0, T ]; L2 (0, 1)) ∩ L2 (0, T ; Ha1 (0, 1)) of (1.2) and there exists a universal positive constant C such that Z T 2 sup ku(t)kL2 (0,1) + ku(t)k2Ha1 (0,1) dt ≤ C(ku0 k2L2 (0,1) + khk2L2 (QT ) ). (2.3) t∈[0,T ]
0
Moreover, if u0 ∈ Ha1 (0, 1), then u ∈ H 1 (0, T ; L2 (0, 1)) ∩ C([0, T ]; Ha1 (0, 1)) ∩ L2 (0, T ; Ha2 (0, 1)), and there exists a universal positive constant C such that Z T 2 2 kut kL2 (0,1) + k(aux )x kL2 (0,1) dt sup ku(t)k2Ha1 (0,1) + t∈[0,T ]
0
≤C
ku0 k2Ha1 (0,1)
+
khk2L2 (QT )
(2.4)
(2.5)
.
Moreover, A generates an analytic semigroup on L2 (0, 1). So far we have introduced all the tools which will let us deal with solutions of problem (1.2), also with additional regularity. Now, we conclude this section with an essential tool for proving Carleman estimates and observability inequalities, that is a new weighted Hardy–Poincar´e inequality for functions which may not be globally absolutely continuous in the domain, but whose irregularity point is “controlled” by the fact that the weight degenerates exactly there. Proposition 2.3 (Hardy–Poincar´e inequality). Assume that p ∈ C([0, 1]), p > 0 on [0, 1] \ {x0 }, p(x0 ) = 0 and there exists q ∈ (1, 2) such that the function x −→
p(x) is nonincreasing on the left of x = x0 |x − x0 |q and nondecreasing on the right of x = x0 . 9
Then, there exists a constant CHP > 0 such that for any function w, locally absolutely continuous on [0, x0 ) ∪ (x0 , 1] and satisfying 1
Z
p(x)|w0 (x)|2 dx < +∞ ,
w(0) = w(1) = 0 and 0
the following inequality holds: Z 1 Z 1 p(x) 2 w (x) dx ≤ C p(x)|w0 (x)|2 dx. HP 2 (x − x ) 0 0 0
(2.6)
Proof. Fix any β ∈ (1, q) and ε > 0 small. Since w(1) = 0, applying H¨older’s inequality and Fubini’s Theorem, we have Z 1 p(x) w2 (x) dx 2 (x − x ) 0 x0 +ε Z 1 Z 2 p(x) 1 β/2 0 −β/2 = ((y − x ) w (y))(y − x ) dy dx 0 0 2 x0 +ε (x − x0 ) x Z 1 Z Z 1 p(x) 1 β 0 2 −β ≤ (y − x ) |w (y)| dy (y − x ) dy dx 0 0 2 x0 +ε (x − x0 ) x x Z 1 Z 1 p(x) 1 β 0 2 (y − x ) |w (y)| dy dx ≤ 0 β − 1 x0 +ε (x − x0 )1+β x Z 1 Z y 1 p(x) = dx dy (y − x0 )β |w0 (y)|2 1+β β − 1 x0 +ε x0 +ε (x − x0 ) Z 1 Z y 1 p(x) = (x − x0 )q−1−β dx dy. (y − x0 )β |w0 (y)|2 q β − 1 x0 +ε x0 +ε (x − x0 ) Now, thanks to our hypothesis, we find p(x) p(y) ≤ , q (x − x0 ) (y − x0 )q
∀ x, y ∈ [x0 + ε, 1], x < y.
Thus Z 1
p(x) w2 (x) dx (x − x0 )2 x0 +ε Z 1 Z y 1 p(y) β 0 2 q−1−β ≤ (y − x ) |w (y)| (x − x ) dx dy 0 0 β − 1 x0 +ε (y − x0 )q x0 +ε (2.7) Z 1 1 p(y) q−β β 0 2 = (y − x ) (y − x ) |w (y)| dy 0 0 (β − 1)(q − β) x0 +ε (y − x0 )q Z 1 1 = p(y)|w0 (y)|2 dy. (β − 1)(q − β) x0 +ε
10
Now, proceeding as before, and using the fact that w(0) = 0, one has Z x0 −ε p(x) w2 (x) dx 2 (x 0 − x) 0 Z Z x Z x0 −ε p(x) x β 0 2 −β (x − y) |w (y)| dy (x − y) dy dx ≤ 0 0 (x0 − x)2 0 0 0 Z x0 −ε Z x 1 p(x) β 0 2 ≤ (x − y) |w (y)| dy dx 0 β−1 0 (x0 − x)1+β 0 Z x0 −ε Z x0 −ε p(x) 1 q−1−β (x0 − y)β |w0 (y)|2 (x − x) dx dy. = 0 β−1 0 (x0 − x)q y By assumption p(x) p(y) ≤ , (x0 − x)q (x0 − y)q
∀ x, y ∈ [0, x0 − ε], y < x.
Hence, Z x0 −ε
p(x) w2 (x) dx (x0 − x)2 0 Z x0 −ε Z x0 −ε 1 p(y) β 0 2 ≤ (x − y) |w (y)| (x0 − x)q−1−β dx dy 0 q β−1 0 (x0 − y) y Z x0 −ε 1 p(y)|w0 (y)|2 dy. = (β − 1)(q − β) 0 (2.8) Passing to the limit as ε → 0 and combining (2.7) and (2.8), the conclusion follows.
3
Carleman Estimate for Degenerate Parabolic Problems
In this section we prove a crucial estimate of Carleman type, that will be useful to prove an observability inequality for the adjoint problem of (1.2) in both the weakly and the strongly degenerate cases. Thus, let us consider the problem ( vt + (avx )x = h, (t, x) ∈ (0, T ) × (0, 1), (3.1) v(t, 0) = v(t, 1) = 0, t ∈ (0, T ), where a satisfies the following assumption: Hypothesis 3.1. The function a satisfies Hypothesis 1.1 or Hypothesis 1.2 and there exists ϑ ∈ (0, K] such that the function x −→
a(x) is nonincreasing on the left of x = x0 |x − x0 |ϑ and nondecreasing on the right of x = x0 . 11
(3.2)
Here K is the constant appearing in Hypothesis 1.1 or 1.2, respectively. Remark 2. Observe that if x0 = 0, Hypothesis 3.1 is the same introduced in [3] only for the strongly degenerate case. On the other hand, here we have to require this additional assumption also in the weakly degenerate case. This is due to the fact that in this case we don’t know if u(x0 ) = 0 for all u ∈ Ha1 (0, 1), as it happens when x0 = 0 and one imposes homogeneous Dirichlet boundary conditions, as in [3]; indeed, the choice of homogeneous Dirichlet boundary conditions in [3] helps in controlling the function at the degeneracy point, while here we don’t require the corresponding condition u(x0 ) = 0, so that some other condition is needed. However, in both cases, Hypothesis 3.1 is satisfied if a(x) = |x − x0 |K , with K ∈ (0, 2). Now, let us introduce the function ϕ(t, x) := Θ(t)ψ(x), where Z x y − x0 1 and ψ(x) := c dy − c (3.3) Θ(t) := 1 2 , [t(T − t)]4 x0 a(y) x20 (1 − x0 )2 , and c1 > 0. A more precise rewith c2 > max a(1)(2 − K) a(0)(2 − K) striction on c1 will be needed later. Observe that Θ(t) → +∞ as t → 0+ , T − , and by Lemma 2.1 we have that, if x > x0 , Z x 1 (y − x0 )K ψ(x) ≤ c1 dy − c2 a(y) (y − x0 )K−1 x0 (x − x0 )K (x − x0 )2−K (1 − x0 )K (1 − x0 )2−K ≤ c1 − c2 ≤ c1 − c2 a(x) 2−K a(1) 2−K (1 − x0 )2 − c2 < 0. = c1 (2 − K)a(1) (3.4) In the same way one can treat the case x ∈ [0, x0 ), so that ψ(x) < 0
for every x ∈ [0, 1].
