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Observable set, observability, interpolation inequality and spectral inequality for the heat equation in Rn
J. Math. Pures Appl. 126 (2019) 144–194
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Journal de Mathématiques Pures et Appliquées www.elsevier.com/locate/matpur
Observable set, observability, interpolation inequality and spectral inequality for the heat equation in Rn ✩ Gengsheng Wang a,∗ , Ming Wang b , Can Zhang c , Yubiao Zhang a a b c
Center for Applied Mathematics, Tianjin University, Tianjin 300072, China School of Mathematics and Physics, China University of Geosciences, Wuhan 430074, China School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China
a r t i c l e
i n f o
Article history: Received 24 November 2017 Available online 3 April 2019 MSC: 35K05 93B05 93B07 Keywords: Characteristic of observable sets Observability inequality Hölder-type interpolation inequality Spectral inequality
a b s t r a c t This paper studies connections among observable sets, the observability inequality, the Hölder-type interpolation inequality and the spectral inequality for the heat equation in Rn . We present the characteristic of observable sets for the heat equation. In more detail, we show that a measurable set in Rn satisfies the observability inequality if and only if it is γ-thick at scale L for some γ > 0 and L > 0. We also build up the equivalence among the above-mentioned three inequalities. More precisely, we obtain that if a measurable set in Rn satisfies one of these inequalities, then it satisfies others. Finally, we get some weak observability inequalities and weak interpolation inequalities where observations are made over a ball. © 2019 Elsevier Masson SAS. All rights reserved.
r é s u m é Cet article étudie les liens entre les ensembles observables, l’inégalité d’observabilité, l’inégalité d’interpolation de Hölder-type et l’inégalité spectrale pour l’équation de la chaleur dans Rn . Nous présentons la caractéristique des ensembles observables pour l’équation de la chaleur. Plus en détail, nous montrons qu’un ensemble mesurable dans Rn satisfait l’inégalité d’observabilité si et seulement s’il est γ-épais à l’échelle L pour certains γ > 0 et L > 0. Nous construisons également l’équivalence entre les trois inégalités mentionnées au-dessus. Plus précisément, on obtient que si un ensemble mesurable dans Rn satisfait une de ces inégalités, alors il satisfait les autres. Enfin, nous obtenons des inégalités d’observabilité faibles et des inégalités d’interpolation faibles dans lesquelles les observations sont effectuées sur une boule. © 2019 Elsevier Masson SAS. All rights reserved.
✩ This work was partially supported by the National Natural Science Foundation of China under grants 11571264, 11701535, 11501424 and 11801408. * Corresponding author. E-mail addresses: [email protected] (G. Wang), [email protected] (M. Wang), [email protected] (C. Zhang), [email protected] (Y. Zhang).
https://doi.org/10.1016/j.matpur.2019.04.009 0021-7824/© 2019 Elsevier Masson SAS. All rights reserved.
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1. Introduction In this paper, we consider the heat equation: ∂t u − Δu = 0
in (0, ∞) × Rn ,
u(0, ·) ∈ L2 (Rn ).
(1.1)
The above equation (1.1) is well-posed.1 For this equation, we will characterize the observable sets and build up connections among several important inequalities which are introduced in the next subsection. Notation. Write C(· · · ) for a positive constant that depends on what are included in the brackets and may vary in different contexts. The same can be said about C (· · · ), C1 (· · · ) and so on. Use Vn to denote the volume of the unit ball in Rn . Let Br (x), with x ∈ Rn and r > 0, be the open ball in Rn , centered at x and of radius r. (Simply write Br = Br (0).) Let Sn−1 be the unit spherical surface in Rn . Let N := {0, 1, 2, . . . } and N + = N\{0}. Denote by Q the open unit cube in Rn , centered at the origin. Let x + LQ, with x ∈ Rn and L > 0, be the set {x + Ly : y ∈ Q}. For each measurable set D ⊂ Rn , denote by |D| and Dc its Lebesgue measure and complement set respectively. For any set G, we write χG for the characteristic function of G. Given f ∈ L2 (Rn ), write f for its Fourier transform.2 Given a measurable function f over Rn , we denote by supp f the support of f , which is the set of all points (in Rn ) where f does not vanish. n 2 1/2 Given x = (x1 , . . . , xn ) ∈ Rn , let |x| := and x := 1 + |x|2 . i=1 xi 1.1. Thick sets and several inequalities We start with introducing sets of γ-thickness at scale L. Sets of γ-thickness at scale L. A measurable set E ⊂ Rn is said to be γ-thick at scale L for some γ > 0 and L > 0, if (1.2) E (x + LQ) ≥ γLn for each x ∈ Rn . About sets of γ-thickness at scale L, several remarks are given in order. (a1 ) To our best knowledge, this definition arose from studies of the uncertainty principle. We quote it from [5] (see Page 5 in [5]). Before [5], some very similar concepts were proposed. For instance, the definition of relative dense sets was given in [18] (see also Page 113 in [15]); the definition of thick sets was introduced in [19]. (a2 ) Let E be a set of γ-thickness at scale L. Then, in each cube with the side length L, |E| is bigger than or equals to γLn . So E distributes almost equally in Rn , and parameters γ and L characterize the distribution of E. (a3 ) The structure of a set of γ-thickness at scale L may be very complicated. Here, we give two examples: 1 1 1 First, the set n∈Z [n − 1000 , n + 1000 ] is 500 -thick at scale 1 in R1 ; Second, the set n∈Z j∈N + [n + 2−j , n + 2−j + 2−j−1 ] is 12 -thick at scale 1 in R1 . 1 First, the Laplacian operator Δ, with its domain D(Δ) = H 2 (Rn ), generates a C0 -semigroup {etΔ : t ≥ 0} in L2 (Rn ); Second, each solution u to (1.1) can be expressed as:
u(t, x) = (e
tΔ
n
u(0, ·))(x) =
K(t, x − y)u(0, y) dy, (t, x) ∈ (0, ∞) × R , Rn
where K(t, x) = (4πt)−n/2 e−|x| /4t , t > 0, x ∈ Rn , is the heat kernel. 2 Given f in the Schwartz class S(Rn ), its Fourier transform is as: f(ξ) = (2π)−n/2 Rn e−ix·ξ f (x) dx, ξ ∈ Rn . Since S(Rn ) is 2 n 2 n dense in L (R ), by a standard way, we can define f for each f ∈ L (R ). 2
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Next, we introduce an observability inequality for the equation (1.1). The observability inequality. A measurable set E ⊂ Rn is said to satisfy the observability inequality for the equation (1.1), if for any T > 0 there exists a positive constant Cobs = Cobs (n, T, E) so that when u solves (1.1),
T
|u(T, x)| dx ≤ Cobs
|u(t, x)|2 dx dt.
2
(1.3)
0 E
Rn
When a measurable E ⊂ Rn satisfies (1.3), it is called an observable set for (1.1). Several notes on the observability inequality (1.3) are given in order. (b1 ) By treating the integral on the left hand side as a recovering term, and the integral on the right hand side as an observation term, we can understand the inequality (1.3) as follows: one can recover a solution of (1.1) at time T , through observing it on the set E and in the time interval (0, T ). From perspective of control theory, the inequality (1.3) is equivalent to the following null controllability: For any u0 ∈ L2 (Rn ) and T > 0, there exists a control f ∈ L2 ((0, T ) × Rn ) driving the solution u to the controlled equation: ∂t u − Δu = χE f in (0, T ) × Rn , from the initial state u0 to the state 0 at time T . (b2 ) Let E be γ-thick at scale L. Corollary 2.1 of this paper gives an explicit expression of Cobs (n, T, E) in terms of γ, L and T . Furthermore, we can see from Remark 2.3 that when L, T and n are given, Cobs (n, T, E) can be taken as a constant depending only on the average measure of E: inf x∈Rn |E (x+LQ)| , but independent of the shape and the position of E. Ln (b3 ) We can compare (1.3) with the observability inequality for the heat equation on a bounded physical domain. Let Ω be a bounded C 2 (or Lipschitz and locally star-shaped, see [2]) domain in Rn . Consider the equation: ⎧ ⎪ ⎪ ⎨∂t u − Δu = 0 u=0 ⎪ ⎪ ⎩u(0, ·) ∈ L2 (Ω).
in (0, ∞) × Ω, on (0, ∞) × ∂Ω,
(1.4)
We say that a measurable set ω ⊂ Ω satisfies the observability inequality for (1.4), if given T > 0, there is a constant C(n, T, ω, Ω) so that when u solves (1.4),
T
|u(T, x)| dx ≤ C(n, T, ω, Ω)
|u(t, x)|2 dx dt.
2
Ω
0
(1.5)
ω
When a measurable set ω ⊂ Ω satisfies (1.5), it is called an observable set for (1.4). The inequality (1.5) has been widely studied. See [14,20,25] for the case where ω is open; [1,2,12] for the case when ω is measurable. (b4 ) When Ω is an unbounded domain and ω is a bounded and open subset of Ω, the inequality (1.5) may not be true. This was showed in [27] for the heat equation in the physical domain R+ . Similar results have been obtained for higher dimension cases in [28]. For the heat equation in an unbounded domain, [29] imposed a condition, in terms of the Gaussian kernel, on the set ω so that the observability inequality does not hold. In particular, [29] showed that the observability inequality fails when Ω is unbounded and |ω| < ∞. Notice that any set E ⊂ Rn of finite measure does not have the characteristic on observable sets of (1.1). This characteristic is indeed the γ-thickness at scale L for some γ > 0 and L > 0. (See Theorem 1.1 of this paper.)
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About works on sufficient conditions of observable sets for heat equations in unbounded domains, we would like to mention the work [6]. It showed that, for some parabolic equations in an unbounded domain Ω ⊂ Rn , the observability inequality holds when observations are made over a subset E ⊂ Ω, with Ω\E bounded. For other similar results, we refer the reader to [30]. When Ω = Rn , such a set E has the characteristic on observable sets of (1.1) mentioned before. (b5 ) An interesting phenomenon is that some potentials (growing at infinity) in heat equations may change the above-mentioned characteristic on observable sets for the heat equations with potentials. In [31,8], the authors realized the following fact: Let A = Δ + V , where V (x) := −|x|2k , x ∈ Rn , with 2 ≤ k ∈ N. Write etA t≥0 for the semigroup generated by the operator A. Let r0 ≥ 0 and let Θ0 be an open subset of Sn−1 . Let Γ = {x ∈ Rn : |x| ≥ r0 , x/|x| ∈ Θ0 }. Then there is C(n, T, Θ0 , r0 , k) so that
T A 2 e u0 dx ≤ C(n, T, Θ0 , r0 , k)
T
0
Rn
tA 2 e u0 dx dt for all u0 ∈ L2 (Rn ).
(1.6)
Γ
The cone Γ does not have the characteristic on observable sets mentioned before, but still holds the observability inequality (1.6). The main reason is as follows: The unbounded potential V changes the behaviour of the solution of the pure heat equation (1.1). This plays an important role in the proof of (1.6) (see [31,8]). It should be pointed out that when V (x) = −|x|2 , x ∈ Rn (which means that the potential grows more slowly at infinity), (1.6) does not hold for the above cone. We refer the readers to [31,8] for more details on this issue. Besides, we also would like to mention [3] for this subject. An interesting question now arises: How do potentials influence characteristics of observable sets? We wish to answer this question in our future studies. We then introduce an interpolation inequality for the equation (1.1). The Hölder-type interpolation inequality. A measurable set E ⊂ Rn is said to satisfy the Hölder-type interpolation inequality for the heat equation (1.1), if for any θ ∈ (0, 1), there is CHold = CHold (n, E, θ) so that for each T > 0 and each solution u to the equation (1.1),
1
|u(T, x)|2 dx ≤ eCHold (1+ T ) Rn
|u(T, x)|2 dx
θ
|u(0, x)|2 dx
1−θ .
(1.7)
Rn
E
Several remarks on the Hölder-type interpolation inequality (1.7) are given in order. (c1 ) The above Hölder-type interpolation inequality is equivalent to what follows: There is θ = θ(n, E) ∈ (0, 1) and CHold = CHold (n, E) so that (1.7) holds for all T > 0 and solutions u to (1.1). This can be verified by the similar way used in the proof of [37, Theorem 2.1]. (c2 ) The inequality (1.7) is a kind of quantitative unique continuation for the heat equation (1.1). It provides a Hölder-type propagation of smallness for solutions of the heat equation (1.1). In fact, if
|u(T, x)|2 dx = δ, then we derive from (1.7) that E
|u(T, x)|2 dx is bounded by Cδ θ for some conRn
stant C > 0. Consequently, u(T, ·) = 0 over Rn provided that it is zero over E. (c3 ) From perspective of control theory, the inequality (1.7) implies the approximate null controllability with cost for impulse controlled heat equations, i.e., given T > τ > 0, ε > 0, there is C = C(n, E, T, τ, ε) such that for each u0 ∈ L2 (Rn ), there is f ∈ L2 (Rn ) so that f L2 (Rn ) ≤ C u0 L2 (Rn ) and u(T, ·) L2 (Rn ) ≤ ε u0 L2 (Rn ) ,
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where u is the solution to the impulse controlled equation: ∂t u − Δu = δ{t=τ } χE f in (0, T ) × Rn , with the initial condition u(0, x) = u0 (x), x ∈ Rn . (See [37, Theorem 3.1].) (c4 ) The Hölder-type interpolation inequality (1.7) can imply the observability inequality (1.3). Moreover, it leads to the following stronger version of (1.3):
|u(T, x)| dx ≤ Cobs
|u(t, x)|2 dx dt, with Cobs = Cobs (n, T, E, F ),
2
Rn
(1.8)
F E
where F ⊂ (0, T ) is a subset of positive measure. This will be presented in Lemma 2.4. We derive (1.8) from (1.7), through using the telescoping series method developed in [36] (see also [38,2]) for heat equations in bounded domains. (c5 ) We can compare (1.7) with an interpolation inequality for the heat equation (1.4). A measurable set ω ⊂ Ω is said to satisfy the Hölder-type interpolation inequality for the equation (1.4), if for any θ ∈ (0, 1), there is C = C(n, Ω, ω, θ) so that for any T > 0 and any solution u to (1.4),
C(1+ T1 )
|u(T, x)| dx ≤ e 2
|u(T, x)| dx 2
θ
ω
Ω
|u(0, x)|2 dx
1−θ .
(1.9)
Ω
In [35], the authors proved that any open and nonempty subset ω ⊂ Ω satisfies the Hölder-type interpolation inequality (1.9) for heat equations with potentials in bounded and convex domains. The frequency function method used in [35] was partially borrowed from [10]. In [2], the authors proved that any subset ω of positive measure satisfies the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) where Ω is a bounded Lipschitz and locally star-shaped domain in Rn . More about this inequality for heat equations in bounded domains, we refer the reader to [36–38]. Finally, we will introduce a spectral inequality for some functions in L2 (Rn ). The spectral inequality. A measurable set E ⊂ Rn is said to satisfy the spectral inequality, if there is a positive constant Cspec = Cspec (n, E) so that for each N > 0,
|f (x)| dx ≤ e 2
Rn
Cspec (1+N )
|f (x)|2 dx for all f ∈ L2 (Rn ) with supp f ⊂ BN .
(1.10)
E
Several notes on the spectral inequality (1.10) are given in order. (d1 ) Recall the Lebeau-Robbiano spectral inequality (see [20,21]): Let Ω be a bounded smooth domain in Rn and let ω be a nonempty open subset of Ω. Write ΔΩ for the Laplacian operator on L2 (Ω) with Domain(ΔΩ ) = H01 (Ω) H 2 (Ω). Let {λj }j≥1 (with 0 < λ1 < λ2 ≤ · · · ) be the eigenvalues of −ΔΩ and let {φj }j≥1 be the corresponding eigenfunctions. Then there is a positive constant C(n, Ω, ω) so that for each λ > 0,
|f (x)|2 dx ≤ eC(n,Ω,ω)(1+ Ω
√ λ)
|f (x)|2 dx for all f ∈ span{φj : λj < λ}.
(1.11)
ω
This inequality was extended to the case where Ω is a bounded C 2 domain via a simpler way in [26]. Then it was extended to the case that Ω is a bounded Lipschitz and locally star-shaped domain; ω is a subset of positive measure so that ω ⊂ BR (x0 ) ⊂ B4R (x0 ) ⊂ Ω for some R > 0 and x0 ∈ Ω; and C(n, Ω, ω) = C(n, Ω, |ω|/|BR |) (see [2, Theorem 5 and Theorem 3]).
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By our understanding, the inequality (1.10) is comparable to (1.11) from two perspectives as follows: First, the inequality (1.10) is satisfied by functions in the subspace: EN f ∈ L2 (Rn ) : supp f ⊂ BN with N > 0, while the inequality (1.11) is satisfied by functions in the subspace: Fλ f = aj φj ∈ L2 (Ω) : {aj }j≥1 ⊂ R with λ > 0. λj <λ
From the definition of the spectral projection in the abstract setting given in [39] (see Pages 262-263 in [39]), we can define two spectral projections: χ[0,N 2 ) (−Δ) and χ[0,λ) (−ΔΩ ) on L2 (Rn ) and L2 (Ω), respectively. Then after some computations, we find that EN and Fλ are the ranges of χ[0,N 2 ) (−Δ) and χ[0,λ) (−ΔΩ ), respectively. Second, the square root of the integral of χ[0,N 2 ) over R is√N which corresponds to the √ N in (1.10), while the square root of the integral of χ[0,λ) over R is λ which corresponds to the λ in (1.11). (d2 ) Though the inequality (1.10) was first named as the spectral inequality in [23] (to our best knowledge), it has been extensively studied for long time. (See, for instance, [5,15,18,19,32,24,33,34,40].) In [19], the author announced that if E is γ-thick at scale L for some γ > 0 and L > 0, then E satisfies the spectral inequality (1.10), and further proved this announcement for the case when n = 1. Earlier, the authors of [24] (see also [15]) proved that E is γ-thick at scale L for some γ > 0 and L > 0 if and only if E satisfies the following inequality: For each N > 0, there is a positive constant C(n, E, N ) so that
|f (x)| dx ≤ C(n, E, N ) 2
Rn
|f (x)|2 dx for each f ∈ L2 (Rn ) with supp f ⊂ BN .
(1.12)
E
This result is often referred as the Logvinenko-Sereda theorem. Before [24], the above equivalence was proved by B.P. Paneyah for the case that n = 1 (see [34,33,15]). In [18], the author claimed (1.12), with C(n, E, N ) = eCspec (1+N ) , and proved this claim for the case when n = 1. In the proof of our main theorem of this paper, the expression C(n, E, N ) = eCspec (1+N ) will play an important role. From this point of view, (1.12) is weaker than the spectral inequality (1.10). (d3 ) The inequality (1.12) is also important. It is closely related to the uncertainty principle (which is an extensive research topic in the theory of harmonic analysis and says roughly that a nonzero function and its Fourier transform cannot be both sharply localized, see [13]). In fact, a measurable set E satisfies the inequality (1.12) if and only if it satisfies the following uncertainty principle:
Rn
⎛
⎜ |f (x)|2 dx ≤ C (n, E, N ) ⎝
E
|f (x)|2 dx +
⎞ ⎟ |f(ξ)|2 dξ ⎠ for all f ∈ L2 (Rn ).
c BN
We refer the interested readers to [5,15,17,32] for the proof of the above result, as well as more general c uncertainty principle, where E and BN are replaced by more general sets. It deserves mentioning what follows: The uncertainty principle can help us to get the exact controllability for the Schrödinger equation with controls located outside of two balls and at two time points. This was realized in [43]. (See [16] for more general cases.) (d4 ) By using a global Carleman estimate, the authors in [23] proved the spectral inequality (1.10) for such an open subset E that satisfies the property: there exists δ > 0 and r > 0 so that ∀y ∈ Rn , ∃ y ∈ E such that Br (y ) ⊂ E and |y − y | ≤ δ.
(1.13)
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It is clear that a set with the above property (1.13) is a set of γ-thick at scale L for some γ > 0 and L > 0.3 (d5 ) With the aid of the spectral inequality (1.10), one can use the same strategy given in [20] to derive the null controllability described in the note (b1 ). 1.2. Aim, motivation and main result Aim According to the note (d2 ) in the previous subsection, the characteristic of a measurable set holding the spectral inequality (1.10) is the γ-thickness at scale L for some γ > 0 and L > 0. Natural and interesting questions are as follows: What is the characteristic of observable sets for (1.1)? How to characterize a measurable set E satisfying the Hölder-type interpolation inequality (1.7)? What are the connections among inequalities (1.3), (1.7) and (1.10)? The aim of this paper is to answer the above questions. Motivation The motivations of our studies are given in order. (i) The first motivation arises from two papers [4] and [1]. In [4], the authors gave, for the wave equation in a bounded physical domain Ω ⊂ Rn , a sufficient and almost necessary condition to ensure an open subset Γ ⊂ ∂Ω to be observable (i.e., Γ satisfies the observability inequality for the wave equation with observations on Γ). This condition is exactly the well known Geometric Control Condition (GCC for short).4 Thus, we can say that the GCC condition is a characteristic of observable open sets on ∂Ω, though this condition is not strictly necessary (see [4] or [22]). The authors in [1] presented a sufficient and necessary condition to ensure a measurable subset ω ⊂ Ω satisfying (1.5). This condition is as: |ω| > 0. Hence, the characteristic of observable sets for the equation (1.4) is as: |ω| > 0. Analogically, it should be very important to characterize observable sets for the heat equation (1.1). However, it seems for us that there is no any such result in the past publications. These motivate us to find the characteristic of observable sets for the equation (1.1). (ii) For the heat equation (1.4), the observability inequality (1.5), the Hölder-type interpolation inequality (1.9) and the spectral inequality (1.11) are equivalent. More precisely, we have that if ω ⊂ Ω is a measurable set, then |ω| > 0 ⇐⇒ ω satisfies (1.11) ⇐⇒ ω satisfies (1.9) ⇐⇒ ω satisfies (1.5).
(1.14)
The proof of (1.14) was hidden in the paper [2]. (See Theorem 5, Theorem 6, as well as its proof, Theorem 1, as well as its proof, in [2].) However, for the heat equation (1.1), the equivalence among these three inequalities has not been touched upon. These motivate us to build up the equivalence among inequalities (1.3), (1.7) and (1.10). It deserves mentioning that for heat equations with lower terms in bounded physical domains, we do not know if (1.14) is true. Main Result The main result of the paper is the next Theorem 1.1. Theorem 1.1. Let E ⊂ Rn be a measurable subset. Then the following statements are equivalent: (i) The set E is γ-thick at scale L for some γ > 0 and L > 0. (ii) The set E satisfies the spectral inequality (1.10). In fact, one can choose L = 2(δ + r), γ = r n (2(δ + r))−n Vn . An open subset Γ ⊂ ∂Ω is said to satisfy the GCC if there exists T0 > 0 such that any geodesic with velocity one meets Γ within time T0 at a nondiffractive point. 3 4
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(iii) The set E satisfies the Hölder-type interpolation inequality (1.7). (iv) The set E satisfies the observability inequality (1.3). Several remarks about Theorem 1.1 are given in order. (e1 ) The equivalence of statements (i) and (iv) in Theorem 1.1 tells us: the characteristic of observable sets for the heat equation (1.1) is the γ-thickness at scale L for some γ > 0 and L > 0. This seems to be new for us. (e2 ) The equivalence among statements (ii), (iii) and (iv) in Theorem 1.1 presents closed connections of the three inequalities. This seems also to be new for us. (e3 ) We find the following way to prove Theorem 1.1: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). We prove (i) ⇒ (ii) by some ideas from [19]. Indeed, this result was announced in [19] and then proved for the case that n = 1 in the same reference. We prove (ii) ⇒ (iii) ⇒ (iv), though using some ideas and techniques from [2,36]. Finally, we show (iv) ⇒ (i) via the structure of a special solution to the equation (1.1). (e4 ) We noticed that four days after we put our current work in arXiv, the paper [9] appeared in arXiv. In [9], the authors independently got the equivalence (i) and (iv) in Theorem 1.1. 1.3. Extensions to bounded observable sets From Theorem 1.1, we see that in order to have (1.3) or (1.7), the set E has to be γ-thick at scale L for some γ > 0 and L > 0. Then a natural and interesting question arises: What are possible substitutions of (1.3) or (1.7), when E is replaced by a ball in Rn ? (It deserves to mention that any ball in Rn does not satisfy the thick condition (1.2).) We try to find the substitutes from two perspectives as follows: (i) We try to add weights on the left hand side and ask ourself if the following inequalities hold for all solutions of (1.1):
T
χBr (x)|u(T, x)| dx ≤ C(T, r , r, n) 2
|u(t, x)|2 dx dt
(1.15)
0 Br
Rn
and
T
ρ(x)|u(T, x)| dx ≤ C(T, ρ, r, n)
|u(t, x)|2 dx dt,
2
Rn
(1.16)
0 Br
where ρ(x) = x−ν or e−|x| . On one hand, we proved that (1.15) is true when r < r, while (1.15) is not true when r > r (see Theorem 3.2 in Subsection 3.2). Unfortunately, we do not know if (1.15) holds when r = r. On the other hand, we showed that (1.16) fails for all r > 0 (see Corollary 3.3 in Subsection 3.2). (ii) We try to find a class of initial data so that (1.3) (where E is replaced by Br ) holds for all solutions of (1.1) with initial data in this class. We have obtained some results on this issue (see Theorem 3.3 in Subsection 3.2). More interesting question is as: what is the biggest class of initial data so that (1.3) (where E is replaced by Br ) holds for all solutions of the heat equation (1.4) with initial data in this class? Unfortunately, we are not able to answer it. We now turn to possible substitutions of (1.7) where E is replaced by B1 . We expect to find b(ε) > 0 for each ε ∈ (0, 1) so that for any T > 0, there is C(n, T ) > 0 such that when u solves (1.1),
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⎛ |u(T, x)|2 dx ≤ C(n, T ) ⎝ε
Rn
⎞
|u(T, x)|2 dx⎠ .
|u(0, x)|2 dx + b(ε)
Rn
(1.17)
B1
Let us explain why (1.17) deserves to be expected. Reason One. Let θ ∈ (0, 1) and T > 0. Then the next two inequalities are equivalent. The first inequality is as: there is C(n, T, θ) so that when u solves (1.1),
θ
1−θ |u(T, x)|2 dx ≤ C(n, T, θ) |u(T, x)|2 dx |u(0, x)|2 dx , Rn
Rn
B1
while the second inequality is as: there is C(n, T, θ) > 0 so that for any ε ∈ (0, 1) and any solution u to (1.1),
⎛ |u(T, x)|2 dx ≤ C(n, T, θ) ⎝ε
Rn
|u(0, x)|2 dx + ε
Rn
− 1−θ θ
⎞ |u(T, x)|2 dx⎠ .
(1.18)
B1
However, (1.18) is not true, for otherwise, we can use the same method developed in [36] (see also [2]) to derive (1.3) (where E is replaced by B1 ) which contradicts the equivalence of (i) and (iv) in Theorem 1.1. Thus, b(ε) in (1.17) cannot grow like a polynomial of ε. But it seems not to be hopeless for us to find some kind of b(ε) so that (1.17) holds. Reason Two. The space-like strong unique continuation of the heat equation (1.1) (see [10]) yields that if u(T, ·) = 0 on the ball B1 , then u(T, ·) = 0 over Rn . The inequality (1.17) is a quantitative version of the aforementioned unique continuation. Though we have not found any b(ε) so that (1.17) is true, we obtained some b(ε) so that (1.17) holds for all solutions to (1.1) with initial data having some slight decay (see Theorem 3.1 in Subsection 3.1). Finally, we would like to mention what follows: With the aid of an abstract lemma (i.e., Lemma 5.1 in [43]), each of extended inequalities mentioned above corresponds to a kind of controllability for controlled heat equations. In Subsection 3.3 of this paper, we will present some applications of some extended inequalities mentioned above to some kinds of controllability for some controlled heat equations. 1.4. Plan of the paper The paper is organized as follows: In Section 2, we prove Theorem 1.1. In Section 3, we present several weak observability inequalities and weak interpolation inequalities (where observations are made in a ball of Rn ) and some applications of these inequalities to controllability. 2. Proof of Theorem 1.1 We are going to prove Theorem 1.1 in the following way: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). The above steps are based on several lemmas: Lemmas 2.1, 2.3, 2.4 and 2.5. We begin with Lemma 2.1 connecting the spectral inequality with sets of γ-thickness at scale L. Lemma 2.1. Suppose that a measurable set E ⊂ Rn is γ-thick at scale L for some γ > 0 and L > 0. Then E satisfies the spectral inequality (1.10), with 1 for some C = C(n). Cspec (n, E) = C(1 + L) 1 + ln γ
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Remark 2.1. The manner that the√ constant eCspec (n,E)(1+N ) (in (1.10)) depends on N is comparable with the manner that the constant eC λ in (1.11) depends on λ. (This has been explained in the remark (d1 ) in Subsection 1.1.) The latter one played an important role in the proof of the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) (see [2]). Analogically, the previous one will play an important role in the proof of Theorem 1.1. Remark 2.2. In [19], the author announced the result in Lemma 2.1 and proved it for the case when n = 1. For the completeness of the paper, we give a detailed proof for Lemma 2.1, based on some ideas and techniques in [19]. To show Lemma 2.1, we need the following result on analytic functions: Lemma 2.2 ([19, Lemma 1]). Let Φ be an analytic function in D5 (0) (the disc in C, centered at origin and ˆ ⊂ I be a subset of positive measure. If of radius 5). Let I be an interval of length 1 such that 0 ∈ I. Let E |Φ(0)| ≥ 1 and M = max|z|≤4 |Φ(z)|, then there exists a generic constant C > 0 such that ln M ˆ ln 2 sup |Φ(x)|. sup |Φ(x)| ≤ C/|E| x∈I
ˆ x∈E
We are now in the position to prove Lemma 2.1. Proof of Lemma 2.1. We only need to prove this lemma for the case when L = 1. In fact, suppose that this is done. Let E be γ-thick at scale L > 0. Define a new set: L−1 E := L−1 x : x ∈ E . One can easily check that L−1 E is γ-thick at scale 1. Given N > 0 and f ∈ L2 (Rn ) with suppf ⊂ BN , let g(x) := f (Lx), x ∈ Rn . One can directly check that g(ξ) = L−n f(L−1 ξ), ξ ∈ Rn ;
supp g ⊂ BLN .
From these, we can apply Lemma 2.1 (with L = 1) to the set L−1 E and the function g to find C = C(n) so that
1 |g(x)|2 dx ≤ e2C(1+ln γ )(1+LN ) |g(x)|2 dx L−1 E
Rn 1
≤ e2C(1+L)(1+ln γ )(1+N )
|g(x)|2 dx.
(2.1)
L−1 E
Meanwhile, by changing variable x → Lx, we deduce that
Rn
|f (x)|2 dx = L−n
|g(x)|2 dx;
Rn
Hence, from (2.1) and (2.2), we find that
E
|f (x)|2 dx = L−n
L−1 E
|g(x)|2 dx.
(2.2)
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|f (x)| dx ≤ e 2
|f (x)|2 dx,
Cspec (n,E)(1+N )
Rn
E
with 1 Cspec (n, E) = 2C(1 + L) 1 + ln . γ This proves the lemma for the general case that L > 0. We now show Lemma 2.1 for the case when L = 1 by several steps. First of all, we arbitrarily fix N > 0 and f ∈ L2 (Rn ) with suppf ⊂ BN . Without loss of generality, we can assume that f = 0. Step 1. Bad and good cubes. For each multi-index j = (j1 , j2 , · · · , jn ) ∈ Zn , let Q(j) := {x = (x1 , . . . , xn ) ∈ Rn : |xi − ji | < 1/2 for all i = 1, 2, · · · , n} . It is clear that Q(j)
Q(k) = ∅ for all j = k ∈ Zn ; Rn =
Q(j),
j∈Zn
where Q(j) denotes the closure of Q(j). From these, we have that
|f (x)| dx = 2
|f (x)|2 dx.
(2.3)
j∈Zn
Rn
Q(j)
We will divide {Q(j) : j ∈ Zn } into two called “good cubes”
disjoint parts whose elements are respectively
|f |2 with
and “bad cubes”. And then we compare
|f |2 and
Rn
|f |2 , respectively. First,
Q(j) is bad
Q(j) is good
we define the function: h(s) := sn (s − 1)−n − 1, s ∈ [2, +∞). It is a continuous and strictly decreasing function satisfying that h(2) ≥ 1,
lim h(s) = 0.
s→+∞
Thus we can take A0 as the unique point in [2, +∞) so that h(A0 ) = 1/2. Clearly, A0 depends only on n, i.e., A0 = A0 (n). Given j ∈ Zn , Q(j) is said to be a good cube, if for each β ∈ N n ,
|∂xβ f (x)|2
dx ≤
|β| A0 N 2|β|
Q(j)
|f (x)|2 dx.
(2.4)
Q(j)
When Q(j) is not a good cube, it is called as a bad cube. Thus, when Q(j) is a bad cube, there is at least one β ∈ N n , with |β| > 0, so that
|β|
|∂xβ f (x)|2 dx > A0 N 2|β| Q(j)
Q(j)
|f (x)|2 dx.
(2.5)
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Using the Plancherel theorem and the assumption that suppf ⊂ BN (0), we obtain that for each β ∈ N n ,
|∂xβ f (x)|2
2 2 β β dx = ∂x f (ξ) dξ = (iξ) f (ξ) dξ =
Rn
Rn
≤ N 2|β|
Rn
|f(ξ)|2 dξ = N 2|β|
|ξ|≤N
β 2 ξ f (ξ) dξ
|ξ|≤N
|f (x)|2 .
(2.6)
Rn
Meanwhile, it follows by (2.5) that when Q(j) is a bad cube,
|f (x)|2 dx ≤
−|β|
A0
β∈N n ,|β|>0
Q(j)
N −2|β|
|∂xβ f (x)|2 dx.
(2.7)
Q(j)
Since Q(j), j ∈ Zn , are disjoint, by taking the sum in (2.7) for all bad cubes, we find that
|f (x)|2 dx ≤
−|β|
A0
β∈N n ,|β|>0
Q(j) is bad
≤
−|β|
A0
N −2|β|
N −2|β|
β∈N n ,|β|>0
|∂xβ f (x)|2 dx Q(j) is bad
|∂xβ f (x)|2 dx.