Moreover, it is also easy to see that ψ ≥ −c1 c2 . Our main result is thus the following. Theorem 3.1. Assume Hypothesis 3.1 and let T > 0. Then, there exist two positive constants C and s0 , such that every solution v of (3.1) in V := L2 0, T ; Ha2 (0, 1) ∩ H 1 0, T ; Ha1 (0, 1) (3.5) satisfies, for all s ≥ s0 , Z T Z 1 2 2 2 3 3 (x − x0 ) v e2sϕ dxdt sΘa(vx ) + s Θ a 0 0 ! Z TZ 1 Z T x=1 2 2sϕ 2sϕ 2 ≤C |h| e dxdt + sc1 aΘe (x − x0 )(vx ) dt x=0 , 0
0
0
where c1 is the constant introduced in (3.3). 12
3.1
Proof of Theorem 3.1
For s > 0, define the function w(t, x) := esϕ(t,x) v(t, x), where v is any solution of (3.1) in V; observe w ∈ V. Of course, w satisfies −sϕ w)t + (a(e−sϕ w)x )x = h, (e w(t, 0) = w(t, 1) = 0, w(T − , x) = w(0+ , x) = 0,
that, since v ∈ V and ψ < 0, then (t, x) ∈ (0, T ) × (0, 1), t ∈ (0, T ), x ∈ (0, 1).
(3.6)
The previous problem can be recast as follows. Set Lv := vt + (avx )x
and Ls w = esϕ L(e−sϕ w),
Then (3.6) becomes sϕ Ls w = e h, w(t, 0) = w(t, 1) = 0, w(T − , x) = w(0+ , x) = 0,
s > 0.
t ∈ (0, T ), x ∈ (0, 1).
Computing Ls w, one has − Ls w = L+ s w + Ls w,
where 2 2 L+ s w := (awx )x − sϕt w + s a(ϕx ) w,
and L− s w := wt − 2saϕx wx − s(aϕx )x w. Moreover, − + − + 2 − 2 2hL+ s w, Ls wi ≤ 2hLs w, Ls wi + kLs wkL2 (QT ) + kLs wkL2 (QT )
= kLs wk2L2 (QT ) = khesϕ k2L2 (QT ) ,
(3.7)
where h·, ·i denotes the usual scalar product in L2 (QT ). As usual, we will − separate the scalar product hL+ s w, Ls wi in distributed terms and boundary terms. Lemma 3.1. The following identity holds: − hL+ s w, Ls wi Z Z Z TZ 1 s T 1 2 3 0 2 2 = ϕtt w dxdt + s (2aϕxx + a ϕx )a(ϕx ) w dxdt {D.T.} 2 0 0 0 0 Z TZ 1 Z TZ 1 − 2s2 aϕx ϕtx w2 dxdt + s (2a2 ϕxx + aa0 ϕx )(wx )2 dxdt 0
0
0
0
(3.8) 13
Z Z Z T s2 1 s 1 2 t=T x=1 [w ϕt ]t=0 dx + [a(ϕx )2 w2 ]t=T + [awx wt ]x=0 dt − t=0 dt 2 0 2 0 0 Z T {B.T.} + [−sϕx (awx )2 + s2 aϕt ϕx w2 − s3 a2 (ϕx )3 w2 ]x=1 x=0 dt 0 Z T Z iT 1 1h 2 + dx. a(w ) [−sa(aϕx )x wwx ]x=1 dt − x x=0 2 0 0 0 − Proof. First, let us note that all integrals which appear in hL+ s w, Ls wi are well defined both in the weakly and the strongly degenerate case by Lemma 2.1, as simple calculations show, recalling that w = esϕ v. Moreover, we remark that all the following integrations by parts are justified by Proposition 2.2 and the fact that w ∈ V. Hence Z TZ 1 Z TZ 1 + Ls wwt dxdt = {(awx )x − sϕt w + s2 a(ϕx )2 w}wt dxdt 0 0 0 0 Z 1 Z T Z T 1 d x=1 2 = [awx wt ]x=0 dt − a(wx ) dx dt 0 0 2 dt 0 Z Z Z 1 Z 1 s T s2 T − dt dt ϕt (w2 )t dx + a(ϕx )2 (w2 )t dx 2 0 2 0 0 0 (3.9) Z Z Z T s2 1 s 1 2 t=T 2 2 t=T = [w ϕ ] dx + [a(ϕ ) w ] dt [awx wt ]x=1 dt − t t=0 x t=0 x=0 2 0 2 0 0 Z 1 Z TZ 1 Z T s 1 d a(wx )2 dx dt + − ϕtt w2 dxdt 2 dt 2 0 0 0 0 Z TZ 1 − s2 aϕx ϕxt w2 dxdt. 0
0
In addition, we have Z
T
1
Z
0
L+ s w(−2saϕx wx )dxdt
0 T
Z
1
Z
aϕt ϕx
0
Z
Z
= −2s
0
+ 2s2
T
Z
0
w2 2
dxdt − 2s3
Z 0
x
1
ϕx 0 T Z 1
(awx )2 2
dxdt x
a2 (ϕx )3 wwx dxdt
0
T
[−sϕx (awx )2 + s2 aϕt ϕx w2 − s3 a2 (ϕx )3 w2 ]x=1 x=0 dt
= 0
Z
T
Z
1
ϕxx (awx )2 dxdt − s2
+s 0
+ s3
0
Z 0
T
T
Z 0
Z
Z
1
(aϕt ϕx )x w2 dxdt
0
1
{(a(ϕx )2 )x aϕx + a(ϕx )2 (aϕx )x }w2 dxdt.
0
At this point, note that (aϕx )x = c1 Θ, so that (aϕx )xx = 0.
14
(3.10)
Moreover Z TZ 1
L+ s w(−s(aϕx )x w)dxdt
0 Z T
0
T
Z
[−sawx w(aϕx )x ]x=1 x=0 dt
= 0
1
Z
(3.11)
awx (aϕx )x wx dxdt
+s 0
0 T
Z
+ s2
1
Z
(aϕx )x ϕt w2 dxdt − s3
1
Z
a(ϕx )2 (aϕx )x w2 dxdt.
0
0
0
0
T
Z
Adding (3.9)-(3.11), (3.8) follows immediately. Now, the crucial step is to prove the following estimate: Lemma 3.2. Assume Hypothesis 3.1. Then there exists a positive constant s0 such that for all s ≥ s0 the distributed terms of (3.8) satisfy the estimate s 2
T
Z
1
Z
0
ϕtt w2 dxdt + s3
0
− 2s2
0
C ≥ s 2
Z 0
1
Z
0 T
Z
T
Z
(2aϕxx + a0 ϕx )a(ϕ2x )w2 dxdt
0
1
Z
Z
aϕx ϕtx w2 dxdt + s
0 T
Z 0
T
Z
0 1
C3 3 s Θa(wx )2 dxdt + 2
Z
1
0 T
0
(2a2 ϕxx + aa0 ϕx )(wx )2 dxdt Z 0
1
Θ3
(x − x0 )2 2 w dxdt, a
for a positive constant C. RT R1 − Proof. Using the definition of ϕ, the distributed terms of 0 0 L+ s wLs wdxdt take the form Z Z Z TZ 1 s T 1¨ Θψw2 dxdt + s3 Θ3 a(2aψ 00 + a0 ψ 0 )(ψ 0 )2 w2 dxdt 2 0 0 0 0 (3.12) Z TZ 1 Z TZ 1 0 2 2 ˙ − 2s2 ΘΘa(ψ ) w dxd + s Θa(2aψ 00 + a0 ψ 0 )(wx )2 dxdt. 0
0
0
0
Because of the choice of ψ(x), one has 2a(x)ψ 00 (x) + a0 (x)ψ 0 (x) = c1
2a(x) − a0 (x)(x − x0 ) . a(x)
Thus (3.12) becomes s 2
Z 0
T
Z
1 2 ¨ Θψw dxdt − 2s2
0
T
Z
0 Z T
1
Z
0 2 2 ˙ ΘΘa(ψ ) w dxdt
0 Z 1
+ sc1 0
+ s3 c1
Θ(2a(x) − a0 (x)(x − x0 ))(wx )2 dxdt
0
Z 0
T
Z
1
Θ3 (ψ 0 )2 (2a(x) − a0 (x)(x − x0 ))w2 dxdt.