(2.8)
Rn
From (2.6) and (2.8), we have that
|f (x)|2 dx ≤
−|β|
β∈N n ,|β|>0
Q(j) is bad
A0
|f (x)|2 dx
Rn
= An0 (A0 − 1)−n − 1
|f (x)|2 dx.
(2.9)
Rn
Since h(A0 ) = 12 , it follows from (2.9) that
|f (x)|2 dx ≤
1 2
|f (x)|2 dx.
(2.10)
|f (x)|2 dx.
(2.11)
Rn
Q(j) is bad
By (2.3) and (2.10), we obtain that
1 |f (x)| dx ≥ 2
2
Rn
Q(j) is good
Step 2. Properties on good cubes. Arbitrarily fix a good cube Q(j). We will prove some properties related to Q(j). First of all, we claim that there is C0 (n) > 0 so that ∂xβ f
L∞ (Q(j))
≤ C0 (n)(1 + N )n
|β| A0 N f L2 (Q(j)) for all β ∈ N n .
In fact, according to (2.4), there is C1 (n) > 0 so that
(2.12)
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⎛
∂xβ f W n,2 (Q(j)) =
⎜ ⎝
μ∈N n ,|μ|≤n
≤
⎞1/2
⎟ |∂xβ+μ f (x)|2 dx⎠
Q(j) |β+μ| 2
N |β+μ| f L2 (Q(j))
A0
μ∈N n ,|μ|≤n
≤ C1 (n)(1 + N )n
|β| A0 N f L2 (Q(j)) for all β ∈ N n .
(2.13)
Meanwhile, because Q(j) satisfies the cone condition, we can apply the Sobolev embedding theorem W n,2 (Q(j)) → L∞ (Q(j)) to find C2 (n) > 0 so that ϕ L∞ (Q(j)) ≤ C2 (n) ϕ W n,2 (Q(j)) for all ϕ ∈ W n,2 (Q(j)). This, along with (2.13), leads to (2.12). Next, we let y ∈ Q(j) satisfy that f L∞ (Q(j)) = |f (y)|.
(2.14)
(Due to the continuity of |f | over Rn , such y exists.) Because the diameter of Q(j) is spherical coordinates centered at y to obtain that |E
χE Q(j) (x)dσ
dr |x−y|=r
0 √
n
=
χE Q(j) (x)dσ
dr |x−y|=r
0
=
n, we can use the
∞ Q(j)| =
√
√
1 n
χE Q(j) (x)dσ.
dr
(2.15)
√ |x−y|= nr
0
In (2.15), we change the variable: x=y+
√
nrw with w ∈ Sn−1 ,
and then obtain that |E
Q(j)| =
√
1 n
√ ( nr)n−1 dr
0
χE Q(j) (y +
√
nrw)dσ
Sn−1
1 ≤n
n/2
dr 0
χE Q(j) (y +
√
nrw)dσ.
(2.16)
Sn−1
For each w ∈ Sn−1 , let √ Iw r ∈ [0, 1] : y + nrw ∈ E Q(j) . Since
(2.17)
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157
n
|Sn−1 | =
2π 2 , where Γ(·) is the Gamma function, Γ( n2 )
it follows from (1.2) and (2.16) that Γ( n2 ) |E Q(j)| |Iω0 | ≥ n/2 n−1 ≥ γ for some w0 ∈ Sn−1 . n |S | 2(nπ)n/2
(2.18)
Then we define a function φ(·) over [0, 1] by √ f (y + ntw0 ) φ(t) = , t ∈ [0, 1]. f L2 (Q(j))
(2.19)
(Since f ∈ L2 (Rn ) satisfies that supp f ⊂ BN , we have that f is analytic over Rn . Consequently, f L2 (Q(j)) = 0 because we assumed that f = 0 over Rn .) We claim that φ(t) can be extended to an entire function in the complex plane. In fact, by (2.19), one can directly check that 3
(k) n 2 k max|β|=k Dβ f φ (0) ≤ f L2 (Q(j))
L∞ (Q(j))
for all k ≥ 0.
(2.20)
k (k) φ (0) ≤ C0 (n)(1 + N )n n 32 A0 N for all k ≥ 0.
(2.21)
By (2.20) and (2.12), we see that
From (2.21), we find that φ(t) = φ(0) + φ (0)t + · · · +
φ(k) (0) k t + ··· , k!
t ∈ [0, 1],
(2.22)
and that the series in (2.22), with t being replaced by any z ∈ C, is convergent. Thus, the above claim is true. From now on, we will use φ(z) to denote the extension of φ(t) over C. Step 3. Recovery of the L2 (Rn ) norm. We will finish our proof in this step. Applying Lemma 2.2, where ˆ = Iw (defined by (2.17) and (2.18)) and Φ = φ, I = [0, 1], E 0 and then using (2.18), we can find C3 = C3 (n) so that ln M
sup |φ(t)| ≤ (C/|Iw0 |) ln 2 sup |φ(t)| t∈Iw0
t∈[0,1]
! ≤
2C(nπ)n/2 γΓ( n2 )
≤M
C3 1+ln
1 γ
" lnlnM2 sup |φ(t)|
t∈Iw0
sup |φ(t)|,
(2.23)
t∈Iw0
where M = max |φ(z)|. |z|≤4
(2.24)
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Two facts are given in order. First, it follows from (2.14) and (2.19) that f L∞ (Q(j)) |f (y)| = = |φ(0)|. f L2 (Q(j)) f L2 (Q(j)) Second, it follows by the definition of Iw0 (see (2.17) and (2.18)) that sup |φ(t)| ≤
t∈Iw0
f L∞ (E Q(j)) . f L2 (Q(j))
The above two facts, along with (2.23), yield that f L∞ (Q(j)) ≤ M
C3 1+ln
1 γ
f L∞ (E Q(j)) .
(2.25)
We next define E x ∈ E Q(j) : |f (x)| ≤
2 |E Q(j)| E
|f (x)| dx .
Q(j)
By the Chebyshev inequality, we have that
|E | ≥
|E
γ Q(j)| ≥ . 2 2
By the same argument as that used in the proof of (2.25), one can obtain that f L∞ (Q(j)) ≤ M
C4 1+ln
1 γ
f L∞ (E Q(j)) for some C4 = C4 (n).
(2.26)
Meanwhile, it follows by the definition of E that f L∞ (E Q(j)) ≤
2 |E Q(j)| E
|f (x)| dx.
(2.27)
|f |2 dx.
(2.28)
Q(j)
From (2.26), (2.27) and the Hölder inequality, we find that
|f |2 dx ≤
4 2C4 M γ
1+ln
1 γ
Q(j)
E
Q(j)
The term M (given by (2.24)) can be estimated by (2.21) as follows: M = max |φ(z)| |z|≤4
≤ max
|z|≤4
∞ φ(k) (0) k=0
k!
|z|k
k 3√ ∞ 4n 2 A0 N ≤ C2 (n)(1 + N ) k! n
k=0
≤e
C5 (1+N )
for some C5 = C5 (n).
(2.29)
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159
Finally, combining (2.28) and (2.29) leads to that
C6 (1+N ) 1+ln
|f |2 dx ≤ e
1 γ
Q(j)
|f |2 dx for some C6 = C6 (n). E
(2.30)
Q(j)
Taking the sum in (2.30) for all good cubes, using (2.11), we see that
|f | dx ≤ 2 2
|f |2 dx
Q(j) is good Q(j)
Rn
≤2
e
C6 (1+N ) 1+ln
1 γ
Q(j) is good
≤ 2e
C6 (1+N ) 1+ln
1 γ
|f |2 dx E
Q(j)
|f |2 dx, E
which leads to (1.10), where 1 Cspec (n, E) = (C6 + 1)(1 + L) 1 + ln , with L = 1. γ This ends the proof of Lemma 2.1.
2
Lemma 2.3 and Lemma 2.4 deal with connections among the spectral inequality (1.10), the Hölder-type interpolation inequality (1.7) and the observability inequality (1.8). In their proofs, we borrowed some ideas and techniques from [2,36]. Lemma 2.3. Suppose that a measurable set E ⊂ Rn satisfies the spectral inequality (1.10). Then E satisfies the Hölder-type interpolation (1.7), with CHold =
1 (Cspec + 1)2 + ln 12. 1−θ
Proof. Let E ⊂ Rn satisfy the spectral inequality (1.10). Arbitrarily fix T > 0, θ ∈ (0, 1) and a solution u to (1.1). Write u0 (x) = u(0, x), x ∈ Rn . Then we have that u(T, x) = eT Δ u0 (x) for all x ∈ Rn . Given N > 0, write respectively χ≤N (D) and χ>N (D) for the multiplier operators with the symbols χ{|ξ|≤N } and χ{|ξ|>N } . Namely, for each g ∈ L2 (Rn ), χ≤N (D)g(ξ) = χ{|ξ|≤N } g(ξ) and χ>N (D)g(ξ) = χ{|ξ|>N } g(ξ) for a.e. ξ ∈ Rn . Then we can express u0 as: u0 = χ≤N (D)u0 + χ>N (D)u0 .
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160
From this and (1.10), we can easily check that
|u(T, x)|2 dx Rn
≤2
T Δ e χ≤N (D)u0 (x)2 dx + 2
Rn
≤ 2eCspec (1+N ) E
≤ 4eCspec (1+N )
T Δ e χ>N (D)u0 (x)2 dx
Rn
T Δ e χ≤N (D)u0 (x)2 dx + 2
T Δ e χ>N (D)u0 (x)2 dx
Rn
T Δ 2 e u0 (x) dx + 2 + 4eCspec (1+N )
T Δ e χ>N (D)u0 (x)2 dx.
(2.31)
Rn
E
Since
T Δ e χ>N (D)u0 (x)2 dx =
Rn
2 −T |ξ|2 e χ>N (ξ)# u0 (ξ) dξ
Rn
≤ e−T N
2
2 u #0 (ξ) dξ
Rn
= e−T N
2
|u0 (x)|2 dx, Rn
it follows from (2.31) that
|u(T, x)|2 dx ≤ 4eCspec (1+N )
Rn
⎛
2 |u(T, x)|2 dx + 2 + 4eCspec (1+N ) e−T N
|u0 (x)|2 dx Rn
E
≤ 6eCspec ⎝eCspec N
|u(T, x)|2 dx + eCspec N −T N
2
⎞
|u0 (x)|2 dx⎠ .
(2.32)
Rn
E
Given ε ∈ (0, 1), choose N = N (ε) so that $ % exp Cspec N − T N 2 = ε. (This can be done since the set: {Cspec s − T s2 : s > 0} contains (−∞, 0].) With the above choice of N , we have that Cspec + N=
& 2 Cspec + 4T ln 1ε 2T
1 ≤ T
' Cspec +
(
1 T ln ε
) .
Thus, with θ ∈ (0, 1) fixed before, we see that + * + ( 2 Cspec Cspec 1 exp √ exp [Cspec N ] ≤ exp ln T ε T + * + * 2 2 Cspec 1−θ 1 θ Cspec exp ln + ≤ exp T θ ε 1−θ T *
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
*
2 Cspec = exp (1 − θ)T
161
+ ε−
1−θ θ
.
From this and (2.32), we find that for every ε ∈ (0, 1), ⎛
2 Cspec
|u(T, x)|2 dx ≤ 6eCspec ⎝e (1−θ)T ε
− 1−θ θ
Rn
⎞
|u0 (x)|2 dx⎠ .
|u(T, x)|2 dx + ε Rn
E
Choosing in the above "θ ! |u(T, x)|2 dx E ε= , |u0 (x)|2 dx Rn we obtain that
⎛
⎞θ ⎛
|u(T, x)|2 dx ≤ 12eCspec e (1−θ)T ⎝
|u(T, x)|2 dx⎠ ⎝
2 Cspec
Rn
E
≤e
2 1 1−θ (Cspec +1) +ln
12 1+ T1
⎞1−θ |u0 (x)|2 dx⎠
Rn
⎛
⎝
⎞θ ⎛
|u(T, x)|2 dx⎠ ⎝
⎞1−θ |u0 (x)|2 dx⎠
,
Rn
E
which leads to (1.7) with CHold = This ends the proof of Lemma 2.3.
1 2 (Cspec + 1) + ln 12. 1−θ
2
Lemma 2.4. Suppose that a measurable set E ⊂ Rn has the property: there is a positive constant CHold = CHold (n, E) so that for any T > 0,
|u(T, x)|2 dx ≤ eCHold
1+ T1
Rn
|u(T, x)|2 dx
1/2
|u(0, x)|2 dx
1/2 ,
(2.33)
Rn
E
when u solves the equation (1.1). Then for each T > 0 and each subset F ⊂ (0, T ) of positive measure, there is a positive constant Cobs = Cobs (n, T, F, CHold ) so that when u solves (1.1),
|u(T, x)| dx ≤ Cobs
|u(s, x)|2 dxds.
2
Rn
(2.34)
F E
In particular, if F = (0, T ) then the constant Cobs in (2.34) can be expressed as: Cobs = exp [36(1 + 3CHold )(1 + 1/T )] . Proof. Suppose that E ⊂ Rn satisfies (2.33). Arbitrarily fix T > 0 and F ⊂ (0, T ) of positive measure. Applying Cauchy’s inequality to (2.33), we find that for all t > 0 and ε > 0,
1 1 |u(t, x)| dx ≤ e2CHold 1+ t ε
Rn
|u(t, x)| dx + ε
2
2
E
Rn
|u0 (x)|2 dx.
(2.35)
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By a translation in time, we find from (2.35) that for all 0 < t1 < t2 and ε > 0,
1 2CHold e ε
|u(t2 , x)|2 dx ≤
1+ t
1 2 −t1
|u(t2 , x)|2 dx + ε
Rn
|u(t1 , x)|2 dx.
(2.36)
Rn
E
Let l be a Lebesgue density point of F . Then according to [36, Proposition 2.1], for each λ ∈ (0, 1), there + is a sequence {lm }∞ l=1 ⊂ (l, T ) so that for each m ∈ N , lm+1 − l = λm (l1 − l)
(2.37)
1 F (lm+1 , lm ) ≥ (lm − lm+1 ). 3
(2.38)
and
Arbitrarily fix m ∈ N + . Take s so that 0 < lm+2 < lm+1 ≤ s < lm < T. Using (2.36) (with t1 = lm+2 and t2 = s) and noting that
2
|u (lm , x)| dx ≤ Rn
|u(s, x)|2 dx and lm+1 − lm+2 ≤ s − lm+2 , Rn
we see that
1 2C |u (lm , x)| dx ≤ e Hold ε 2
1 m+1 −lm+2
1+ l
2
|u(s, x)| dx + ε
|u (lm+2 , x)| dx.
2
Rn
(2.39)
Rn
E
Integrating with s over F (lm+1 , lm ) in (2.39) implies that
2
2
|u (lm , x)| dx ≤ ε Rn
1 1 2C e Hold + ε |F (lm+1 , lm )|
|u (lm+2 , x)| dx
Rn
1 m+1 −lm+2
1+ l
|u(s, x)|2 dxds.
F
(2.40)
(lm+1 ,lm ) E
Since it follows by (2.38) that 1 1 1 − (lm+1 , lm ) ≥ (lm − lm+1 ) ≥ e lm −lm+1 , F 3 3 we obtain from (2.40) that
2
|u (lm , x)| dx ≤ ε Rn
2
|u (lm+2 , x)| dx
Rn
+
3 lm −l1m+1 +2CHold e ε
(2.41) 1 m+1 −lm+2
|u(s, x)|2 dxds.
F
Meanwhile, it follows by (2.37) that
1+ l
(lm+1 ,lm ) E
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lm − lm+1 =
163
1 (lm − lm+2 ) 1+λ
(2.42)
λ (lm − lm+2 ) . 1+λ
(2.43)
and lm+1 − lm+2 = Inserting (2.42) and (2.43) into (2.41), we find that
2
2
|u (lm , x)| dx ≤ ε Rn
|u (lm+2 , x)| dx
Rn
C 1 + 3e2CHold e lm −lm+2 ε
|u(s, x)|2 dxds
F
(2.44)
(lm+1 ,lm ) E
with C = 1 + λ +
2CHold (1 + λ) . λ
(2.45)
Rewrite (2.44) as εe
C − lm −l
2
Rn
C 2 − lm −lm+2
|u (lm , x)| dx − ε e
m+2
2
|u (lm+2 , x)| dx Rn
|u(s, x)|2 dxds.
≤ 3e2CHold F
(2.46)
(lm+1 ,lm ) E
√ Next, we fix λ ∈ (1/ 2, 1). Let μ :=
1 2−λ−2 .
Then μ > 1. Setting, in (2.46),
, (μ − 1)C , ε = exp − lm − lm+2 we have that e
μC − lm −l
2
|u (lm , x)| dx − e
m+2
Rn
− l(2μ−1)C m −l
2
|u (lm+2 , x)| dx
m+2
Rn
|u(s, x)|2 dxds.
≤ 3e2CHold F
(2.47)
(lm+1 ,lm ) E
Meanwhile, one can easily check that , , μC (2μ − 1)C = exp − 2 . exp − lm − lm+2 λ (lm − lm+2 ) Because lm+2 − lm+4 = λ2 (lm − lm+2 ) , we deduce from (2.47) and (2.48) that
(2.48)
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164
e
μC − lm −l
2
|u (lm , x)| dx − e
m+2
Rn
2
|u (lm+2 , x)| dx Rn
≤ 3e2CHold
μC m+2 −lm+4
−l
|u(s, x)|2 dxds. F
(lm+1 ,lm ) E
Summing the above inequality for all odd m derives that
e
− l μC −l 1
2
|u (l1 , x)| dx ≤ 3e
3
2CHold
m=1 F
Rn
∞
|u(s, x)|2 dxds
(lm+1 ,lm ) E
|u(s, x)|2 dxds
≤ 3e2CHold F
(l,l1 ) E
≤ 3e2CHold
|u(s, x)|2 dxds. F E
Thus, we have that
|u(T, x)| dx ≤
|u(l1 , x)| dx ≤ 3e
2
Rn
2
2CHold
e
μC l1 −l3
Rn
|u(s, x)|2 dxds,
(2.49)
F E
which leads to (2.34) with Cobs
, μC . = 3 exp 2CHold + l 1 − l3
Finally, in the case when F = (0, T ), we set T 2T , l= l1 = 3 3
( and λ =
2 . 3
Then we have that (see (2.45)) l 1 − l3 =
T , μ = 2 and C ≤ 2 + 6CHold . 9
Now, we derive from (2.49) that
|u(T, x)| dx ≤ 3e 2
2CHold
e
36 1+3CHold T
T
|u(s, x)|2 dxds 0 E
Rn 36(1+3CHold ) 1+ T1
T
≤e
|u(s, x)|2 dxds. 0 E
This completes the proof of Lemma 2.4. 2 The next lemma seems to be new to our best knowledge. The key of its proof is the structure of a special solution to the equation (1.1). This structure is based on the heat kernel.
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Lemma 2.5. Suppose that a measurable set E ⊂ Rn satisfies the observability inequality (1.3). Then the set E is γ-thick at scale L for some γ > 0 and L > 0. Proof. Let E ⊂ Rn be a measurable set satisfying the observability inequality (1.3). Recall that the heat kernel is as: K(t, x) = (4πt)−n/2 e−|x|
2
/4t
, t > 0, x ∈ Rn .
Given u0 ∈ L2 (Rn ), the function defined by
(t, x) −→
K(t, x − y)u0 (y) dy, (t, x) ∈ (0, ∞) × Rn ,
(2.50)
Rn
is a solution to the equation (1.1) with the initial condition u(0, x) = u0 (x), x ∈ Rn . Arbitrarily fix x0 ∈ Rn . By taking u0 (x) = (4π)−n/2 e−|x−x0 |
2
/4
, x ∈ Rn ,
in (2.50), we get the following solution to the equation (1.1): v(t, x) = (4π(t + 1))− 2 e− n
|x−x0 |2 4(t+1)
,
t ≥ 0, x ∈ Rn .
(2.51)
From (2.51), we obtain by direct computations that
|v(1, x)|2 dx = 4−n π − 2 . n
(2.52)
Rn
From (2.51), we also find that for an arbitrarily fixed L > 0, L2
v(t, x) ≤ (4π)− 2 e− 16 e− n
|x−x0 |2 16
for all 0 ≤ t ≤ 1 and x ∈ Rn , with |x − x0 | ≥ L.
(2.53)
By (2.53), the above solution v satisfies that
1
|v(t, x)|2 dx dt ≤ (2π)− 2 e− n
L2 8
.
(2.54)
0 |x−x0 |≥L
Meanwhile, by taking T = 1 and u = v in the observability inequality (1.3), we see that
1
|v(1, x)| dx ≤ C
|v(t, x)|2 dx dt.
2
(2.55)
0 E
Rn
Here and in what follows, C stands for the constant Cobs (n, 1, E) in (1.3). Now, it follows from (2.52), (2.55) and (2.54) that
4
−n − n 2
π
1
≤C 0 E
BL (x0 )
|v(t, x)|2 dx dt + C(2π)− 2 e− n
L2 8
.
(2.56)
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166
Choose L > 0 in such a way that C(2π)− 2 e− n
L2 8
≤
1 −n − n 4 π 2. 2
Then by (2.56) and (2.51), we obtain that 1 −n − n 4 π 2 ≤C 2
1
|v(t, x)|2 dx dt
0 E
1
BL (x0 )
(4π(t + 1))−n e−
=C 0 E
1
(4π)−n dx dt ≤ C(4π)−n |E
≤C
dx dt
BL (x0 )
0 E
|x−x0 |2 2(t+1)
BL (x0 )|,
BL (x0 )
from which, it follows that |E
BL (x0 )| ≥ (2C)−1 π 2 . n
(2.57)
Since BL (x0 ) ⊂ (x0 + 2LQ), we see from (2.57) that n E (x0 + 2LQ) ≥ (2C)−1 π 2 . From this, as well as the choice of L, we can find L > 0 and γ > 0, which are independent of the choice of x0 , so that E (x0 + L Q) ≥ γ(L )n .
(2.58)
Notice that x0 in (2.58) was arbitrarily taken from Rn . Hence, the set E is γ-thick at scale L . This ends the proof of Lemma 2.5. 2 We now on the position to prove Theorem 1.1. Proof of Theorem 1.1. We can prove it in the following way: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). Indeed, the conclusions (i) ⇒ (ii), (ii) ⇒ (iii), (iii) ⇒ (iv) and (iv) ⇒ (i) follow respectively from Lemma 2.1, Lemma 2.3, Lemma 2.4 and Lemma 2.5. This ends the proof of Theorem 1.1. 2 Tracking the constants in Lemma 2.1, Lemma 2.3 and Lemma 2.4, we can easily get the following consequences of Theorem 1.1: Corollary 2.1. Let E ⊂ Rn be a set of γ-thick at scale L for some γ > 0 and L > 0. Then the following conclusions are true for a constant C = C(n) > 0:
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
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(a) The set E satisfies the Hölder-type interpolation (1.7) with CHold (n, E, θ) =
C 1 2 (1 + L)2 1 + ln . 1−θ γ
(b) The set E satisfies the observability inequality (1.3) with 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
2
1 γ
1+ T1
.
Remark 2.3. Let E be γ-thick at scale L. From (b) of Corollary 2.1, we can take the observable constant as: 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
1 γ
2
1+ T1
,
which depends only on γ, when L and T and n are given. Meanwhile, we define the average measure of E as: γ0 γ0 (E) infn
|E
x∈R
(x + LQ)| . Ln
(2.59)
One can easily check that E is also γ0 -thick at scale L. Then by making use of (b) of Corollary 2.1 again, we can take the observable constant as: 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
1 γ0
2
1+ T1
.
(2.60)
From (2.60) and (2.59), we see that when L, T and n are given, Cobs (n, E, T ) can be taken as a constant depending only on the average measure of E, but independent of the shape and the position of E. In particular, if E1 , E2 ⊂ Rn satisfy that |Ei
(x + LQ)| = γLn for all x ∈ Rn with γ ∈ (0, 1], i = 1, 2,
then Cobs (n, E1 , T ) and Cobs (n, E2 , T ) can be taken as the same. 3. Weak interpolation and observability inequalities In this section, we introduce several weak observability inequalities and interpolation inequalities, where observations are made over a ball in Rn . On the one hand, these inequalities can be viewed as extensions of (1.3) and (1.7) in some senses, while on the other hand, they are independently interesting. 3.1. Weak interpolation inequalities with observation on the unit ball We begin with introducing two spaces. Given a > 0 and ν > 0, we set ν
L2 (ea|x| dx) := {f : Rn → R : f is measurable and f L2 (ea|x|ν
dx)
< +∞},
equipped with the norm: ⎛ f L2 (ea|x|ν
dx)
:= ⎝
Rn
⎞1/2 |f (x)|2 ea|x| dx⎠ ν
ν
, f ∈ L2 (ea|x| dx).
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168
Given ν > 0, we set L2 (xν dx) := {f : Rn → R : f is measurable and f L2 ( x ν dx) < +∞}, equipped with the norm: ⎛ f L2 ( x ν dx) := ⎝
⎞1/2
|f (x)|2 xν dx⎠
, f ∈ L2 (xν dx).
Rn
Notice that any function in one of the above spaces decays along the radical direction. In this subsection, we will build up some interpolation inequalities for solutions to (1.1), with initial data ν in L2 (ea|x| dx) (or L2 (xν dx)). In these inequalities, observations are made over the unit ball in Rn and at one time point. The purpose to study such observability has been explained in Subsection 1.3. Our main results about this subject are included in the following theorem: Theorem 3.1. (i) There is θ = θ(n) ∈ (0, 1) and C = C (n) such that for any ε > 0, T > 0 and a > 0,
4| ln θ| − |u(T, x)|2 dx ≤ C1 (a, T ) ε |u0 (x)|2 ea|x| dx + ε−1 eε a |u(T, x)|2 dx ,
Rn
Rn
B1
when u solves (1.1) with the initial condition u(0, ·) = u0 (·) ∈ L2 (ea|x| dx). Here, C1 (a, T ) = eC
1+ T1 +a+a2 T
(
1 + a−n Γ
a . 2| ln θ|
(ii) There is θ = θ(n) ∈ (0, 1) and C = C (n) so that for any ε ∈ (0, 1), T > 0 and ν ∈ (0, 1],
1
(3| ln θ|+1) 1 ν ε 2 ν e |u(T, x)| dx ≤ C2 (ν, T ) ε |u0 (x)| x dx + e |u(T, x)|2 dx ,
2
Rn
Rn
B1
when u solves (1.1) with the initial condition u(0, ·) = u0 (·) ∈ L2 (xν dx). Here, ν
C2 (ν, T ) = (1 + T 2 )eC
1+ T1
.
Remark 3.1. (a) The condition that ν ≤ 1 in (ii) of Theorem 3.1 is not necessary. We make this assumption only for the brevity of the statement of the theorem. Indeed, from the definition of L2 (xν dx), we see that L2 (xν dx) → L2 (x dx) for any ν ≥ 1. From this and (ii) of Theorem 3.1, one can easily check that when ν > 1, any solution of (1.1) satisfies that
1
(3| ln θ|+1) 1 1 ε |u(T, x)|2 dx ≤ 1 + T 2 eC (n) 1+ T ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx .
Rn
Rn
B1
(b) [11, Theorem 1] contains the following result: There is a universal constant C > 0 so that for each T > 0 and R > 0, sup 0≤t≤T
e
2 t|x| 4 t2 +R2
u(t)
L2 (Rn )
≤ C u(0) L2 (Rn ) + e
2 T |x| 4 T 2 +R2
u(T )
L2 (Rn )
,
(3.1)
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169
when u solves (1.1). The first inequality in Theorem 3.1 is comparable to the above inequality (3.1). By our understanding, these two inequalities can be viewed as different versions of Hardy uncertainty principle. On one hand, the inequality (3.1) can be understood as follows: From some information on a solution to (1.1) at infinity in Rn at two time points 0 and T , one can know the behaviour of this solution at infinity in Rn at each time t ∈ [0, T ]. On the other hand, the first inequality in Theorem 3.1 can be explained in the following way: From some information on a solution to (1.1) at infinity in Rn at time 0, and in the ball B1 in Rn at time T , one can know the behaviour of this solution at infinity in Rn at time T . Similarly, we can compare the second inequality in Theorem 3.1 with (3.1). It deserves to mention that we can only prove inequalities in Theorem 3.1 for the pure heat equation (1.1), while [11, Theorem 1] gave the inequality (3.1) for heat equations with general potentials. (c) The first inequality in Theorem 3.1 can also be understood as follows: If we know in advance that the initial datum of a solution to (1.1) is in the unit ball of L2 (ea|x| dx), then by observing this solution in the unit ball of Rn at time T , we can approximately recover this solution over Rn at the same time T , with the error C1 (a, T )ε. The second inequality in Theorem 3.1 can be explained in a very similar way. Before proving Theorem 3.1, we give a consequence of its conclusion (ii). This consequence can be viewed as an extension of the spectral inequality (1.10) when the γ-thickness set E is replaced by a unit ball. Corollary 3.1. There are two constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for each ε ∈ (0, 1) and each f ∈ L2 ((1 + |x|2 ) dx), with supp f ⊂ BN ,
⎛ |f (x)|2 dx ≤ C ⎝ε(2 + 8N 2 )e2N
2
Rn
where cε = ee
|f (x)|2 (1 + |x|2 ) dx + cε
Rn
(3| ln θ|+1) 1 ε
⎞ |f (x)|2 dx⎠ ,
(3.2)
B1
.
Proof. First of all, according to (ii) of Theorem 3.1 and (a) of Remark 3.1 (with T = 1 and ν = 2), there exists θ = θ(n) ∈ (0, 1) and C = C(n) so that for each ε ∈ (0, 1) and each u0 ∈ L2 ((1 + |x|2 ) dx), ⎛
|eΔ u0 (x)|2 dx ≤ C ⎝ε
Rn
|u0 (x)|2 (1 + |x|2 ) dx + cε
Rn
⎞ |eΔ u0 (x)|2 dx⎠ ,
(3.3)
B1
(3| ln θ|+1) 1
ε where cε = ee . Next we arbitrarily fix ε ∈ (0, 1) and f ∈ L2 ((1 + |x|2 ) dx), with supp f ⊂ BN for some N > 0. Define a function h in the following manner:
2 h(ξ) = e|ξ| f(ξ) for a.e. ξ ∈ Rn .
(3.4)
Then we have two facts: First, f = eΔ h; Second, by the Plancherel theorem and (3.4), using the fact that supp f ⊂ BN , after some computations, we obtain that
|h(x)|2 (1 + |x|2 ) dx = Rn
| h(ξ)|2 + |∇ξ h(ξ)|2 dξ
Rn
≤ (2 + 8N 2 )e2N
2
Rn
|f(ξ)|2 + |∇ξ f(ξ)|2 dξ
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170
2N 2
2
|f (x)|2 (1 + |x|2 ) dx < ∞.
= (2 + 8N )e
Rn
From which, it follows that h ∈ L2 ((1 + |x|2 ) dx). Finally, by the above two facts, we can use (3.3), with u0 = h, to get (3.2). This ends the proof of Corollary 3.1. 2 To show Theorem 3.1, we need some preliminaries. We begin with some auxiliary lemmas on the persisν tence of the heat semigroup in the spaces L2 (ea|x| dx) and L2 (xν dx). ν
Lemma 3.1. Let a > 0 and 0 < ν ≤ 1. Then when u0 ∈ L2 (ea|x| dx), etΔ u0
n
L2 (ea|x|ν
dx)
≤ 2 2 ea
2 2−ν
ν
u0 L2 (ea|x|ν
t 2−ν
for all t > 0.
dx)
ν
Proof. Arbitrarily fix a > 0, 0 < ν ≤ 1 and u0 ∈ L2 (ea|x| dx). Using the fundamental solution of (1.1) and ν the definition of L2 (ea|x| dx), we have that etΔ u0
L2 (ea|x|ν
=
dx)
a|x|ν 2
e
(4πt)−n/2
Rn
e−
|x−y|2 4t
2 u0 (y) dy
dx
12
.
(3.5)
Rn
Since |x|ν ≤ (|x − y| + |y|)ν ≤ |x − y|ν + |y|ν for all x, y ∈ Rn , (Here, we used the elementary inequality: (τ + s)ν ≤ τ ν + sν , τ, s > 0.) it follows from (3.5) that e
tΔ
u0
L2 (ea|x|ν
dx)
≤
−n/2
e−
(4πt) Rn
≤
|x−y|2 4t
Rn
(4πt)−n/2 e−
|x|2 4t
ν
+ a|x| 2
dx ·
Rn
=
ν
+ a|x−y| 2
e
e
a|y|ν 2
a|y|ν 2
|u0 (y)| dy
2
2 u0 (y)
dy
dx
12
12
Rn
(4πt)−n/2 e 4t (−|x| 1
2
ν
+2ta|x| )
dx · u0 L2 (ea|x|ν
dx) .
(3.6)
Rn
Meanwhile, by the Young inequality: 2ta|x| ≤ ν
|x|2 2 ν
2
+
(2ta) 2−ν 2 2−ν
≤
2 1 2 |x| + (2ta) 2−ν , 2
we obtain that
−n/2
(4πt) Rn
e
2 ν 1 4t (−|x| +2ta|x| )
dx ≤
e
2 (2ta) 2−ν 4t
(4πt)−n/2 e−
|x|2 8t
dx
Rn n 2
=2 e
2 (2ta) 2−ν 4t
n
≤ 2 2 ea
2 2−ν
ν
t 2−ν
.
Now, the desired inequality follows from (3.6) and (3.7). This ends the proof of Lemma 3.1.
(3.7) 2
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171
Remark 3.2. The inequality in Lemma 3.1 does not hold for the case when ν > 1. Indeed, given ν > 1, ν ν 1 let u0 (x) = e− 2 |x| x−n , x ∈ Rn . It is clear that u0 ∈ L2 (e|x| dx). However, we have that for any t > 0, ν etΔ u0 ∈ / L2 (e|x| dx). This can be proved as follows: Arbitrarily fix t > 0. By some direct calculations, we find that when |x| ≥ 2,
1 1 1 ν etΔ u0 (x) ≥ C(4πt)−n/2 e− 4t e− 2 |x|− 2 |x| − 1/2−n for some C = C(n).
This leads to that etΔ u0
≥ C(4πt)−n/2 e− 4t 1
L2 (e|x|ν dx)
e|x|
ν
− |x|− 12 ν
|x| − 1/2−2n dx
1/2 .