0
15
By assumption, one can estimate the previous terms in the following way: s 2
T
Z
1
Z
2 ¨ Θψw dxdt − 2s2
0
0
0
T
Z
T
Z
1 0 2 2 ˙ ΘΘa(ψ ) w dxdt
0
Θ(2a(x) − a0 (x)(x − x0 ))(wx )2 dxdt 1
Z
Θ3 (ψ 0 )2 (2a(x) − a0 (x)(x − x0 ))w2 dxdt
+ s c1 0
0 T
Z T
1
Z
s 2
0 2 2 ˙ ΘΘa(ψ ) w dxdt +
0
0
Z
Z
0
0
≥ −2s2
T
1
Z
+ sc1 3
Z
1
Z
Θa(wx )2 dxdt + s3 C 3
+ sC 0
0
T
Z
T
Z
Z
1
1
Z
Θ3
0
0
2 ¨ Θψw dxdt
0
0
(x − x0 )2 2 w dxdt, a
˙ ≤ where C > 0 is some universal positive constant. Observing that |ΘΘ| 9/4 3 3/2 ¨ cΘ ≤ cΘ and |Θ| ≤ cΘ , for a positive constant c, we conclude that, for s large enough, Z TZ 1 Z TZ 1 2 0 2 2 ˙ Θ3 a(ψ 0 )2 w2 dxdt ΘΘa(ψ ) w dxdt ≤ 2cs2 −2s 0 0 0 0 Z TZ 1 Z Z C 3 3 T 1 3 (x − x0 )2 2 (x − x0 )2 2 = 2cs2 w dxdt ≤ s w dxdt. Θ3 c21 Θ a 4 a 0 0 0 0 Moreover, Z Z Z Z T 1 s T 1 s 2 ¨ Θψw dxdt ≤ c1 c Θ3/2 bw2 dxdt 0 0 2 0 0 2 Z Z c1 c2 T 1 3/2 2 c +s Θ w dxdt , 2 0 0 Z
x
T
Z
(3.13)
y − x0 |x − x0 |K dy ≥ 0. Now, since the function x 7→ is a(x) x0 a(y) nonincreasing on [0, x0 ) and nondecreasing on (x0 , 1] (see Lemma 2.1), one has (x − x0 )2 b(x) ≤ , see (3.4). Hence (2 − K)a(x)
where b(x) =
s c1 c 2
Z 0
0
1
Θ3/2 bw2 dxdt ≤
C3 3 s 8
Z 0
T
Z 0
1
Θ3
(x − x0 )2 2 w dxdt, a
for s large enough. R R T 1 It remains to bound the term 0 0 Θ3/2 w2 dxdt . Using the Young inequal-
16
ity, we find Z c1 c2 1 3/2 2 Θ w dx c 2 0 Z 3/4 1/4 2 c1 c2 1 a1/3 2 3 |x − x0 | 2 =s Θ c w Θ w 2 0 a |x − x0 |2/3 Z 1 Z 1 3c1 c2 a1/3 c1 c2 |x − x0 |2 2 ≤s c Θ w2 dx + s c Θ3 w dx. 2/3 4 4 a |x − x0 | 0 0
s
(3.14)
Now, consider p(x) = (a(x)|x)− x0 |4 )1/3 . It is clear that, set(the function 2/3 2/3 2 (1 − x0 )2 x0 , , by Lemma 2.1 we have p(x) = ting C1 := max a(0) a(1) 2/3 (x − x0 )2 a1/3 p(x) a(x) ≤ C1 a(x) and . Moreover, using = a(x) (x − x0 )2 |x − x0 |2/3 p(x) 4+ϑ Hypothesis 3.1, one has that the function ∈ (1, 2), , where q := q |x − x0 | 3 is nonincreasing on the left of x = x0 and nondecreasing on the right of x = x0 . The Hardy-Poincar´e inequality (see Proposition 2.3) implies Z
1
Θ 0
a1/3 w2 dx = |x − x0 |2/3
Z
1
p w2 dx ≤ CHP 2 (x − x ) 0 0 Z 1 ≤ CHP C1 Θa(wx )2 dx,
Z
1
Θ
0
Θp(wx )2 dx (3.15)
0
where CHP and C1 are the Hardy-Poincar´e constant and the constant introduced before, respectively. Thus, for s large enough, by (3.14) and (3.15), we have c1 c2 s c 2
Z
1
Θ 0
3/2
C w dx ≤ s 2 2
Z 0
1
C3 3 Θa(wx ) dx + s 8 2
Z 0
1
Θ3
(x − x0 )2 2 w dxdt, a
for a positive constant C. Using the estimates above, from (3.13) we finally obtain Z Z s T 1 C Z TZ 1 2 ¨ Θψw dxdt ≤ s Θa(wx )2 dxdt 2 0 0 2 0 0 Z Z C 3 3 T 1 3 (x − x0 )2 2 + s Θ w dxdt. 4 a 0 0
17
Summing up, we obtain Z TZ 1 Z Z s T 1 2 a(aϕx )xx wwx dxdt ϕtt w dxdt + s 2 0 0 0 0 Z TZ 1 Z TZ 1 (2a2 ϕxx + aa0 ϕx )(wx )2 dxdt aϕx ϕtx w2 dxdt + s − 2s2 + s3
0 T
Z
C s 2
0
0
(2aϕxx + a0 ϕx )a(ϕx )2 w2 dxdt
0
0
≥
0 1
Z
Z
T
Z
1
Θa(wx )2 dxdt +
0
0
C3 3 s 2
Z
T
Z
1
Θ3
0
0
(x − x0 )2 2 w dxdt. a
For the boundary terms in (3.8), it holds: Lemma 3.3. The boundary terms in (3.8) reduce to Z T x=1 −s Θ(awx )2 ψ 0 x=0 dt. 0
Proof. Using the definition of ϕ, we have that the boundary terms become Z T 0 2 ˙ (B.T.) = [awx wt − saΘ(aψ 0 )0 wwx + s2 ΘΘaψψ w 0
− s3 a2 Θ3 (ψ 0 )3 w2 − sΘ(awx )2 ψ 0 ]x=1 (3.16) x=0 dt t=T Z 1 2 s ˙ + s aw2 (ψ 0 )2 Θ2 − 1 a(wx )2 dx. + − w2 ψ Θ 2 2 2 0 t=0 Since w ∈ V, w ∈ C [0, T ]; Ha1 (0, 1) . Thus w(0, x), w(T, x), wx (0, x), wx (T, x) T R1 and 0 a(wx )2 0 dx are indeed well defined. Using the boundary conditions of w and the definition of w, we get that t=T Z 1 s 2 ˙ s2 2 0 2 2 1 2 − w ψ Θ + aw (ψ ) Θ − a(wx ) dx = 0. 2 2 2 0 t=0 Moreover, since w ∈ V, we have that wt (t, 0) and wt (t, 1) make sense. Therefore, also a(0)wx (t, 0) and a(1)wx (t, 1) are well defined. In fact w(t, ·) ∈ Ha2 (0, 1) RT and a(·)wx (t, ·) ∈ W 1,2 (0, 1) ⊂ C([0, 1]). Thus 0 [awx wt ]x=1 x=0 dt is well defined and actually equals 0, as we get using the boundary conditions on w. Now, consider the second, the third and the fourth terms of (3.16). By definition of ψ and using the hypothesis on a, the functions (aψ 0 )0 , aψψ 0 and a2 (ψ 0 )3 are bounded on [0, 1]. Thus, by the boundary conditions on w, one has Z T Z Th ix=1 x=1 0 2 ˙ s [aΘ(aψ 0 )0 wwx ]x=0 dt = s2 ΘΘaψψ w dt 0
x=0
0
= s3
Z 0
18
T
2 3 0 3 2 x=1 a Θ (ψ ) w x=0 dt = 0.
From Lemma 3.1, Lemma 3.2, and Lemma 3.3, we deduce immediately that there exist two positive constants C and s0 , such that all solutions w of (3.6) satisfy, for all s ≥ s0 , T
Z
1
Z
− L+ s wLs wdxdt ≥ Cs
0
0
T
Z
T
1
(x − x0 )2 2 w dxdt a 0 0 2 x=1 Θa (wx )2 ψ 0 x=0 dt.
Z
T
Z
Θa(wx )2 dxdt
0
0
+ Cs3
1
Z
−s 0
Z
Θ3
(3.17)
Thus, a straightforward consequence of (3.7) and of (3.17) is the next result. Proposition 3.1. Assume Hypothesis 3.1 and let T > 0. Then, there exist two positive constants C and s0 , such that all solutions w of (3.6) in V satisfy, for all s ≥ s0 , Z
T
1
Z
2
s
3
T
Z
Z
1
Θa(wx ) dxdt + s 0
0
0
Z
T
Z
≤C
|h| e 0
0
1 2 2sϕ(t,x)
Θ3 Z
dxdt + s
0
0
T
(x − x0 )2 2 w dxdt a ! 2 2 0 x=1 Θa (wx ) ψ x=0 dt. .
Recalling the definition of w, we have v = e−sϕ w and vx = −sΘψ 0 e−sϕ w + wx . Thus, by the Cauchy–Schwarz inequality Z T Z 1 (x − x0 )2 2 2sϕ sΘa(vx )2 + s3 Θ3 v e dxdt a 0 0 Z T Z 1 2 2 2 3 2 3 (x − x0 ) ≤ w dxdt, sΘa(wx ) dxdt + s c1 Θ a 0 0
−sϕ
e
and by Proposition 3.1, Theorem 3.1 follows.
4
Application of Carleman estimates to observability inequalities
In this section we provide a possible application of the Carleman estimates established in the previous section, considering the control problem (1.2). In particular, we consider the situation in which x0 is inside the control interval x0 ∈ ω = (α, β) ⊂ (0, 1).
19
(4.1)
Now, we associate to the linear problem (1.2) the homogeneous adjoint problem vt + (avx )x = 0, (t, x) ∈ QT , (4.2) v(t, 0) = v(t, 1) = 0, t ∈ (0, T ), v(T, x) = v (x) ∈ L2 (0, 1), T where T > 0 is given. By the Carleman estimate in Theorem 3.1, we will deduce the following observability inequality for both the weakly and the strongly degenerate cases: Proposition 4.1. Assume Hypothesis 3.1 and (4.1). Then there exists a positive constant CT such that every solution v ∈ C([0, T ]; L2 (0, 1))∩L2 (0, T ; Ha1 (0, 1)) of (4.2) satisfies Z
1
v 2 (0, x)dx ≤ CT
Z
0
4.1
0
T
Z
v 2 (t, x)dxdt.
(4.3)
w
Proof of Proposition 4.1
In this subsection we will prove, as a consequence of the Carleman estimate proved in Section 3, the observability inequality (4.3). For this purpose, we will give some preliminary results. As a first step, we consider the adjoint problem with more regular final–time datum vt + (avx )x = 0, (t, x) ∈ QT , (4.4) v(t, 0) = v(t, 1) = 0, t ∈ (0, T ), v(T, x) = v (x) ∈ D(A2 ), T where
n o D(A2 ) = u ∈ D(A) Au ∈ D(A)
and Au := (aux )x . Observe that D(A2 ) is densely defined in D(A) (see, for example, [8, Lemma 7.2]) and hence in L2 (0, 1). As in [11], [12] or [22], letting VT vary in D(A2 ), we define the following class of functions: n o W := v is a solution of (4.4) . Obviously (see, for example, [8, Theorem 7.5]) W ⊂ C 1 [0, T ] ; Ha2 (0, 1) ⊂ V ⊂ U, where, V is defined in (3.5) and U := C([0, T ]; L2 (0, 1)) ∩ L2 (0, T ; Ha1 (0, 1)). We start with 20
(4.5)
Proposition 4.2 (Caccioppoli’s inequality). Let ω 0 and ω two open subintervals of (0, 1) such that ω 0 ⊂⊂ ω ⊂ (0, 1) and x0 6∈ ω ¯ 0 . Let ϕ(t, x) = Θ(t)Υ(x), where Θ is defined in (3.3) and Υ ∈ C([0, 1], (−∞, 0)) ∩ C 1 ([0, 1] \ {x0 }, (−∞, 0)) is such that
c |Υx | ≤ √ in [0, 1] \ {x0 } (4.6) a for some c > 0. Then, there exist two positive constants C and s0 such that every solution v ∈ W of the adjoint problem (4.4) satisfies Z TZ Z TZ (vx )2 e2sϕ dxdt ≤ C v 2 dxdt, (4.7) 0
ω0
0
ω
for all s ≥ s0 . Remark 3. Of course, our prototype for Υ is the function ψ defined in (3.3). Indeed, s x − x0 |x − x0 |2 1 1 0 p |ψ (x)| = c1 = c1 ≤ cp a(x) a(x) a(x) a(x) by Lemma 2.1. Proof of Proposition 4.2. Let us consider a smooth function ξ : [0, 1] → R such that 0 ≤ ξ(x) ≤ 1, for all x ∈ [0, 1], ξ(x) = 1, x ∈ ω0 , ξ(x) = 0, x ∈ [ 0, 1] \ ω. Since v solves (4.4) and has homogeneous boundary conditions, by the choice of ϕ, we have Z 1 Z T Z TZ 1 d 0= ξ 2 e2sϕ v 2 dx dt = (2sξ 2 ϕt e2sϕ v 2 + 2ξ 2 e2sϕ vvt )dxdt 0 dt 0 0 0 Z TZ 1 Z TZ 1 2 2sϕ 2 = 2s ξ ϕt e v dxdt + 2 ξ 2 e2sϕ v(−(avx )x )dxdt 0
Z
0 T
0 1
Z
ξ 2 ϕt e2sϕ v 2 dxdt + 2
= 2s 0
Z
0 T
ξ 2 ϕt e2sϕ v 2 dxdt + 2
= 2s 0 T
Z
0 1
Z
0
Z
T
Z
0 T
ξ 2 ϕt e2sϕ v 2 dxdt + 2
ω
Z
+2 0
(ξ 2 e2sϕ )x avvx dxdt
0
0
= 2s Z
1
Z
ξ 2 e2sϕ a(vx )2 dxdt
+2 0
(ξ 2 e2sϕ v)x avx dxdt
0 T
Z
1
Z
0 1
Z
0 T
Z
Z
T
Z
0
ω
ξ 2 e2sϕ a(vx )2 dxdt.
ω
21
(ξ 2 e2sϕ )x avvx dxdt
Hence, by definition of ξ and the Cauchy–Schwartz inequality, the previous identity gives T
Z
Z
ξ 2 e2sϕ a(vx )2 dxdt = −2s
2 0
ω T
Z
Z
2
≤ −2s
ξ ϕt e 0
T
Z
ω
T
Z
2sϕ 2
√
Z
v dxdt +
ω
0
aξesϕ vx
2
dxdt
ω
√ (ξ 2 e2sϕ )x 2 dxdt a v ξesϕ Z Z 2 2sϕ 2 ξ ϕt e v dxdt +
Z
+ 0
ω T
Z = −2s
ω 2 2sϕ
0 T
Z
ξ 2 ϕt e2sϕ v 2 dxdt
(ξ 2 e2sϕ )x avvx dxdt
−2 0
Z
0
ω T Z
Z
T
Z
Z
ξ 2 e2sϕ a(vx )2 dxdt
ω
0
[(ξ e )x ]2 2 av dxdt. ξ 2 e2sϕ
Z
+ 0
T
ω
Thus, T
Z
Z
0
ξ 2 e2sϕ a(vx )2 dxdt ≤ −2
T
Z
ω
Z
0
Z
T
ω
Z
+ 0
ξ 2 sϕt e2sϕ v 2 dxdt
ω
[(ξ 2 e2sϕ )x ]2 2 av dxdt. ξ 2 e2sϕ
Since x0 6∈ ω ¯ 0 , then T
Z
Z
inf a(x)
x∈ω 0
Z
e
Z
Z
Z
ξ 2 e2sϕ a(vx )2 dxdt
(vx ) dxdt ≤ ω ¯0
0
ξ 2 e2sϕ a(vx )2 dxdt
≤ 0
T
Z
2
ω0
0 T
2sϕ
ω T
Z
≤ −2 0
ξ 2 sϕt e2sϕ v 2 dxdt +
ω
Z
T
0
Z ω
[(ξ 2 e2sϕ )x ]2 2 av dxdt. ξ 2 e2sϕ
Calculations show that sϕt e2sϕ is uniformly bounded if s ≥ s0 > 0, since Υ is strictly negative, a rough estimate being |sϕt e2sϕ | ≤ c
1 1/4 s0 (− max Υ)1/4
.