(3.8)
|x|≥2
Meanwhile, one can easily find a constant M > 2 so that ν
|x|ν − (|x| − 1/2) ≥
ν ν−1 |x| , when |x| ≥ M. 4
From this and (3.8), we obtain that etΔ u0
≥ C(4πt)−n/2 e− 4t 1
L2 (e|x|ν dx)
e 4 |x| ν
ν−1
|x| − 1/2−2n dx
1/2
= ∞.
|x|≥M
Lemma 3.2. Let ν ≥ 0. Then for any u0 ∈ L2 (xν dx), ν etΔ u0 L2 ( x ν dx) ≤ 4ν+2 Γ(ν/2 + n) 1 + t 4 u0 L2 ( x ν dx) for all t > 0. Proof. The proof is similar to that of Lemma 3.1. (See also [41, Lemma B.6.1].)
2
s Lemma 3.3. Given s > 0, there is C = C(n, s) so that when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ) for some a > 0,
Dα f L∞ (Rn ) ≤ C |α|+1 a−
2|α|+3n 2s
1 (α!) s f(ξ)
L2 (ea|ξ|s dξ)
for each α ∈ N n .
(Here, we adopt the convention that α! = α1 !α2 ! · · · αn !.) s Remark 3.3. From Lemma 3.3, we see that if f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ), with s > 0 and a > 0, then f is analytic, when s = 1, while f is ultra-analytic, when s > 1. s Proof of Lemma 3.3. Arbitrarily fix s > 0, a > 0 and f ∈ L2 (Rn ), with f ∈ L2 (ea|ξ| dξ). Then arbitrarily fix α = (α1 , . . . , αn ) ∈ N n , β = (β1 , . . . , βn ) ∈ N n and γ = (γ1 , . . . , γn ) ∈ N n , with |γ| ≤ n. Several facts are given in order.
Fact One: By direct computations, we see that
Rn
2β −a|ξ|s ξ e dξ ≤
n . 2β −a(Σn |ξ |s /n) ξ e i=1 i dξ =
|ξi |
2βi
e−a|ξi |
s
/n
dξi
i=1 Rξi
Rn
∞ n n n 2βis+1 ∞ 2βi +1 . . 2βi −ar s /n = 2 r e dr = 2 t s −1 e−t dt a i=1 i=1 0
n 2β + 1 n 2|β|+n . s i Γ = 2n . a s i=1
0
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172
From this, we obtain that ⎛ ⎝
⎞1/2
|ξ
2(α+γ)
|e
−a|ξ|s
dξ ⎠
≤2
n/2
Rn
n ( n 2|α|+3n . 2αi + 2γi + 1 2s . Γ a s i=1
(3.9)
Fact Two: By the Sobolev embedding H n (Rn ) → L∞ (Rn ), we can find C1 (n) > 0 so that Dα f L∞ (Rn ) ≤ C1 (n)
Dα+γ f
L2 (Rn )
.
(3.10)
γ∈N n ,|γ|≤n
Fact Three: By the Plancheral theorem and the Hölder inequality, we obtain that ⎛ Dα+γ f
L2 (Rn )
= ξ α+γ f(ξ)
L2 (Rn )
≤ f(ξ)ea|ξ|
s
⎝
/2
L2 (Rn )
⎞1/2 |ξ 2(α+γ) |e−a|ξ| dξ ⎠ s
.
(3.11)
Rn
Fact Four: There exists C2 = C2 (n, s) so that ( 1 2αi + 2γi + 1 Γ ≤ C2αi +1 Γ1/s (αi ) = C2αi +1 (αi !) s . s
(3.12)
The proof of (3.12) is as follows: From the Stirling formula, we have that lim √
x→+∞
Γ(x) 1 = 1. 2πe−x xx+ 2
(3.13)
From (3.13), we can find constants M1 = M1 (s) and C3 = C3 (n, s) so that for all αi > M1 , i +1 + 1 2α + 2γ + 1 2α + 2γ + 1 2αi +2γ √ 2αi +2γi +1 s 2 i i i i s ≤ 2 2πe− s s 2α + 2γ + 1 2γis+1 + 12 2α + 2γ + 1 2αs i √ 2αi +2γi +1 i i i i s = 2 2π · e− · s s 2α + 2γ + 1 2αs i √ 2γi +1 i i −x + 12 s ≤ 2 2π sup e x · s x>0 2γ +1 2α √ 2γ +1 1 is + 12 2αi + 2n + 1 s i −( is + 12 ) 2γi + 1 + = 2 2π · e · s 2 s
Γ
2αi
≤ C3αi αi s .
(3.14)
From (3.13), we can also find an absolute constant M2 ≥ 1 so that for all αi > M2 , √ 1 Γ(αi ) ≥ 2−1 2πe−αi αi αi + 2 .
(3.15)
According to (3.14) and (3.15), there is a constant C4 (n, s) so that & i +1 Γ 2αi +2γ s Γ1/s (αi )
≤ [C4 (n, s)]
αi
for all αi > M := max{M1 , M2 }.
Meanwhile, it is clear that there is a constant C5 (n, s) so that
(3.16)
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& i +1 Γ 2αi +2γ s Γ1/s (αi )
173
≤ C5 (n, s) for all αi ≤ M.
(3.17)
(Here we agree that Γ(0) = ∞.) Combining (3.16) and (3.17) leads to (3.12). Inserting (3.12) into (3.9), noticing that |γ| ≤ n, we find that for some C6 = C6 (n, s), ⎞1/2
n n 2|α|+3n . 1 s 2s ⎝ ξ 2(α+γ) e−a|ξ| dξ ⎠ ≤ 2n/2 C2αi +1 (αi !) s a i=1 ⎛
Rn
|α|+1 − 2|α|+3n 2s
≤ C6
a
1
(α!) s .
(3.18)
Finally, it follows from (3.10), (3.11) and (3.18) that for some C = C(n, s),
Dα f L∞ (Rn ) ≤ C1 (n)
s f(ξ)ea|ξ| /2
γ∈N n ,|γ|≤n
≤ C |α|+1 a−
2|α|+3n 2s
(α!) s f(ξ)
|α|+1
L2 (Rn )
C6
1
L2 (ea|ξ|s dξ)
1
(α!) s
.
2
This ends the proof of Lemma 3.3.
The next corollary is a consequence of Lemma 3.3. 2 Corollary 3.2. There is C = C(n) > 0 so that when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ) for some a > 0,
2
Dα f L∞ (Rn ) ≤ eC(1+b
1 |α|! ) 1+ a b|α|
f(ξ)
L2 (ea|ξ|2 dξ)
for all b > 0 and α ∈ N n .
Remark 3.4. Let u0 ∈ L2 (Rn ) be arbitrarily given. Set u(t, x) = etΔ u0 (x), (t, x) ∈ (0, ∞) × Rn . Then u is the solution of (1.1) with u(0, ·) = u0 (·). Arbitrarily fix t > 0. By applying Corollary 3.2 (where f (·) = u(t, ·) and a = 2t), we see that the radius of analyticity of u(t, ·) (which is treated as a function of x) is independent of t. It is an analogy result for solutions of the heat equation in a bounded domain with an analytic boundary (see [2,12]). This property plays a very important role in the proof of the observability estimates from measurable sets when using the telescoping series method developed in [2,12]. Proof of Corollary 3.2. Arbitrarily fix a > 0 and f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ). Then arbitrarily fix b > 0 and α ∈ N n . According to Lemma 3.3 (with s = 2), there is C = C (n) so that 2
Dα f L∞ (Rn ) ≤ C
|α|+1 − 2|α|+3n 4
≤ g(|α|)
a
|α|! f (ξ) b|α|
1 (α!) 2 f(ξ)
L2 (ea|ξ|2 dξ)
L2 (ea|ξ|2 dξ)
,
(3.19)
where r −1/2 g(r) = C a−3n/4 bC a−1/2 (r!) , r ∈ N. To estimate g(r) pointwisely, we use (3.15) to find that when r > M2 ≥ 1 (where M2 is given by (3.15)),
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r −1/2 √ 1 g(r) ≤ C a−3n/4 bC a−1/2 2−1 2πe−r rr+ 2 r ≤ 21/4 π −1/4 C a−3n/4 be1/2 C a−1/2 r−r/2 ≤ 21/4 π −1/4 C a−3n/4 sup (be1/2 C a−1/2 )r r−r/2 r>0
= 21/4 π −1/4 C a−3n/4 e
b2 C 2 2a
.
(3.20)
Meanwhile, it is clear that when r ≤ M2 , M2 g(r) ≤ C a−3n/4 bC a−1/2 + 1 .
(3.21)
From (3.20) and (3.21), it follows that g(r) ≤ eC
1 1+b2 1+ a
for all r = 0, 1, 2, . . . ,
which, together with (3.19), yields the desired inequality. This ends the proof of Corollary 3.2. 2 To prove Theorem 3.1, we also need the decomposition: Rn =
Ωj , with Ωj := {x ∈ Rn : j − 1 ≤ |x| < j}.
(3.22)
j≥1
The next lemma concerns with the propagation of smallness for some real-analytic functions with respect to the above decomposition of Rn . Lemma 3.4. There are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0, j ≥ 1 and 2 f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ),
1
|f |2 dx ≤ j (n−1)(1−θ) eC(1+ a ) Ωj+1
|f |2 dx
θ
|f|2 ea|ξ| dξ 2
1−θ .
Rn
Ωj
The proof of Lemma 3.4 needs Corollary 3.2 and the next lemma which is quoted from [1] (see also [2, Theorem 4]), but is originally from [42]. Lemma 3.5 ([1, Theorem 1.3]). Let R > 0 and let f : B2R ⊂ Rn → R be real analytic in B2R verifying |Dα f (x)| ≤ M (ρR)−|α| |α|!, when x ∈ B2R and α ∈ N n for some positive numbers M and ρ ∈ (0, 1]. Let ω ⊂ BR be a subset of positive measure. Then there are two constants C = C(ρ, |ω|/|BR |) > 0 and θ = θ(ρ, |ω|/|BR |) ∈ (0, 1) so that f L∞ (BR ) ≤ CM 1−θ
θ 1
|f (x)| dx . |ω| ω
We are now in the position to show Lemma 3.4. Proof of Lemma 3.4. Let a > 0 and j ≥ 1 be arbitrarily given. Arbitrarily fix f ∈ L2 (Rn ) with f ∈ 2 L2 (ea|ξ| dξ). The rest proof is divided into the following several steps.
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Step 1. The decompositions of Ωj and Ωj+1 in the polar coordinates. In the polar coordinate system, we have that /
Ωj = (r, ϑ1 , · · · , ϑn−1 ) ∈ [j − 1, j) × [0, 2π)n−1 , Ωj+1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ [j, j + 1) × [0, 2π)n−1 .
(3.23)
When j is large, the distance between two points in Ωj can be very large. This makes our studies on the propagation from Ωj to Ωj+1 harder. To pass this barrier, we need to build up a suitable refinement for each Ωj . We set [0, 2π] =
,
Δl , with Δl :=
1≤l≤j
" l l−1 2π, 2π . j j
(3.24)
Given (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, · · · , n − 1), we set /
Ωj;k1 ,··· ,kn−1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ Ωj : ϑ1 ∈ Δk1 , · · · , ϑn−1 ∈ Δkn−1 , Ωj+1;k1 ,··· ,kn−1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ Ωj+1 : ϑ1 ∈ Δk1 , · · · , ϑn−1 ∈ Δkn−1 .
(3.25)
Then one can easily check that for each ˆj ∈ {j, j + 1}, Ωˆj;k1 ,··· ,kn−1
Ωˆj;k1 ,··· ,k
n−1
= ∅, if (k1 , · · · , kn−1 ) = (k1 , · · · , kn−1 ),
(3.26)
and Ωˆj =
Ωˆj;k1 ,··· ,kn−1 ,
(3.27)
where the union is taken over all different (k1 , . . . , kn−1 ) ∈ N n−1 , with 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. In what follows, we write d Ωj;k1 ,··· ,kn−1 for the diameter of Ωj;k1 ,··· ,kn−1 . Step 2. To prove the following three properties: (O1) There are constants c1 = c1 (n) and c2 = c2 (n) so that for any (k1 , . . . , kn−1 ) ∈ N n−1 , with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), c1 Vn ≤ Ωj;k1 ,··· ,kn−1 ≤ c2 Vn .
(3.28)
(O2) We have that for any (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), d Ωj;k1 ,··· ,kn−1 = d (Ωj;1,··· ,1 ) :=
sup
x,x ∈Ωj;1,··· ,1
|x − x |.
(3.29)
(O3) We have that for any fix (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), d Ωj;k1 ,··· ,kn−1 ≤ 2π
0
3
i2 ≤ 2πn 2 ;
(3.30)
1≤i≤n
j+1 d Ωj+1;k1 ,··· ,kn−1 ≤ 2π j
0 1≤i≤n
3
i2 ≤ 4πn 2 .
(3.31)
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To see (O1), we use the definitions of Ωj;k1 ,··· ,kn−1 and Ωj to find that Ωj;k
1 ,···
,kn−1
=
1 j n−1
|Ωj | = Vn
j n − (j − 1)n , j n−1
which leads to (3.28). The conclusion (O2) follows immediately from the definitions of Ωj;k1 ,··· ,kn−1 and d Ωj;k1 ,··· ,kn−1 . To show (3.30) in (O3), we let (r, ϑ1 , · · · , ϑn−1 ) and (r, ϑ1 , · · · , ϑn−1 ) be the polar coordinates of x = (x1 , . . . , xn ) and x = (x1 , . . . , xn ), respectively. Then we have that x, x ∈ Ωj;1,··· ,1 ⇐⇒ j − 1 ≤ r < j, 0 ≤ ϑl , ϑl <
2π for all l = 1, · · · , n − 1. j
(3.32)
Notice that the connection between (x1 , . . . , xn ) and (r, ϑ1 , . . . , ϑn−1 ) is as: ⎧ x1 = r cos ϑ1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x = r sin ϑ1 cos ϑ2 , ⎪ ⎪ ⎨ 2 ··· ⎪ ⎪ ⎪ ⎪ = r sin ϑ1 sin ϑ2 · · · sin ϑn−2 cos ϑn−1 , x ⎪ ⎪ ⎪ n−1 ⎪ ⎩ xn = r sin ϑ1 sin ϑ2 · · · sin ϑn−2 sin ϑn−1 . Then, by the mean value theorem, we have that for some ζ ∈ (0, 2π/j), |x1 − x1 | = r| cos ϑ1 − cos ϑ1 | = r| sin ζ · (ϑ1 − ϑ1 )| ≤ j| sin ζ|
2π ≤ 2π. j
By inserting suitable terms and using the mean value theorem, we have that |x2 − x2 | ≤ r | sin ϑ1 cos ϑ2 − sin ϑ1 cos ϑ2 | + | sin ϑ1 cos ϑ2 − sin ϑ1 cos ϑ2 | ≤ 2π · 2. Similarly, we can verify that |xi − xi | ≤ 2π · i for all i = 3, . . . , n. These, along with (3.29), lead to (3.30). The inequality (3.31) in (O3) can be proved in the same way. The reason that the factor j+1 j appears in (3.31) is as follows: Since 1 ≤ ki ≤ j (i = 1, . . . , n − 1), we see from the definition of Ωj+1;1,··· ,1 that x, x ∈ Ωj+1;1,··· ,1 ⇐⇒ j ≤ r < j + 1, 0 ≤ ϑl , ϑl <
2π for all l = 1, · · · , n − 1. j
(The above is comparable with (3.32).) Step 3. To prove that there are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) (both independent of a, j and f ) so that
1 C(1+ a )
|f | dx ≤ e 2
Ωj+1;k1 ,··· ,kn−1
Since Ωj;k1 ,··· ,kn−1
Ωj;k1 ,··· ,kn−1
|f | dx 2
θ
|f|2 ea|ξ| dξ 2
1−θ .
Rn
Ωj+1;k1 ,··· ,kn−1 is connected (see (3.25)), it follows from (3.30) and (3.31) that
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3 Ωj+1;k1 ,··· ,kn−1 ≤ d Ωj;k1 ,··· ,kn−1 + d Ωj+1;k1 ,··· ,kn−1 ≤ 6πn 2 . d Ωj;k1 ,··· ,kn−1 Thus, there exists x 1 ∈ Rn such that Ωj;k1 ,··· ,kn−1
3
Ωj+1;k1 ,··· ,kn−1 ⊂ BR0 (1 x), with R0 = 7πn 2 .
(3.33)
= C(n) According to Corollary 3.2 where b = R0 , there is C > 0 such that for all α ∈ N n , 1+ 1 a
Dα f L∞ (Rn ) ≤ eC
|α|! |α|
f(ξ)
R0
L2 (ea|ξ|2 dξ)
.
(3.34)
By (3.34), as well as (3.33), we can apply Lemma 3.5, where 1+ 1 a
ρ = 1, R = R0 , BR = BR0 (1 x), B2R = B2R0 (1 x), ω = Ωj;k1 ,··· ,kn−1 , M = eC
f(ξ)
L2 (ea|ξ|2 dξ)
,
to find constants C0 = C0 (n) > 0 and θ = θ(n) ∈ (0, 1) so that ! f L2 (BR0 (1x)) ≤ C0 f θ 2 L
Ωj;k1 ,··· ,kn−1
1+ 1 a
≤ C0 eC
1+ 1 a
eC
f θ 2 L
Ωj;k1 ,··· ,kn−1
"1−θ
f(ξ)
L2 (ea|ξ|2 dξ)
1−θ L2 ea|ξ|2 dξ
f(ξ)
.