˙ ≤ cΘ5/4 and Indeed, |Θ| |sϕt e2sϕ | ≤ cs(−Υ)Θ5/4 e2sϕ ≤
c 5/4 s(−Υ)
for some constants c > 0 which may vary at every step.
22
On the other hand,
[(ξ 2 e2sϕ )x ]2 can be estimated by ξ 2 e2sϕ C e2sϕ + s2 (ϕx )2 e2sϕ .
Of course, e2sϕ < 1, while s2 (ϕx )2 e2sϕ can be estimated with c c (Υx )2 ≤ (− max Υ)2 a by (4.6), for some constants c > 0. In conclusion, we can find a positive constant C such that Z TZ Z TZ [(ξ 2 e2sϕ )x ]2 2 2 2sϕ 2 ξ sϕt e v dxdt + av dxdt −2 ξ 2 e2sϕ 0 ω 0 ω Z TZ ≤C v 2 dxdt, 0
ω
and the claim follows. We shall need the following lemma: Lemma 4.1. Assume Hypothesis 3.1 and (4.1). Then there exist two positive constants C and s0 such that every solution v ∈ W of (4.4) satisfies, for all s ≥ s0 , Z T Z 1 Z TZ (x − x0 )2 2 2sϕ v e dxdt ≤ C v 2 dxdt. sΘavx2 + s3 Θ3 a 0 w 0 0 Here Θ and ϕ are as in (3.3). For the proof of the previous lemma we need the following classical Carleman estimate (see, for example [3, Proposition 4.4]): Proposition 4.3 (Classical Carleman estimates). Let z be the solution of ( zt + (azx )x = h ∈ L2 (0, T ) × (A, B) , (4.8) z(t, A) = z(t, B) = 0, t ∈ (0, T ), where a ∈ C 1 [A, B] is a strictly positive function. Then there exist positive constants c, r and s0 such that for any s ≥ s0 Z TZ B Z TZ B sΘerζ (zx )2 e−2sΦ dxdt + s3 Θ3 e3rζ z 2 e−2sΦ dxdt 0 A 0 A (4.9) Z TZ B Z Th i e−2sΦ h2 dxdt − c
≤c
0
A
σ(t, ·)e−2sΦ(t,·) |zx (t, ·)|2
0
Here the functions ζ, σ and Φ are defined in the following way: Z B 1 p ζ(x) := dy, σ(t, x) := rsΘ(t)erζ(x) , a(y) x 23
x=B
dt.
x=A
Φ(t, x) := Θ(t)Ψ(x) and Ψ(x) := e2rζ(A) − erζ(x) > 0, where (t, x) ∈ [0, T ] × [A, B] and Θ is defined in (3.3). (Observe that Φ > 0 and Φ(t, x) → +∞, as t ↓ 0, t ↑ T .) Proof of Lemma 4.1. By assumption, we can find two subintervals ω1 ⊂ (0, x0 ), ω2 ⊂ (x0 , 1) such that (ω1 ∪ ω2 ) ⊂⊂ ω \ {x0 }. Now, set λi := inf ωi and βi := sup ωi , i = 1, 2 and consider a smooth function ξ : [0, 1] → R such that 0 ≤ ξ(x) ≤ 1, for all x ∈ [0, 1], ξ(x) = 1, x ∈ [λ1 , β2 ], ξ(x) = 0, x ∈ [0, 1] \ ω. Define w := ξv, where v is the solution of (4.4). Hence, w satisfies ( wt + (awx )x = (aξx v)x + ξx avx =: f, (t, x) ∈ (0, T ) × (0, 1), w(t, 0) = w(t, 1) = 0, t ∈ (0, T ).
(4.10)
Applying Theorem 3.1 and using the fact that w = 0 in a neighborhood of x = 0 and x = 1, we have Z TZ 1 Z T Z 1 2 2 2sϕ 2 3 3 (x − x0 ) w e dxdt ≤ C e2sϕ f 2 dxdt sΘa(wx ) + s Θ a 0 0 0 0 (4.11) for all s ≥ s0 . Then, using the definition of ξ and in particular the fact that ξx and ξxx are supported in ω ˜ , where ω ˜ := [inf ω, λ1 ] ∪ [β2 , sup ω], we can write f 2 = ((aξx v)x + ξx avx )2 ≤ C(v 2 + (vx )2 )χω˜ , since the function a0 is bounded on ω ˜ . Hence, applying Proposition 4.2 and (4.11), we get Z T Z β2 (x − x0 )2 2 2sϕ v e dxdt sΘa(vx )2 + s3 Θ3 a 0 λ1 Z T Z β2 (x − x0 )2 2 2sϕ = w e sΘa(wx )2 + s3 Θ3 dxdt a 0 λ1 (4.12) Z T Z 1 2 2 2sϕ 2 3 3 (x − x0 ) w e dxdt ≤ sΘa(wx ) + s Θ a 0 0 Z TZ Z TZ ≤C e2sϕ (v 2 + (vx )2 )dxdt ≤ C v 2 dxdt, 0
ω ˜
0
ω
for a positive constant C. Now, consider a smooth function η : [0, 1] → R such that 0 ≤ η(x) ≤ 1, for all x ∈ [0, 1], η(x) = 1, x ∈ [β h 2 , 1], i η(x) = 0, x ∈ 0, λ2 +2β2 . 3
24
Define z := ηv, where v is the solution of (4.4). Then z satisfies (4.8) and (4.9), with ih := (aηx v)x + ηx avx , A = λ2 and B = 1. Since h is supported in h λ2 +2β2 , β2 3
, by Propositions 4.2 and 4.3 with Z
1
ζ(x) = ζ1 (x) := x
1 p dy, a(y)
(4.13)
we get T
Z
1
Z
0
Z
sΘerζ1 (zx )2 e−2sΦ dxdt +
T
Z
1
Z
e−2sΦ h2 dxdt ≤ C
≤c λ2
0 T
0
Z
T
s3 Θ3 e3rζ1 z 2 e−2sΦ dxdt
Z
v 2 dxdt + C
Z
T
0
ω ˜1
Z
e−2sΦ (vx )2 dxdt
ω ˜1
v 2 dxdt,
≤C 0
Z
1
λ2
0
λ2 T
Z
Z
ω
(4.14) where ω ˜ 1 = (λ2 , β2 ). Now, choose the constant c1 in (3.3) so that e2rζ1 (λ2 ) − 1 2rζ1 (λ2 ) e −1 c1 ≥ max , x20 c2 − (1−x0 )2 c2 − a(1)(2−K) a(0)(2−K) where ζ1 is defined as before. Then, by definition of ϕ, the choice of c1 and by Lemma 2.1, one can prove that there exists a positive constant k, for example (1 − x0 )2 k = max max a, , a(1) [λ2 ,1] such that a(x)e2sϕ(t,x) ≤ kerζ1 (x) e−2sΦ(t,x) and
(x − x0 )2 2sϕ(t,x) e ≤ kerζ1 (x) e−2sΦ(t,x) ≤ ke3rζ1 (x) e−2sΦ(t,x) a(x)
for every (t, x) ∈ [0, T ] × [λ2 , 1]. Thus, by (4.14), one has Z
T
0
1
Z
(x − x0 )2 2 2sϕ z e dxdt a Z 1 Z TZ 1 rζ1 2 −2sΦ sΘe (zx ) e dxdt + k s3 Θ3 e3rζ1 z 2 e−2sΦ dxdt
λ2 T
Z ≤k
sΘa(zx )2 + s3 Θ3
0
λ2
Z
T
Z
≤ kC 0
0
v 2 dxdt,
ω
25
λ2
for a positive constant C. As a trivial consequence, Z T Z 1 (x − x0 )2 2 2sϕ sΘa(vx )2 + s3 Θ3 v e dxdt a β2 0 Z T Z 1 (x − x0 )2 2 2sϕ sΘa(zx )2 + s3 Θ3 = z e dxdt a β2 0 Z T Z 1 (x − x0 )2 2 2sϕ sΘa(zx )2 + s3 Θ3 ≤ z e dxdt a λ2 0 Z TZ ≤ kC v 2 dxdt, 0
(4.15)
ω
for a positive constant C. Thus (4.12) and (4.15) imply Z TZ Z T Z 1 (x − x0 )2 2 2sϕ v e dxdt ≤ C v 2 dxdt, (4.16) sΘa(vx )2 + s3 Θ3 a 0 ω 0 λ1 ıfor some positive constant C. To complete the proof it is sufficient to prove a similar inequality on the interval [0, λ1 ]. To this aim, we follow a reflection procedure introducing the functions ( v(t, x), x ∈ [0, 1], (4.17) W (t, x) := −v(t, −x), x ∈ [−1, 0], where v solves (4.4), and ( a ˜(x) :=
a(x), a(−x),
Then W satisfies the problem Wt + (˜ aWx )x = 0,
x ∈ [0, 1], x ∈ [−1, 0].
(4.18)
(t, x) ∈ (0, T ) × (−1, 1), (4.19)
W (t, −1) = W (t, 1) = 0,
t ∈ (0, T ).
Now, consider a cut off function ρ : [−1, 1] → R such that 0 ≤ ρ(x) ≤ 1, for all x ∈ [−1, 1], ρ(x) = 1, x ∈ (−λ h 1 , λ1 ), i h i ρ(x) = 0, x ∈ −1, − λ1 +2β1 ∪ λ1 +2β1 , 1 . 3
3
Define Z := ρW , where W is the solution of (4.19). Then Z satisfies (4.8) and (4.9), with h := (˜ aρx W )x + ρx a ˜Wx , A = −β1 and B = β1 . Now define Z β1 1 p ζ(x) = ζ2 (x) := dy, a ˜(y) x 26
Using Proposition 4.3 with ˜ x) := Θ(t)(e2rζ2 (−β1 ) − erζ2 (x) ), Φ(t,
(4.20)
the fact that Zx (t,h −β1 ) = Zx (t, iβ1 ) h= 0, the definition of W and the fact that i λ1 +2β1 λ1 +2β1 ρ is supported in − 3 , −λ1 ∪ λ1 , 3 , give Z
T
β1
Z
˜
sΘerζ2 (Zx )2 e−2sΦ dxdt +
−β1 Z T Z β1
0
Z
β1
˜
s3 Θ3 e3rζ2 Z 2 e−2sΦ dxdt
−β1
0 ˜
e−2sΦ h2 dxdt
≤C −β1
0 T
Z
T
Z
Z
≤C
−λ1
λ +2β − 1 3 1
0
e
˜ −2sΦ
2
Z
2
T
Z
λ1 +2β1 3
(W + (Wx ) )dxdt + C 0
˜
e−2sΦ (W 2 + (Wx )2 )dxdt.
λ1
(4.21) Now, putting Ξ(x) := e2rζ2 (−β1 ) − erζ2 (x) and A :=
e2rζ2 (−β1 ) − erζ2 (−β1 ) Ξ(−β1 ) = ∈ (0, 1), Ξ(β1 ) e2rζ2 (−β1 ) − 1
we note that for any x ∈ [0, β1 ], s ≥ s0 and t ∈ (0, T ) we have e−2sΘ(t)Ξ(−x) ≤ e−2AsΘ(t)Ξ(x) . Hence, using the oddness of the involved functions, Z
T
Z
−λ1
e
−
0
Z
T
˜ −2sΦ
λ1 +2β1 3 λ1 +2β1 3
Z
≤
2
Z
2
0
Z
2
T
Z
λ1 +2β1 3
v dxdt + C Z
T
Z
0
Z
2
≤
T
Z
λ1 +2β1 3
ω
0
˜
e−2AsΦ (W 2 + (Wx )2 )dxdt
λ1
e−2AsθΞ (vx )2 dxdt
λ1
λ1 +2β1 3
v dxdt + C 0
Z
(W + (Wx ) )dxdt ≤
λ1
0
T
e−2AsθΞ (vx )2 dxdt,
λ1
(4.22) for some positive constant C. Now, after relabeling s˜ = As, (4.22) and Proposition 4.2 imply the existence of C > 0 and s1 > 0 such that for all s ≥ s1 we get Z T Z −λ1 Z TZ ˜ e−2sΦ (W 2 + (Wx )2 )dxdt ≤ C v 2 dxdt. (4.23) 0
−
λ1 +2β1 3
0
ω
On the other hand, Proposition 4.2 immediately implies in an easier way that Z T Z λ1 +2β Z TZ 1 3 ˜ e−2sΦ (W 2 + (Wx )2 )dxdt ≤ C v 2 dxdt (4.24) 0
λ1
0
for all s large enough and for a suitable C > 0. 27
ω
In conclusion, (4.21)–(3.3) imply that there exists s0 and C > 0 such that Z
T
Z
β1
sΘe
rζ2
˜ 2 −2sΦ
(Zx ) e
−β1
0
T
Z dxdt+
Z
β1 3
3 3rζ2
s Θ e
˜ 2 −2sΦ
Z e
Z
Z
dxdt ≤ C
−β1
0
T
0
v 2 dxdt
ω
(4.25) for all s ≥ s0 . Now, define ˜ ϕ(t, ˜ x) := Θ(t)ψ(x), where ˜ ψ(x) :=
ψ(x), ψ(−x) = c1
x ≥ 0, Z
x
−x0
t + x0 dt − c2 , a ˜(t)
x < 0.
(4.26)
and choose the constant c1 so that e2rζ1 (λ2 ) − 1 2rζ1 (λ2 ) 2rζ2 (−β1 ) 2rζ2 (−β1 ) e −1 e −1 e −1 c1 ≥ max , , , . 2 2 2 2 (1−x ) (1−x ) x0 x0 0 0 c2 − a(1)(2−K) c2 − a(0)(2−K) c2 − a(1)(2−K) c2 − a(0)(2−K) Thus, by definition of ϕ, ˜ one can prove as before that there exists a positive constant k, for example x20 k = max max a ˜, , a(0) [−β1 ,β1 ] such that
˜
˜ a ˜(x)e2sϕ(t,x) ≤ kerζ2 (x) e−2sΦ(t,x)
and
(x − x0 )2 2sϕ(t,x) ˜ ˜ e ˜ ≤ kerζ2 (x) e−2sΦ(t,x) ≤ ke3rζ2 (x) e−2sΦ(t,x) a ˜(x)
for every (t, x) ∈ [0, T ] × [−β1 , β1 ]. Thus, by (4.25), one has Z
T
0
β1
Z
(x − x0 )2 2 2sϕ˜ Z e dxdt a ˜ Z T Z β1 ˜ ˜ rζ2 2 −2sΦ sΘe (Zx ) e dxdt + k s3 Θ3 e3rζ2 Z 2 e−2sΦ dxdt
sΘ˜ a(Zx )2 + s3 Θ3
−β1 T Z β1
Z ≤k
−β1
0
Z
T
Z
≤ kC 0
0
−β1
v 2 dxdt.
ω
(4.27)
28
Hence, by (4.27) and the definition of W and Z, we get T
λ1
(x − x0 )2 2 v + sΘa(vx )2 e2sϕ dxdt a 0 0 Z T Z λ1 (x − x0 )2 2 s3 Θ3 = W + sΘa(Wx )2 e2sϕ dxdt a 0 0 Z T Z λ1 (x − x0 )2 2 s3 Θ3 W + sΘ˜ a(Wx )2 e2sϕ˜ dxdt ≤ a ˜ −λ1 0 Z T Z λ1 (x − x0 )2 2 s3 Θ3 = Z + sΘ˜ a(Zx )2 e2sϕ˜ dxdt a ˜ −λ1 0 Z T Z β1 (x − x0 )2 2 s3 Θ3 ≤ Z + sΘ˜ a(Zx )2 e2sϕ˜ dxdt a ˜ −β1 0 Z TZ ≤C v 2 dxdt,
Z
Z
0
s3 Θ3
(4.28)
ω
for a positive constant C. Therefore, by (4.16) and (4.28), Lemma 4.1 follows. We shall also use the following Lemma 4.2. Assume Hypothesis 3.1 and (4.1). Then there exists a positive constant CT such that every solution v ∈ W of (4.4) satisfies 1
Z
T
Z
2
Z
v (0, x)dx ≤ CT 0
0
v 2 (t, x)dxdt.