(3.35)
(Here, we used (3.28) and a coordinate translation.) Finally, the desired inequality of this step follows from (3.35) and (3.33). Step 4. To prove the inequality of this lemma. From (3.26) and (3.27), we see that Ωj is the disjoint union of all Ωj;k1 ,··· ,kn−1 with different (k1 , . . . , kn−1 ) ∈ N n−1 satisfying 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. Meanwhile, by (3.23), (3.24) and (3.25), one can also check that Ωj+1 is the disjoint union of all Ωj+1;k1 ,··· ,kn−1 with different (k1 , . . . , kn−1 ) ∈ N n−1 satisfying 1 ≤ ki ≤ j for all i = 1, . . . , n −1. These, along with Lemma 3.4, yield that for some C = C(n) > 0 and θ = θ(n),
|f |2 dx = Ωj+1
≤
|f |2 dx
Ωj+1;k1 ,··· ,kn−1
eC
1
1+ a
|f |2 dx
≤e
|f | dx 2
Ωj;k1 ,··· ,kn−1
= j (n−1)(1−θ) eC
1 1+ a
Ωj
|f|2 ea|ξ| dξ 2
1−θ
Rn
Ωj;k1 ,··· ,kn−1 1 C 1+ a
θ
|f |2 dx
θ
θ
|f|2 ea|ξ| dξ 2
1−θ
Rn
|f|2 ea|ξ| dξ 2
1−θ ,
(3.36)
Rn
where the sums are taken over all different (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. (Notice that there are j n−1 such (k1 , . . . , kn−1 ).). Hence, the desired conclusion follows from (3.36). This ends the proof of Lemma 3.4. 2 Based on Lemma 3.4, we can have the next propagation result which will be used later.
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Lemma 3.6. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0 and j ≥ 1,
|f |2 dx ≤ j n−1 eC
1 1+ a
|f |2 dx
θj
2
1−θj ,
Rn
B1
Ωj+1
|f|2 ea|ξ| dξ
2 when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ). 2 Proof. Arbitrarily fix a > 0 and j ≥ 1. And then arbitrarily fix f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ). From Lemma 3.4, we can use the induction method to verify that
2 j−2 j−1 (n−1)(1−θ) |f |2 dx ≤ j(j − 1)θ (j − 2)θ · · · 2θ 1θ
Ωj+1 1
× eC(1+ a )(1+θ+···+θ C
1
≤ j n−1 e 1−θ (1+ a )
j−1
)
|f |2 dx
θj
2
1−θj
Rn
Ω1
|f |2 dx
|f|2 ea|ξ| dξ
θj
|f|2 ea|ξ| dξ 2
1−θj .
(3.37)
Rn
Ω1
Since Ω1 = B1 and θ = θ(n) ∈ (0, 1), the desired conclusion in the lemma follows from (3.37). This ends the proof of Lemma 3.6. 2 The next proposition plays a very important role in the proof of Theorem 3.1. Proposition 3.1. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0, t > 0 and ε > 0,
e−a|x| |f |2 dx ≤ eC
1+ 1t +a
1 + a−n Γ
Rn
a ε 2| ln θ|
|f|2 et|ξ| dξ + eε 2
Rn
−
2| ln θ| a
|f |2 dx ,
B1
2 when f ∈ L2 (Rn ) satisfies f ∈ L2 (et|ξ| dξ).
To prove Proposition 3.1, we need the following result quoted from [43]: Lemma 3.7 ([43, Lemma 3.1]). Let a > 0, b ∈ (0, 1) and θ ∈ (0, 1). Then ∞
bθ e−ak ≤ k
k=1
a ea a Γ | ln b|− | ln θ| . | ln θ| | ln θ|
Proof of Proposition 3.1. Arbitrarily fix a > 0 and t > 0. And then arbitrarily fix f ∈ L2 (Rn ) satisfies 2 f ∈ L2 (et|ξ| dξ). It suffices to show the inequality in Proposition 3.1 for the above fixed a, t, f and any ε > 0. Without loss of generality, we can assume that
A := Rn
|f|2 et|ξ| dξ = 0 and B := 2
|f |2 dx = 0.
(3.38)
B1
For otherwise, when A = 0, we have that f = 0 over Rn , thus the desired inequality is trivial; while when B = 0, we can use the analyticity of f (which follows from Corollary 3.2) to see that f = 0 over Rn and then the desired inequality is trivial again.
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By (3.22), we have that
e−a|x| |f |2 dx =
Rn
e−a|x| |f |2 dx +
e−a|x| |f |2 dx
j≥1Ω j+1
B1
|f |2 dx +
≤
e−aj |f |2 dx.
(3.39)
j≥1Ω j+1
B1
We now estimate the last term of (3.39). According to Lemma 3.6, there is C1 = C1 (n) > 0 and θ = θ(n) ∈ (0, 1) so that
e−aj |f |2 dx ≤ e
1
C1 1+ 1t
j≥1Ω j+1
⎛
⎞θj ⎛
j n−1 e−aj ⎝
|f |2 dx⎠ ⎝
j≥1
≤ eC1
1+ t
B1
a n θj n! (2/a) e− 2 j A (B/A) ,
⎞1−θj |f|2 et|ξ| dξ ⎠ 2
Rn
(3.40)
j≥1
where A and B are given by (3.38). In the proof of (3.40), we used the inequality: j n e− 2 j ≤ n! (2/a) a
n
for all j ≥ 1.
Meanwhile, by Lemma 3.7 (with b = B/A), we have that
a e2 a Γ A| ln(B/A)|− 2| ln θ| . | ln θ| 2| ln θ| a
e− 2 j A(B/A)θ ≤ j
a
j≥1
(3.41)
About A/B, there are only two possibilities: either A/B > e or A/B ≤ e. In the first case when A/B > e, we claim that −
A| ln(B/A)|− 2| ln θ| ≤ εA + eε a
2| ln θ| a
B for all ε > 0.
In fact, when ε satisfies that A| ln(B/A)|− 2| ln θ| ≤ εA, a
(3.42) is trivial. On the other hand, when ε > 0 satisfies that A| ln(B/A)|− 2| ln θ| > εA, a
we have that A/B < eε
−
2| ln θ| a
.
This, along with the fact that A/B > e, yields that −
A| ln(B/A)|− 2| ln θ| ≤ A ≤ B · A/B ≤ eε a
Thus, we have proved (3.42).
2| ln θ| a
B for all ε > 0.
(3.42)
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Inserting (3.42) into (3.41) leads to
2| ln θ| − e 2 a Γ εA + eε a B for all ε > 0. | ln θ| 2| ln θ| a
e− 2 j A(B/A)θ ≤ j
a
j≥1
(3.43)
Now it follows from (3.39),(3.40) and (3.43) that for some C2 = C2 (n) > 0,
e
−a|x|
C1 (1+ 1t )
|f | dx ≤ e 2
Rn
a a 2| ln θ| 2 − n e Γ n! 1 + (2/a) εA + eε a B | ln θ| 2| ln θ|
1 ≤ eC2 (1+ t +a) 1 + a−n Γ
2| ln θ| − a εA + eε a B for any ε > 0. 2| ln θ|
This proves the desired inequality for the first case that A/B > e. In the second case where A/B ≤ e, we derive directly that
e
−a|x|
|f | dx ≤
|f | dx ≤
2
Rn
2
Rn
Rn
⎛
≤ e ⎝ε
2 |f|2 et|ξ| dξ ≤ e
|f |2 dx
B1 −
|f|2 et|ξ| dξ + eε 2
2| ln θ| a
Rn
⎞ |f |2 dx⎠ for any ε > 0.
B1
This proves the desired inequality for the second case that A/B ≤ e. Hence, we end the proof of Proposition 3.1. 2 We now are on the position to show Theorem 3.1. Proof of Theorem 3.1. (i). Arbitrarily fix u0 ∈ L2 (ea|x| dx). Let u(T, x) = (eT Δ u0 )(x), x ∈ Rn . By the Hölder inequality, we have that
|u(T, x)| dx ≤ 2
Rn
|u(T, x)| e
2 a|x|
1/2
dx
Rn
|u(T, x)|2 e−a|x| dx
1/2 .
(3.44)
Rn
We will estimate the two terms on right side of (3.44) one by one. For the first term, we apply Lemma 3.1 (with ν = 1) to obtain that
|u(T, x)|2 ea|x| dx ≤ 2n e2a Rn
2
|u0 (x)|2 ea|x| dx.
T
(3.45)
Rn
T Δ u ∈ L2 (eT |ξ| dξ), since To estimate the second term (on right side of (3.44)), we first notice that e 0 2 a|x| 2 n u0 ∈ L (e dx) ⊂ L (R ). Thus, we can apply Proposition 3.1 (with f = eT Δ u0 and t = 2T ) to find C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for each ε > 0, 2
e Rn
−a|x|
2| ln θ| 2 ε− a |u(T, x)| dx ≤ C(T, a, n) ε |u0 (x)| dx + e |u(T, x)|2 dx , 2
Rn
B1
where 1 C(T, a, n) = eC(1+ T +a) 1 + a−n Γ
a . 2| ln θ|
(3.46)
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Inserting (3.45) and (3.46) into (3.44), we get that for each ε > 0
|u(T, x)|2 dx ≤ C
Rn
ε1/2 ≤C
|u0 (x)|2 ea|x| dx
|u0 (x)| e
2 a|x|
Rn
ε1/2 ≤ 2−1 C
ε1/2 ≤C
1/2
1/2 2| ln θ| − |u(T, x)|2 dx ε |u0 (x)|2 dx + eε a
Rn
1/2 dx
Rn
B1
|u0 (x)| dx + ε
1/2
2
ε
−1/2
e
Rn
|u0 (x)|2 ea|x| dx + ε1/2
Rn
|u(T, x)|2 dx
1/2
B1 −
|u0 (x)|2 dx + ε−1/2 eε
Rn
|u0 (x)|2 ea|x| dx + ε−1/2 e
2| ln θ| ε− a
2| ln θ| ε− a
Rn
2| ln θ| a
|u(T, x)|2 dx
B1
|u(T, x)|2 dx ,
(3.47)
B1
where 2 = C(T, C a, n) = 2n/2 ea T
C(T, a, n).
Since ε > 0 can be arbitrary taken, we replace ε by ε2 in (3.47) to get the desired conclusion in (i) of Theorem 3.1. (ii). Arbitrarily fix u0 ∈ L2 (xν dx). Let u(T, x) = eT Δ u0 (x), x ∈ Rn . Three facts are given in order. Fact One. Using the inequality: 1
1 ≤ ε|x|ν + e(1/ε) ν e−|x| for all ε > 0 and x ∈ Rn , we find that
1
|u(T, x)|2 dx ≤ ε Rn
|u(T, x)|2 |x|ν dx + e(1/ε) ν
Rn
|u(T, x)|2 e−|x| dx.
(3.48)
Rn
Fact Two. Since 0 < ν ≤ 1, we can apply Lemma 3.2 to find C1 = C1 (n) so that
2 ν |u(T, x)|2 |x|ν dx ≤ 4ν+2 Γ(ν/2 + n) 1 + T 4 u0 L2 ( x ν dx)
Rn
ν ≤ C1 1 + T 2
|u0 (x)|2 xν dx.
(3.49)
Rn
Fact Three. We can use (3.46) (with a = 1 and ε = μ) to find C2 = C2 (n) and θ = θ(n) ∈ (0, 1) so that for all μ > 0,
e−|x| |u(T, x)|2 dx ≤ eC2
1+ T1
μ
Rn
−2| ln θ|
|u0 (x)|2 dx + eμ
Rn
2 |u(T, x)| dx .
(3.50)
B1
To continue the proof, we arbitrarily fix ε ∈ (0, 1). We will first use Fact Three, and then use Fact One 1
and Fact Two. By taking μ = εe−( ε ) ν in (3.50), we obtain that 1
1
e(1/ε) ν
Rn
e−|x| |u(T, x)|2 dx
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182
1
≤e
(1/ε) ν
= eC2
e
C2 1+ T1
1
1+ T
μ
|u0 (x)| dx + e 2
Rn
|u0 (x)|2 dx + bε
ε Rn
μ−2| ln θ|
|u(T, x)|2 dx
B1
|u(T, x)|2 dx ,
(3.51)
B1
where 2 3 1 1 bε = exp (1/ε) ν + (ε−2 e2(1/ε) ν )| ln θ| . Meanwhile, one can directly check the following two inequalities: s2 ≤ es for all s > 0;
(3.52)
s + e3| ln θ|s ≤ e(3| ln θ|+1)s for all s > 0.
(3.53)
Choosing s = ε−1 and s = (1/ε) ν in (3.52) and (3.53) respectively, using 0 < ν ≤ 1, we find that 1
2 3 1 1 bε ≤ exp (1/ε) ν + (e1/ε e2(1/ε) ν )| ln θ| 2 3 1 1 ≤ exp (1/ε) ν + (e3(1/ε) ν )| ln θ| 2 1 3 ≤ exp e(3| ln θ|+1)(1/ε) ν .
(3.54)
Combining (3.51) and (3.54) leads to that 1
e
e−|x| |u(T, x)|2 dx
(1/ε) ν
Rn
≤ eC2 (n)
1+ T1
ε
|u0 (x)|2 dx + ee
1 (3| ln θ|+1)(1/ε) ν
Rn
|u(T, x)|2 dx .
(3.55)
B1
Finally, inserting (3.49) and (3.55) into (3.48), we obtain that for some C3 = C3 (n),
|u(T, x)|2 dx Rn
1
(3| ln θ|+1)(1/ε) ν ν 1 ≤ C1 (1 + T 2 ) + eC2 (1+ T ) ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx
Rn
B1
1
(3| ln θ|+1)(1/ε) ν ν 1 ≤ (1 + T 2 )eC3 (1+ T ) ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx ,
Rn
B1
which leads to the desired conclusion in (ii) of Theorem 3.1. Hence, we end the proof of Theorem 3.1. 2 3.2. Weak observability inequalities with observations on balls According to Theorem 1.1, it is impossible to recover a solution of (1.1) by observing it over a ball. Thus, two interesting questions arise. First, can we recover a solution of (1.1) over a ball by observing it
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183
on another ball? Second, can we have observability inequalities with observations over balls for solutions of (1.1) with some kind of initial values? The answer to the first question is almost negative, while we give partially positive answer for the second question. The first main result of this subsection is stated as follows: Theorem 3.2. (i) There is an absolute positive constant C so that for all T > 0 and 0 < r < r, !
u (T, x) dx ≤ 2
Br
Cn 1 + T (r − r )2
" T
u2 (t, x) dx dt, when u solves (1.1). 0 Br
(ii) Given T > 0 and r > r > 0, there is no constant C = C(T, r , r, n) so that
T
u (T, x) dx ≤ C 2
u2 (t, x) dx dt for any solution u to (1.1). 0 Br
Br
Proof. (i) Arbitrarily fix T > 0 and 0 < r < r. Arbitrarily fix a solution u to (1.1). Let u(0, x) = u0 (x), x ∈ Rn . Choose a C 2 function ϕ on Rn so that for some absolute constant C > 0, −|α|
0 ≤ ϕ(x) ≤ 1 over Rn ; |Dα ϕ(x)| ≤ C (r − r )
for all α ∈ N n , with |α| ≤ 2,
(3.56)
and so that / ϕ(x) =
1, x ∈ Br , 0, x ∈ Brc .
Set v = ϕu. Then v satisfies vt − Δv = −2∇ϕ · ∇u − Δϕu
in R+ × Rn ,
v(0, ·) = ϕ(·)u0 (·) in Rn .
(3.57)
Multiplying (3.57) by tv leads to 1 1 2 tv t − v 2 − tvΔv = −2tu∇ϕ · ∇v + t 2|∇ϕ|2 − ϕΔϕ u2 . 2 2
(3.58)
Integrating (3.58) over (0, T ) × Rn , we have that
1 2
1 T v (T, x) dx − 2
T
2
T
v (t, x) dx dt +
0 Rn
Rn
2
t |∇v(t, x)| dx dt
2
0 Rn
T
−2tu∇ϕ(x) · ∇v(t, x) dx dt
= 0 Rn
T
+
2 t 2 |∇ϕ(x)| − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
0 Rn
Meanwhile, by the Hölder inequality, we find that
(3.59)
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184
T
−2tu(t, x)∇ϕ(x) · ∇v(t, x) dx dt 0 Rn
T
T
2
≤
2
t |∇v(t, x)| dx dt + 0
t |∇ϕ(x)| u2 (t, x) dx dt. 0
Rn
(3.60)
Rn
Inserting (3.60) into (3.59) leads to that
v 2 (T, x) dx
T Rn
T
T
≤
2
v (t, x) dx dt + 2 0 Rn
t 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
(3.61)
0 Rn
Since v = u on Br and v = 0 on Brc , we can use (3.61) and (3.56) to get that
u2 (T, x) dx Br
1 ≤ T
T
2 u (t, x) dx dt + T
T
0 Br
! ≤
2
8Cn 1 + T (r − r )2
t 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt
0 Br \Br
" T
u2 (t, x) dx dt, 0 Br
which leads to the desired conclusion in (i) of Theorem 3.2. (ii) By contradiction, we suppose that there were T > 0, r > r > 0 and C = C(T, r , r, n) > 0 so that
T
u (T, x) dx ≤ C 2
u2 (t, x) dx dt for any solution u to (1.1).