ω
Proof. Multiplying the equation of (4.4) by vt and integrating by parts over (0, 1), one has Z 0=
1
Z
1
(vt + (avx )x )vt dx =
Z −
0 1
(vt2
avx vtx dx = 0
0
1
0
Z
1 d vt2 dx − 2 dt
1
0
1 d a(vx )2 ≥ − 2 dt
x=1
vt2 dx + [avx vt ]x=0
+ (avx )x vt )dx =
0
Z
1
Z
Z
1
a(vx )2 dx.
0
R1 Thus, the function t 7→ 0 a(vx )2 dx is increasing for all t ∈ [0, T ]. In particular, R1 R1 avx (0, x)2 dx ≤ 0 avx (t, x)2 dx. Integrating the last inequality over T4 , 3T 4 , 0 Θ being bounded therein, we find Z 0
1
2 a(vx ) (0, x)dx ≤ T 2
3T 4
Z T 4
a(vx )2 (t, x)dxdt
0 3T 4
Z
1
Z
Z
≤ CT T 4
0
29
1
sΘa(vx )2 (t, x)e2sϕ dxdt.
Hence, by Lemma 4.1 and the previous inequality, there exists a positive constant C such that Z 1 Z TZ 2 a(vx ) (0, x)dx ≤ C v 2 dxdt. (4.29) 0
0
ω
Proceeding again as in the proof of Lemma 3.2 and applying the HardyPoincar´e inequality, by (4.29), one has Z
1
0
a (x − x0 )2
1/3
Z
1
p v 2 (0, x)dx 2 0 (x − x0 ) Z 1 ≤ CHP p(vx )2 (0, x)dx
2
v (0, x)dx =
0 1
Z
a(vx )2 (0, x)dx ≤ C
≤ C1 CHP 0
Z 0
T
Z
v 2 dxdt,
ω
4 1/3
for a positive constant C. Here p(x) − x0 | ) , CHP)is the Hardy( = (a(x)|x 2/3 2/3 2 (1 − x0 )2 x0 , , as before. Poincar´e constant and C1 := max a(0) a(1) a(x) By Lemma 2.1, is nondecreasing on [0, x0 ) and nonincreasing on (x − x0 )2 (x0 , 1], then ( 1/3 1/3 1/3 ) a(1) a(0) a(x) ≥ C2 := min , > 0. (x − x0 )2 (1 − x0 )2 x20 Hence Z
1
v(0, x)2 dx ≤ C
C2 0
T
Z 0
Z
v 2 dxdt
ω
and the thesis follows. Proof of Proposition 4.1. The proof is now standard, but we give it with some precise references: let vT ∈ L2 (0, 1) and let v be the solution of (4.2) associated to vT . Since D(A2 ) is densely defined in L2 (0, 1), there exists a sequence (vTn )n ⊂ D(A2 ) which converges to vT in L2 (0, 1). Now, consider the solution vn associated to vTn . As shown in Theorem 2.1, the semigroup generated by A is analytic, hence A is closed (for example, see [17, Theorem I.1.4] ; thus, by [17, Theorem II.6.7], we get that (vn )n converges to a certain v in C(0, T ; L2 (0, 1)), so that Z 1 Z 1 2 lim vn (0, x)dx = v 2 (0, x)dx, n→+∞
0
and also Z
T
0
Z
lim
n→+∞
0
vn2 dxdt
ω
Z
Z
= 0
30
T
ω
v 2 dxdt.
But, by Lemma 4.2 we know that Z
1
vn2 (0, x)dx ≤ CT
0
Z 0
T
Z
vn2 dxdt.
ω
Thus Proposition 4.1 is now proved.
5
Linear Extension
In this section we want to extend the observability inequality proved in the previous section starting from linear complete problems of the form ut − (a(x)ux )x + c(t, x)u = h(t, x)χω (x), (t, x) ∈ (0, T ) × (0, 1), (5.1) u(t, 1) = u(t, 0) = 0, t ∈ (0, T ), u(0, x) = u0 (x), x ∈ (0, 1), where u0 ∈ L2 (0, 1), h ∈ L2 (QT ), c ∈ L∞ (QT ), ω is as in (4.1) and a satisfies Hypothesis 3.1. Observe that the well-posedness of (5.1) follows by [23, Theorem 4.1]. As for the previous case, we shall prove an observability inequality for the solution of the associated homogeneous adjoint problem vt + (avx )x − cv = 0, (t, x) ∈ (0, T ) × (0, 1), (5.2) v(t, 1) = v(t, 0) = 0, t ∈ (0, T ), v(T ) = vT ∈ L2 (0, 1). To obtain an observability inequality for (5.2) like the one in Proposition 4.1, we consider the problem ( vt + (a(x)vx )x − cv = h, (t, x) ∈ (0, T ) × (0, 1), (5.3) v(t, 1) = v(t, 0) = 0, t ∈ (0, T ), and we prove the following Carleman estimate as a corollary of Theorem 3.1: Corollary 5.1. Assume Hypothesis 3.1 and let T > 0. Then, there exist two positive constants C and s0 , such that every solution v in V of (5.3) satisfies, for all s ≥ s0 , Z T Z 1 (x − x0 )2 2 2sϕ sΘa(vx )2 + s3 Θ3 v e dxdt a 0 0 ! Z TZ 1 Z T x=1 2 2sϕ 2sϕ 2 ≤C h e dxdt + sc1 aΘe (x − x0 )(vx ) dt x=0 , 0
0
0
where c1 is the constant introduced in (3.3).
31
¯ where h ¯ := h + cv. Proof. Rewrite the equation of (5.3) as vt + (avx )x = h, Then, applying Theorem 3.1, there exists two positive constants C and s0 > 0, such that Z T Z 1 2 2 3 3 (x − x0 ) 2 sΘa(vx ) + s Θ v e2sϕ dxdt a 0 0 ! (5.4) Z T Z TZ 1 x=1 2 2sϕ 2sϕ 2 ¯ h e dxdt + sc1 aΘe (x − x0 )(vx ) dt x=0 ≤C 0
0
0
¯ the term for all s ≥ s0 . Using the definition of h, estimated in the following way T
Z
1
Z
0
¯ 2 e2sϕ dxdt ≤ 2 h
0
T
Z
0
0
¯ 2 e2sϕ(t,x) dxdt can be |h|
1
Z
0
RT R1
h2 e2sϕ dxdt + 2kck2L∞ (QT )
0
Z
T
0
Z
1
e2sϕ v 2 dxdt.
0
(5.5) Applying the Hardy-Poincar´e inequality (see Proposition 2.3) to w(t, x) := esϕ(t,x) v(t, x) and proceeding as in (3.14), recalling that 0 < inf Θ ≤ Θ ≤ cΘ2 , one has Z 1 Z 1 Z 1 Z s 1 (x − x0 )2 2 e2sϕ v 2 dx = w2 dx ≤ C a(wx )2 dx + w dx 2 0 a 0 0 0 Z 1 Z 1 (x − x0 )2 ≤ CΘ ae2sϕ (vx )2 dx + CΘ3 s2 e2sϕ v 2 dx. a 0 0 Using this last inequality in (5.5), we have Z
T
0
1
Z
¯ 2 e2sϕ dxdt ≤ 2 h
Z
0
0
+
T
Z
1
|h|2 e2sϕ dxdt
0
kck2L∞ (QT ) C
T
Z
Z
0 T Z 1
Z 0
+ kck2L∞ (QT ) Cs2
0
1
Θae2sϕ (vx )2 dxdt
0
Θ3 e2sϕ
(5.6)
(x − x0 )2 2 v dxdt, a
for a positive constant C. Using this inequality in (5.4), we obtain Z TZ 1 (x − x0 )2 2 2sϕ v e dxdt ≤ C 2 |h|2 e2sϕ dxdt a 0 0 0 0 Z TZ 1 Z TZ 1 2 (x − x ) 0 + Θae2sϕ (vx )2 dxdt + s2 e2sϕ Θ3 v 2 dxdt a 0 0 0 0 Z T x=1 + sc1 aΘe2sϕ (x − x0 )(vx )2 dt x=0 .