(3.62)
0 Br
Br
We would use a constructive method to derive a contradiction with (3.62). For this purpose, we define, for each k ≥ 1, uk (t, x) =
|x1 −k|2 +|x |2 1 e− 4(t+1) , n/2 (4π(t + 1))
(t, x) = (t, x1 , x ) ∈ [0, ∞) × R × Rn−1 .
One can easily check that uk is the solution of (1.1) with initial value: uk (0, x) =
|x −k|2 +|x |2 1 − 1 4 e , (4π)n/2
x = (x1 , x ) ∈ R × Rn−1 .
It is clear that {uk (0, ·)}k≥1 is uniformly bounded in L2 (Rn ). We next show that when k is large enough, uk does not satisfy (3.62) (which leads to a contradiction). To this end, we need two estimates:
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
T
u2k (t, x) dx dt ≤
185
(k−r)2 r n T Vn − 2(T +1) , when k > r + e 2n(T + 1); (4π(T + 1))n
(3.63)
0 Br
u2k (T, x) dx ≥
(k−r− σ )2 3 1 − 2(T +1) e (σ/3)n Vn , with σ = r − r, when k > r . (4π(T + 1))n
(3.64)
Br
To show (3.63), we obtain from a direct computation that ∂t (ln uk (t, x)) = −
|x1 − k|2 + |x |2 n 1 + for all t ≥ 0, x = (x1 , x ) ∈ Rn , 2 t+1 4(t + 1)2
from which, it follows that when t ≥ 0 and x = (x1 , x ) ∈ Rn , ∂t (ln uk (t, x)) > 0 ⇐⇒ |x1 − k|2 + |x |2 > 2n(t + 1). This implies that k>r+
2n(T + 1) =⇒ ∂t (ln uk (t, x)) > 0 for all (t, x) × (0, T ) × Br .
From the above, we see that when k > r + 2n(T +1), we have that for each x ∈ Br , uk (t, x) is an increasing function of t on [0, T ]. Hence, when k > r + 2n(T + 1),
T
T
u2k (t, x) dx dt
≤
0 Br
T
u2k (T, x) dx dt
0 Br
=
T (4π(T + 1))n
=
|x −k|2 +|x |2 1 − 1 2(T +1) e dx dt (4π(T + 1))n
0 Br
e
|x −k|2 +|x |2 − 1 2(T +1)
dx ≤
T (4π(T + 1))n
Br
≤
n
r T Vn e (4π(T + 1))n
|k−r|2
e− 2(T +1) dx
Br
(k−r)2 − 2(T +1)
,
which leads to (3.63). We next show (3.64). Given b > 0, use Br (b, 0 ) to denote the ball centered at (b, 0, 0, · · · , 0) and of radius r, namely Br (b, 0 ) := x ∈ Rn : |x1 − b|2 + |x |2 ≤ r2 . Let k > r and σ = r − r. Then
u2k (T, x) dx ≥ Br
Br
u2k (T, x) dx Bk−r− σ (k,0 )
3
|x1 −k|2 +|x |2 1 e− 2(T +1) dx n (4π(T + 1))
= Br
Bk−r− σ (k,0 ) 3
1 e− ≥ (4π(T + 1))n Meanwhile, it is clear that
2 k−r− σ 3 2(T +1)
Bk−r− σ3 (k, 0 ) . Br
(3.65)
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B σ3 (r + 2σ/3, 0 ) ⊂ Br
Bk−r− σ3 (k, 0 ),
which leads to that Bk−r− σ3 (k, 0 ) ≥ B σ3 (r + 2σ/3, 0 ) = (σ/3)n Vn . Br
(3.66)
Now (3.64) follows from (3.65) and (3.66) at once. From (3.63) and (3.64), we find that when k > max{r + 2n(T + 1), r },
T
u2k (t, x) dx dt 0 Br
u2k (T, x) dx
(k−r)2
− 2(T +1) r n T Vn (4π(T +1))n e
≤
− 1 (4π(T +1))n e
2 k−r− σ 3 2(T +1)
σ n 3
Vn
Br
=T =T
3r n σ 3r n
1
k−r− σ 3
e 2(T +1) 1
e 2(T +1)
2
−(k−r)2
2 2σ (σ 3 ) − 3 (k−r)
σ 3r n r −r 2 2(r −r) 1 =T e 2(T +1) 3 − 3 (k−r) . r −r
(3.67)
Since (
r − r 2 2(r − r) ) − (k − r) → −∞, as k → ∞, 3 3
we see from (3.67) that
T
u2k (t, x) dx dt lim
0 Br
k→+∞
= 0, u2k (T, x) dx
Br
from which, it follows that uk does not satisfy (3.62), when k is large enough. This shows the conclusion in (ii) of Theorem 3.2. Hence, we end the proof of Theorem 3.2. 2 The next corollary is a direct consequence of Theorem 3.2. Corollary 3.3. Given ν > 0, T > 0 and r > 0, there is no constant C = C(T, r, ν, n) > 0 so that
T
u (T, x)ρ(x) dx ≤ C 2
Rn
u2 (t, x) dx dt for any u solves (1.1), 0 Br
where either ρ(x) = x−ν , x ∈ Rn , or ρ(x) = e−|x| , x ∈ Rn .
(3.68)
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Remark 3.5. It was announced in [7, p. 384] (without proof) that given a bounded interval E, there is no positive weight function ρ such that
∞
T
|u(T, x)| ρ(x) dx ≤ C
|u(t, x)|2 dx dt
2
0
0 E
for all solutions of the heat equation in the physical space (0, ∞). The above Corollary 3.2 presents a similar result for the heat equation in the physical space Rn . The second main result of this subsection is stated as follows: Theorem 3.3. (i) There is a generic constant C so that for any T > 0, M > r > 0 and u0 ∈ L2 (Rn ) with supp u0 ⊂ Br , !
|u(T, x)|2 dx ≤ Rn
Cn 1 + T (M − r)2
" T
|u(t, x)|2 dx dt, 0 BM
where u is the solution to (1.1) with u(0, ·) = u0 (·). (ii) Assume that 0 ≤ u0 ∈ L1 (Rn ) so that
u0 (x) dx ≥ μ
u0 (x) dx for some r > 0 and μ ∈ (0, 1).
Rn
Br
Then for any T > 0, M > 0 and any solution u to (1.1) with u(0, ·) = u0 (·),
2 2 +1 π 2 T 2 −1 4r2 |u(T, x)| dx ≤ e T Vn (r ∧ M )n μ2 n
n
n
T
|u(t, x)|2 dx dt.
2
Rn
0 BM
Here, r ∧ M := min{r, M }. Proof. (i) The proof is similar to that of (i) of Theorem 3.2. Arbitrarily fix T > 0, M > r > 0 and u0 ∈ L2 (Rn ) with supp u0 ⊂ Br . Write u for the solution to (1.1) with u(0, ·) = u0 (·). Choose a C 2 function ϕ over Rn so that for some absolute constant C > 0, 0 ≤ ϕ(x) ≤ 1 over Rn ; |Dα ϕ(x)| ≤ C(M − r)−|α| for all α ∈ N n , with |α| ≤ 2
(3.69)
and so that / ϕ(x) =
0, x ∈ Br , c . 1, x ∈ BM
Set v = ϕu. Multiplying (3.57) by v, we find that 1 2 v t − vΔv = −2u∇ϕ · ∇v + 2|∇ϕ|2 − ϕΔϕ u2 . 2 Integrating (3.70) over (0, T ) × Rn , we obtain that
(3.70)
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1 2
1 v (T, x) dx − 2
T
2
Rn
|∇v(t, x)|2 dx dt
2
v (0, x) dx +
T
T
−2u(t, x)∇ϕ(x) · ∇v(t, x) dx dt +
=
(3.71)
0 Rn
Rn
0 Rn
2|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
0 Rn
Since the support of u0 is contained in Br , we have that v(0, ·) = 0 over Rn . Then by the Hölder inequality, we deduce from (3.71) that
T
2 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (s, x) dx ds.
v (T, x) dx ≤ 2
0
Rn
(3.72)
Rn
Note that (3.72) is still true if we replace T by any t ∈ (0, T ). This implies that
T
T t
v (t, x) dx dt ≤ 2
0
0
Rn
0
2 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dτ dt.
(3.73)
Rn
c Since v = u on BM , it follows from (3.69) and (3.73) that
T
CnT u (t, x) dx ≤ (M − r)2
T
2
0 |x|≥M
u2 (t, x) dx dt.
(3.74)
0 r≤|x|≤M
Meanwhile, it is clear that
1 u (T, x) dx ≤ T
T
2
u2 (t, x) dx dt.
(3.75)
0 Rn
Rn
Combining (3.74) and (3.75), we obtain that
1 u (T, x) dx ≤ T
T
2
Rn
1 ≤ T ! ≤
1 u (t, x) dx dt + T
T
2
0 |x|≤M
T
u2 (t, x) dx dt 0 |x|≥M
1 CnT u (t, x) dx dt + T (M − r)2
T
2
0 |x|≤M
Cn 1 + T (M − r)2
u2 (t, x) dx dt 0 r≤|x|≤M
" T
u2 (t, x) dx dt, 0 |x|≤M
which leads to the conclusion (i) of Theorem 3.3. (ii) Let T > 0 and M > 0 be arbitrarily given. Arbitrarily fix u0 so that
0 ≤ u0 ∈ L1 (Rn );
u0 (x) dx ≥ μ
Br
Rn
u0 (x) dx for some r > 0 and μ ∈ (0, 1).
(3.76)
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189
Write u for the solution to (1.1) with u(0, ·) = u0 (·). We first prove that when 0 < M ≤ r,
2 2 +1 π 2 T 2 −1 4r2 |u(T, x)| dx ≤ e T Vn M n μ2 n
n
n
T
|u(t, x)|2 dx dt.
2
(3.77)
0 BM
Rn
For this purpose, we need the following two estimates: ⎛
T
T 2
u2 (t, x) dx dt ≥ 2−1 (4π)−n Vn M n μ2 T −(n−1) e
2 − 4r T
|x|≤M
⎝
⎞2 u0 (x) dx⎠ ;
(3.78)
Rn
⎛
u2 (T, x) dx ≤ 2
− 3n 2
(πT )−n/2 ⎝
Rn
⎞2
u0 (x) dx⎠ .
(3.79)
Rn
To show (3.78), we observe that
u(t, x) =
(4πt)−n/2 e−
|x−y|2 4t
u0 (y) dy,
(t, x) ∈ (0, ∞) × Rn .
(3.80)
Rn
By (3.80) and (3.76), we find that when t > 0, |x| ≤ r,
|x−y|2 r2 u(t, x) ≥ (4πt)−n/2 e− 4t u0 (y) dy ≥ (4πt)−n/2 e− t u0 (y) dy |y|≤r
≥ μ(4πt)−n/2 e
|y|≤r
2 − rt
u0 (x) dx.
(3.81)
Rn
Since M ≤ r, it follows from (3.81) that
T
⎛ u (t, x) dx dt ≥ Vn M μ ⎝ 2
T 2
n 2
|x|≤M
⎞2 u0 (x) dx⎠
Rn
T
(4πt)−n e−
T 2
≥ 2−1 (4π)−n Vn M n μ2 T −(n−1) e
⎛ 2 − 4r T
⎝
2r 2 t
dt ⎞2
u0 (x) dx⎠ ,
Rn
which leads to (3.78). We now show (3.79). By (3.80) and the Young inequality, we have that
|x|2 − 3n 4 (πT )−n/4 u(T, x) L2 (Rn ) ≤ (4πT )−n/2 e− 4T u (x) dx = 2 u0 (x) dx, 0 2 n L (R ) Rn
Rn
which leads to (3.79). Next, by (3.78) and (3.79), we see that
Rn
2 2 +1 π 2 T 2 −1 4r2 e T Vn M n μ2 n
u2 (T, x) dx ≤
n
n
T
u2 (t, x) dx dt T 2
|x|≤M
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190
2 2 +1 π 2 T 2 −1 4r2 ≤ e T Vn M n μ2 n
n
n
T
u2 (t, x) dx dt, 0 |x|≤M
which leads to (3.77) for the case when 0 < M ≤ r. Finally, when M > r, we apply (3.77) (with M = r) to obtain that
2 2 +1 π 2 T 2 −1 4r2 u (T, x) dx ≤ e T Vn rn μ2 n
n
n
T
2
Rn
2 2 +1 π 2 T 2 −1 4r2 e T ≤ Vn rn μ2 n
n
n
u2 (t, x) dx dt 0 |x|≤r
T
u2 (t, x) dx dt, 0 |x|≤M
which leads to (3.77) for the case that M > r. So the conclusion (ii) in Theorem 3.3 is true. Hence, we end the proof of Theorem 3.3. 2 3.3. Applications to controllability In this subsection, we will use Theorem 3.1 and Theorem 3.2 to derive some kinds of controllability for some kinds of controlled heat equations. We start with introducing the controlled equation for the applications of Theorem 3.1. Let 0 < τ < T . Consider the impulsively controlled heat equation: ⎧ ⎪ in (0, T ) \ {τ } × Rn , ⎨ ∂t y − Δy = 0 (3.82) Rn , y|t=τ = y|t=τ − + χB1 h in ⎪ ⎩ y| n in R , t=0 = y0 where y0 ∈ L2 (Rn ) is an initial state, h ∈ L2 (Rn ) is a control and y|t=τ − denotes the left limit of the solution y (which is treated as a function from R+ to L2 (Rn )) at time τ . Write y(·, ·; y0 , h) for the solution to (3.82). Theorem 3.4. The following two conclusions are true: (i) Let a > 0. Then for any y0 ∈ L2 (Rn ) and ε > 0, there is a control hy0 ,ε ∈ L2 (Rn ) so that
C1 (a, T − τ ) ≥
1 ε
|y0 (x)|2 dx
Rn
e−a|x| |y(T, x; y0 , hy0 ,ε )|2 dx + εe−ε
Rn
−
4| ln θ| a
|hy0 ,ε (x)|2 dx, Rn
where C1 (a, T − τ ) and θ are given by (i) of Theorem 3.1. ˜ y ,ε ∈ L2 (Rn ) so that (ii) Let ν ∈ (0, 1]. Then for any y0 ∈ L2 (Rn ) and ε > 0, there is a control h 0
C2 (ν, T − τ ) 1 ≥ ε
|y0 (x)|2 dx
Rn
x
−ν
˜ y ,ε )|2 dx + e−e |y(T, x; y0 , h 0
Rn
where C2 (ν, T − τ ) and θ are given by (ii) of Theorem 3.1.
1 (3| ln θ|+1) 1 ν ε
˜ y ,ε (x)|2 dx, |h 0 Rn
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191
Proof. (i) We will use [43, Lemma 5.1] and (i) of Theorem 3.1 to prove the conclusion (i). For this purpose, we set up our control problem in the framework of [43, Lemma 5.1]. Arbitrarily fix a > 0. Define the Banach space: ⎧ ⎨ Xa :=
f ∈ L2 (Rn ) :
⎩
ea|x| |f (x)|2 dx < ∞ Rn
⎫ ⎬ ⎭
,
(3.83)
endowed with the norm ⎛ f Xa := ⎝
⎞1/2
ea|x| |f (x)|2 dx⎠
, f ∈ Xa .
Rn
One can directly check that the dual space of Xa is as:
· Xa∗
Xa∗ = C0∞ (Rn )
,
(3.84)
with the norm: ⎛ g Xa∗ := ⎝
⎞1/2 e−a|x| |g(x)|2 dx⎠
, g ∈ Xa∗ .
(3.85)
Rn
We let X := L2 (Rn ) = X ∗ , Y := L2 (Rn ; C) = Y ∗ and Z := Xa ,
(3.86)
where the space Xa is given by (3.83). Define two operators R : Z → X and O : Z → Y by Rz := eΔ(T −τ ) z for each z ∈ Xa ;
Oz := χB1 eΔ(T −τ ) z for each z ∈ Xa .
(3.87)
One can directly check that R∗ f = eΔ(T −τ ) f, ∀ f ∈ L2 (Rn );
O∗ h = eΔ(T −τ ) (χB1 h) = y(T, ·; 0, h), ∀ h ∈ L2 (Rn ).
(3.88)
By (i) of Theorem 3.1, (3.87), (3.86) and (3.83), we can verify that for any ε > 0, 4| ln θ| − Rz 2L2 (Rn ) ≤ C1 (a, T − τ ) ε Oz 2L2 (Rn ) + ε−1 eε a z 2Xa for each z ∈ Xa ,
(3.89)
where C1 (·, ·) and θ are given by (i) of Theorem 3.1. Next, we arbitrarily fix ε > 0 and y0 ∈ L2 (Rn ). According to [43, Lemma 5.1] and (3.89), there exists hy0 ,ε ∈ L2 (Rn ) so that 1 1 hy0 ,ε 2Y ∗ + R∗ (eΔτ y0 ) − O∗ hy0 ,ε 2Z ∗ ≤ eΔτ y0 2X ∗ . 4| ln θ| − C1 (a, T − τ )ε a −1 ε C1 (a, T − τ )ε e Since {eΔt }t≥0 is contractive in L2 (Rn ), from the above, (3.88), (3.86) and (3.85), we obtain the conclusion (i) of this theorem.