Z
T
Z
1
sΘa(vx )2 + s3 Θ3
0
Hence, for all s ≥ s0 , where s0 is assumed sufficiently large, the thesis follows.
32
As a consequence of the previous corollary, one can deduce an observability inequality for the adjoint problem (5.3) (5.2). In fact, without loss of generality we can assume that c ≥ 0 (otherwise one can reduce the problem to this case introducing v˜ := e−λt v for a suitable λ). Using this assumption we can prove that the analogous of Lemma 4.1 and of Lemma 4.7 still hold true. Thus, as before, one can prove the following observability inequality: Proposition 5.1. Assume Hypotheses 3.1 and (4.1). Then there exists a positive constant C such that every solution v ∈ C([0, T ]; L2 (0, 1))∩L2 (0, T ; Ha1 (0, 1)) of (5.2) satisfies Z
1
v 2 (0, x)dx ≤ CT
Z
0
0
T
Z
v 2 (t, x)dxdt.
(5.7)
ω
References [1] E.M. Ait Ben Hassi, F. Ammar Khodja, A. Hajjaj, L. Maniar, Carleman estimates and null controllability of coupled degenerate systems, Evol. Equ. Control Theory 2 (2013), 441–459. [2] E.M. Ait Ben Hassi, F. Ammar Khodja, A. Hajjaj, L. Maniar, Null controllability of degenerate parabolic cascade systems, Port. Math. 68 (2011), 345–367. [3] F. Alabau-Boussouira, P. Cannarsa, G Fragnelli, Carleman estimates for degenerate parabolic operators with applications to null controllability J. Evol. Eqs 6 (2006), 161-204. [4] K. Beauchard, Null controllability of Kolmogorov-type equations, Math. Control Signals Systems, to appear. [5] K. Beauchard, P. Cannarsa, R. Guglielmi, Null controllability of Grushintype operators in dimension two, submitted. [6] K. Beauchard, E. Zuazua, Some controllability results for the 2D Kolmogorov equation, Ann. Inst. H. Poincar´e Anal. Non Lin´eaire 26 (2009), 1793–1815. [7] A. Bensoussan, G. Da Prato, M.C. Delfout, S.K. Mitter, Representation and Control of Infinite Dimensional Systems, Systems and Control: Foundations and applications, Birkh¨auser, 1993. [8] H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer Science+Business Media, LLC 2011. [9] J.M. Buchot, J.P. Raymond, A linearized model for boundary layer equations, in ”Optimal control of complex structures” (Oberwolfach, 2000), Internat. Ser. Numer. Math. 139 (2002), Birkhauser, Basel, 31–42.
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[10] P. Cannarsa, G. Fragnelli, Null controllability of semilinear degenerate parabolic equations in bounded domains, Electron. J. Differential Equations 2006 (2006), 1–20. [11] P. Cannarsa, G. Fragnelli, D. Rocchetti, Controllability results for a class of one-dimensional degenerate parabolic problems in nondivergence form, J. Evol. Equ. 8 (2008), 583–616. [12] P. Cannarsa, G. Fragnelli, D. Rocchetti, Null controllability of degenerate parabolic operators with drift, Netw. Heterog. Media 2 (2007), 693–713. [13] P. Cannarsa, G. Fragnelli, J. Vancostenoble, Regional controllability of semilinear degenerate parabolic equations in bounded domains, J. Math. Anal. Appl. 320 (2006), 804–818. [14] P. Cannarsa, G. Fragnelli, J. Vancostenoble, Linear degenerate parabolic equations in bounded domains: controllability and observability, IFIP Int. Fed. Inf. Process. 202 (2006), 163–173, Springer, New York. [15] I. Chueshov, I. Lasiecka, D. Toundykov, Global attractor for a wave equation with nonlinear localized boundary damping and a source term of critical exponent, J. Dynam. Differential Equations 21 (2009), 269–314. [16] H. Emamirad, G. R. Goldstein, J.A. Goldstein, Chaotic solution for the Black-Scholes equation, Proc. Amer. Math. Soc. 140 (2012), 2043–2052. [17] K.J. Engel, R.Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer-Verlag, New York, 2000. [18] E. Fern´ andez-Cara, M. Gonz´alez-Burgos, S. Guerrero, J.-P. Puel, Null controllability of the heat equation with boundary fourier conditions: the linear case, ESAIM Control Optim. Calc. Var. 12 (2006), 442–465. [19] E. Fern´ andez-Cara, E. Zuazua, Null and approximate controllability for weakly blowing up semilinear heat equations, Ann. Inst. H. Poincar´e Anal. Non Lin´eaire 17 (2000), 583–616. [20] W.H. Fleming, M. Viot, Some measure-valued Markov processes in population genetics theory, Indiana Univ. Math. J. 28 (1979), 817–843. [21] C. Flores, L. De Teresa, Carleman estimates for degenerate parabolic equations with first order terms and applications, C. R. Math. Acad. Sci. Paris 348 (2010), 391–396. [22] G. Fragnelli, Null controllability of degenerate parabolic equations in non divergence form via Carleman estimates, Discrete Contin. Dyn. Syst. - S 6, (2013). [23] G. Fragnelli, G. Ruiz Goldstein, J.A. Goldstein, S. Romanelli, Generators with interior degeneracy on spaces of L2 type, Electron. J. Differential Equations 2012 (2012), 1–30. 34
[24] A. V. Fursikov, O. Yu. Imanuvilov, Controllability of evolution equations, Lecture Notes Series, Research Institute of Mathematics, Global Analysis Research Center, Seoul National University 34, 1996. [25] O Yu. Imanuvilov, M. Yamamoto, An inverse problem and an observability inequality for the Lam system with stress boundary condition, Appl. Anal. 88 (2009), 711–733. [26] J. Le Rousseau, G. Lebeau, On carleman estimates for elliptic and parabolic operators. applications to unique continuation and control of parabolic equations, ESAIM Control Optim. Calc. Var. 18 (2012), 712–747. [27] J. Le Rousseau, L. Robbiano, Carleman estimate for elliptic operators with coefficients with jumps at an interface in arbitrary dimension and application to the null controllability of linear parabolic equations, Arch. Rational Mech. Anal. March 195 (2010), 953–990. [28] G. Lebeau, L. Robbiano, Contrˆole exact de l’´equation de la chaleur, Comm. Partial Differential Equations 20 (1995), 335–356. [29] J.-P. Raymond, M. Vanninathan, Null controllability in a heatsolid structure model, Appl. Math. Optim. 59 (2009), 247–273. [30] M. Salo, L. Tzou, Carleman estimates and inverse problems for Dirac operators, Math. Ann. 344 (2009), 161–184. [31] M. Salo, L. Tzou, Inverse problems with partial data for a Dirac system: a Carleman estimate approach, Adv. Math. 225 (2010), 487–513. [32] N. Shimakura, Partial Differential Operators of elliptic type, Translations of Mathematical Monographs 99 (1992), American Mathematical Society, Providence, RI. [33] A. Stahel, Degenerate semilinear parabolic equations, Differential Integral Equations 5 (1992), 683–691. [34] J. Vancostenoble, E. Zuazua, Hardy inequalities, observability, and control for the wave and Schr¨ odinger equations with singular potentials, SIAM J. Math. Anal. 41 (2009), 1508–1532. [35] B. Wu, Carleman estimate for a strongly damped wave equation and applications to an inverse problem, Math. Methods Appl. Sci. 35 (2012), 427–437. [36] M. Yamamoto, Y. Zhang, Conditional stability in determining a zerothorder coefficient in a half-order fractional diffusion equation by a Carleman estimate, Inverse Problems 28 (2012), 105010, 10 pp.
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