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(ii) We will use [43, Lemma 5.1] and (ii) of Theorem 3.1 to prove the conclusion (ii). To this end, we arbitrarily fix ν ∈ (0, 1]. Define the Banach space: Xν :=
⎧ ⎨ ⎩
f ∈ L2 (Rn ) :
xν |f (x)|2 dx < ∞ Rn
⎫ ⎬ ⎭
,
(3.90)
endowed with the norm ⎛ f Xν := ⎝
⎞1/2
xν |f (x)|2 dx⎠
, f ∈ Xν .
Rn
One can directly check that the dual space of Xν is as:
· Xν∗
Xν∗ = C0∞ (Rn )
,
with the norm ⎛ g Xν∗ := ⎝
⎞1/2 x−ν |g(x)|2 dx⎠
, g ∈ Xν∗ .
(3.91)
Rn
We let X := L2 (Rn ) = X ∗ , Y := L2 (Rn ) = Y ∗ and Z := Xν ,
(3.92)
where the space Xν is given by (3.90). Define two operators R : Z → X and O : Z → Y by Rz := eΔ(T −τ ) z for each z ∈ Xν ;
Oz := χB1 eΔ(T −τ ) z for each z ∈ Xν .
(3.93)
One can directly check that R∗ f = eΔ(T −τ ) f, ∀ f ∈ L2 (Rn );
O∗ h = eΔ(T −τ ) (χB1 h) = y(T, ·; 0, h), ∀ h ∈ L2 (Rn ).
(3.94)
By (ii) of Theorem 3.1, (3.93), (3.92) and (3.90), we can verify that for any ε > 0, Rz 2L2 (Rn )
1 (3| ln θ|+1) 1 ν ε 2 e ≤ C2 (ν, T − τ ) ε Oz L2 (Rn ) + e z 2Xν for each z ∈ Xν ,
(3.95)
where C2 (·, ·) and θ are given by (ii) of Theorem 3.1. Next, we arbitrarily fix ε > 0 and y0 ∈ L2 (Rn ). According to [43, Lemma 5.1] and (3.95), there exists ˜ y ,ε ∈ L2 (Rn ) so that h 0 1 ˜ y ,ε 2 ∗ + h Y 0 C2 (a, T − τ )ε
1 1 (3| ln θ|+1) 1 ν ε e
˜ y ,ε 2 ∗ ≤ eΔτ y0 2 ∗ . R∗ (eΔτ y0 ) − O∗ h Z X 0
C2 (a, T − τ )e
Since {eΔt }t≥0 is contractive in L2 (Rn ), from the above, (3.94), (3.92) and (3.91), we obtain the conclusion (ii) of this theorem. Hence, we end the proof of Theorem 3.4. 2
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Remark 3.6. Theorem 3.4 presents some approximate null controllability with a cost for the impulsively controlled heat equation (3.82). The conclusion (i) in this theorem says that given y0 ∈ L2 (Rn ) and ε > 0, there is a control hy0 ,ε ∈ L2 (Rn ), with the cost: C (a, T − τ ) − 4| ln θ| 12 1 eε a y0 L2 (Rn ) , ε X∗
so that the corresponding solution to (3.82) derives y0 at time 0 into Br a , with r = (C1 (a, T − 1 X∗ τ )ε) 2 y0 L2 (Rn ) , at time T . Here, Br a is the ball in Xa∗ (which is defined by (3.84)), centered at the origin and of radius r. In the similar manner, we can explain the conclusion (ii) in Theorem 3.4. The approximate null controllability with a cost for impulsively controlled heat equations in a bounded domain was studied in [37]. Similar controllability was studied for impulsively controlled Schrödinger equations in [43]. We end this subsection with introducing an application of Theorem 3.2. Let T > 0 and r > 0. Consider the controlled equation: /
∂t z − Δz = χBr v z(0, ·) = z0 (·)
in in
0, T ) × Rn , Rn ,
(3.96)
where z0 ∈ L2 (Rn ) is an initial state and v ∈ L2 ((0, T ) × Rn ) is a control. Write z(·, ·; z0 , v) for the solution to (3.96). By a standard duality method, we can easily obtain the next controllability result, with the aid of Theorem 3.2. (We will omit the detailed proof.) Theorem 3.5. Let T > 0 and r > r > 0. Then for each z0 ∈ L2 (Rn ), with supp z0 ⊂ Br , there is a control vz0 ∈ L2 ((0, T ) × Rn ), with
T
0 Rn
1
Cn + |vz0 (t, x)| dxdt ≤ T (r − r )2 2
|z0 (x)|2 dx,
Rn
where the generic constant C is given by (i) of Theorem 3.2, so that z(T, x; z0 , vz0 ) = 0, for a.e. x ∈ Rn . References [1] J. Apraiz, L. Escauriaza, Null-control and measurable sets, ESAIM Control Optim. Calc. Var. 19 (2013) 239–254. [2] J. Apraiz, L. Escauriaza, G. Wang, C. Zhang, Observability inequalities and measurable sets, J. Eur. Math. Soc. 16 (2014) 2433–2475. [3] V. Barbu, Exact null internal controllability for the heat equation on unbounded convex domains, ESAIM Control Optim. Calc. Var. 20 (2014) 222–235. [4] C. Bardos, G. Lebeau, J. Rauch, Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary, SIAM J. Control Optim. 30 (1992) 1024–1065. [5] A. Bonami, B. Demange, A survey on uncertainty principles related to quadratic forms, Collect. Math. (2006) 1–36. [6] V.R. Cabanillas, S.B. de Menezes, E. Zuazua, Null controllability in unbounded domains for the semilinear heat equation with nonlinearities involving gradient terms, J. Optim. Theory Appl. 110 (2001) 245–264. [7] P. Cannarsa, P. Martinez, J. Vancostenoble, Null controllability of the heat equation in unbounded domains by a finite measure control region, ESAIM Control Optim. Calc. Var. 10 (2004) 381–408. [8] T. Duyckaerts, L. Miller, Resolvent conditions for the control of parabolic equations, J. Funct. Anal. 263 (2012) 3641–3673. [9] M. Egidi, I. Veselić, Sharp geometric condition for null-controllability of the heat equation on Rd and consistent estimates on the control cost, Arch. Math. 111 (2018) 85–99. [10] L. Escauriaza, F.J. Fernández, S. Vessella, Doubling properties of caloric functions, Appl. Anal. 85 (2006) 205–223.
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Contents lists available at ScienceDirect
Journal de Mathématiques Pures et Appliquées www.elsevier.com/locate/matpur
Observable set, observability, interpolation inequality and spectral inequality for the heat equation in Rn ✩ Gengsheng Wang a,∗ , Ming Wang b , Can Zhang c , Yubiao Zhang a a b c
Center for Applied Mathematics, Tianjin University, Tianjin 300072, China School of Mathematics and Physics, China University of Geosciences, Wuhan 430074, China School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China
a r t i c l e
i n f o
Article history: Received 24 November 2017 Available online 3 April 2019 MSC: 35K05 93B05 93B07 Keywords: Characteristic of observable sets Observability inequality Hölder-type interpolation inequality Spectral inequality
a b s t r a c t This paper studies connections among observable sets, the observability inequality, the Hölder-type interpolation inequality and the spectral inequality for the heat equation in Rn . We present the characteristic of observable sets for the heat equation. In more detail, we show that a measurable set in Rn satisfies the observability inequality if and only if it is γ-thick at scale L for some γ > 0 and L > 0. We also build up the equivalence among the above-mentioned three inequalities. More precisely, we obtain that if a measurable set in Rn satisfies one of these inequalities, then it satisfies others. Finally, we get some weak observability inequalities and weak interpolation inequalities where observations are made over a ball. © 2019 Elsevier Masson SAS. All rights reserved.
r é s u m é Cet article étudie les liens entre les ensembles observables, l’inégalité d’observabilité, l’inégalité d’interpolation de Hölder-type et l’inégalité spectrale pour l’équation de la chaleur dans Rn . Nous présentons la caractéristique des ensembles observables pour l’équation de la chaleur. Plus en détail, nous montrons qu’un ensemble mesurable dans Rn satisfait l’inégalité d’observabilité si et seulement s’il est γ-épais à l’échelle L pour certains γ > 0 et L > 0. Nous construisons également l’équivalence entre les trois inégalités mentionnées au-dessus. Plus précisément, on obtient que si un ensemble mesurable dans Rn satisfait une de ces inégalités, alors il satisfait les autres. Enfin, nous obtenons des inégalités d’observabilité faibles et des inégalités d’interpolation faibles dans lesquelles les observations sont effectuées sur une boule. © 2019 Elsevier Masson SAS. All rights reserved.
✩ This work was partially supported by the National Natural Science Foundation of China under grants 11571264, 11701535, 11501424 and 11801408. * Corresponding author. E-mail addresses: [email protected] (G. Wang), [email protected] (M. Wang), [email protected] (C. Zhang), [email protected] (Y. Zhang).
https://doi.org/10.1016/j.matpur.2019.04.009 0021-7824/© 2019 Elsevier Masson SAS. All rights reserved.
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1. Introduction In this paper, we consider the heat equation: ∂t u − Δu = 0
in (0, ∞) × Rn ,
u(0, ·) ∈ L2 (Rn ).
(1.1)
The above equation (1.1) is well-posed.1 For this equation, we will characterize the observable sets and build up connections among several important inequalities which are introduced in the next subsection. Notation. Write C(· · · ) for a positive constant that depends on what are included in the brackets and may vary in different contexts. The same can be said about C (· · · ), C1 (· · · ) and so on. Use Vn to denote the volume of the unit ball in Rn . Let Br (x), with x ∈ Rn and r > 0, be the open ball in Rn , centered at x and of radius r. (Simply write Br = Br (0).) Let Sn−1 be the unit spherical surface in Rn . Let N := {0, 1, 2, . . . } and N + = N\{0}. Denote by Q the open unit cube in Rn , centered at the origin. Let x + LQ, with x ∈ Rn and L > 0, be the set {x + Ly : y ∈ Q}. For each measurable set D ⊂ Rn , denote by |D| and Dc its Lebesgue measure and complement set respectively. For any set G, we write χG for the characteristic function of G. Given f ∈ L2 (Rn ), write f for its Fourier transform.2 Given a measurable function f over Rn , we denote by supp f the support of f , which is the set of all points (in Rn ) where f does not vanish. n 2 1/2 Given x = (x1 , . . . , xn ) ∈ Rn , let |x| := and x := 1 + |x|2 . i=1 xi 1.1. Thick sets and several inequalities We start with introducing sets of γ-thickness at scale L. Sets of γ-thickness at scale L. A measurable set E ⊂ Rn is said to be γ-thick at scale L for some γ > 0 and L > 0, if (1.2) E (x + LQ) ≥ γLn for each x ∈ Rn . About sets of γ-thickness at scale L, several remarks are given in order. (a1 ) To our best knowledge, this definition arose from studies of the uncertainty principle. We quote it from [5] (see Page 5 in [5]). Before [5], some very similar concepts were proposed. For instance, the definition of relative dense sets was given in [18] (see also Page 113 in [15]); the definition of thick sets was introduced in [19]. (a2 ) Let E be a set of γ-thickness at scale L. Then, in each cube with the side length L, |E| is bigger than or equals to γLn . So E distributes almost equally in Rn , and parameters γ and L characterize the distribution of E. (a3 ) The structure of a set of γ-thickness at scale L may be very complicated. Here, we give two examples: 1 1 1 First, the set n∈Z [n − 1000 , n + 1000 ] is 500 -thick at scale 1 in R1 ; Second, the set n∈Z j∈N + [n + 2−j , n + 2−j + 2−j−1 ] is 12 -thick at scale 1 in R1 . 1 First, the Laplacian operator Δ, with its domain D(Δ) = H 2 (Rn ), generates a C0 -semigroup {etΔ : t ≥ 0} in L2 (Rn ); Second, each solution u to (1.1) can be expressed as:
u(t, x) = (e
tΔ
n
u(0, ·))(x) =
K(t, x − y)u(0, y) dy, (t, x) ∈ (0, ∞) × R , Rn
where K(t, x) = (4πt)−n/2 e−|x| /4t , t > 0, x ∈ Rn , is the heat kernel. 2 Given f in the Schwartz class S(Rn ), its Fourier transform is as: f(ξ) = (2π)−n/2 Rn e−ix·ξ f (x) dx, ξ ∈ Rn . Since S(Rn ) is 2 n 2 n dense in L (R ), by a standard way, we can define f for each f ∈ L (R ). 2
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Next, we introduce an observability inequality for the equation (1.1). The observability inequality. A measurable set E ⊂ Rn is said to satisfy the observability inequality for the equation (1.1), if for any T > 0 there exists a positive constant Cobs = Cobs (n, T, E) so that when u solves (1.1),
T
|u(T, x)| dx ≤ Cobs
|u(t, x)|2 dx dt.
2
(1.3)
0 E
Rn
When a measurable E ⊂ Rn satisfies (1.3), it is called an observable set for (1.1). Several notes on the observability inequality (1.3) are given in order. (b1 ) By treating the integral on the left hand side as a recovering term, and the integral on the right hand side as an observation term, we can understand the inequality (1.3) as follows: one can recover a solution of (1.1) at time T , through observing it on the set E and in the time interval (0, T ). From perspective of control theory, the inequality (1.3) is equivalent to the following null controllability: For any u0 ∈ L2 (Rn ) and T > 0, there exists a control f ∈ L2 ((0, T ) × Rn ) driving the solution u to the controlled equation: ∂t u − Δu = χE f in (0, T ) × Rn , from the initial state u0 to the state 0 at time T . (b2 ) Let E be γ-thick at scale L. Corollary 2.1 of this paper gives an explicit expression of Cobs (n, T, E) in terms of γ, L and T . Furthermore, we can see from Remark 2.3 that when L, T and n are given, Cobs (n, T, E) can be taken as a constant depending only on the average measure of E: inf x∈Rn |E (x+LQ)| , but independent of the shape and the position of E. Ln (b3 ) We can compare (1.3) with the observability inequality for the heat equation on a bounded physical domain. Let Ω be a bounded C 2 (or Lipschitz and locally star-shaped, see [2]) domain in Rn . Consider the equation: ⎧ ⎪ ⎪ ⎨∂t u − Δu = 0 u=0 ⎪ ⎪ ⎩u(0, ·) ∈ L2 (Ω).
in (0, ∞) × Ω, on (0, ∞) × ∂Ω,
(1.4)
We say that a measurable set ω ⊂ Ω satisfies the observability inequality for (1.4), if given T > 0, there is a constant C(n, T, ω, Ω) so that when u solves (1.4),
T
|u(T, x)| dx ≤ C(n, T, ω, Ω)
|u(t, x)|2 dx dt.
2
Ω
0
(1.5)
ω
When a measurable set ω ⊂ Ω satisfies (1.5), it is called an observable set for (1.4). The inequality (1.5) has been widely studied. See [14,20,25] for the case where ω is open; [1,2,12] for the case when ω is measurable. (b4 ) When Ω is an unbounded domain and ω is a bounded and open subset of Ω, the inequality (1.5) may not be true. This was showed in [27] for the heat equation in the physical domain R+ . Similar results have been obtained for higher dimension cases in [28]. For the heat equation in an unbounded domain, [29] imposed a condition, in terms of the Gaussian kernel, on the set ω so that the observability inequality does not hold. In particular, [29] showed that the observability inequality fails when Ω is unbounded and |ω| < ∞. Notice that any set E ⊂ Rn of finite measure does not have the characteristic on observable sets of (1.1). This characteristic is indeed the γ-thickness at scale L for some γ > 0 and L > 0. (See Theorem 1.1 of this paper.)
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About works on sufficient conditions of observable sets for heat equations in unbounded domains, we would like to mention the work [6]. It showed that, for some parabolic equations in an unbounded domain Ω ⊂ Rn , the observability inequality holds when observations are made over a subset E ⊂ Ω, with Ω\E bounded. For other similar results, we refer the reader to [30]. When Ω = Rn , such a set E has the characteristic on observable sets of (1.1) mentioned before. (b5 ) An interesting phenomenon is that some potentials (growing at infinity) in heat equations may change the above-mentioned characteristic on observable sets for the heat equations with potentials. In [31,8], the authors realized the following fact: Let A = Δ + V , where V (x) := −|x|2k , x ∈ Rn , with 2 ≤ k ∈ N. Write etA t≥0 for the semigroup generated by the operator A. Let r0 ≥ 0 and let Θ0 be an open subset of Sn−1 . Let Γ = {x ∈ Rn : |x| ≥ r0 , x/|x| ∈ Θ0 }. Then there is C(n, T, Θ0 , r0 , k) so that
T A 2 e u0 dx ≤ C(n, T, Θ0 , r0 , k)
T
0
Rn
tA 2 e u0 dx dt for all u0 ∈ L2 (Rn ).
(1.6)
Γ
The cone Γ does not have the characteristic on observable sets mentioned before, but still holds the observability inequality (1.6). The main reason is as follows: The unbounded potential V changes the behaviour of the solution of the pure heat equation (1.1). This plays an important role in the proof of (1.6) (see [31,8]). It should be pointed out that when V (x) = −|x|2 , x ∈ Rn (which means that the potential grows more slowly at infinity), (1.6) does not hold for the above cone. We refer the readers to [31,8] for more details on this issue. Besides, we also would like to mention [3] for this subject. An interesting question now arises: How do potentials influence characteristics of observable sets? We wish to answer this question in our future studies. We then introduce an interpolation inequality for the equation (1.1). The Hölder-type interpolation inequality. A measurable set E ⊂ Rn is said to satisfy the Hölder-type interpolation inequality for the heat equation (1.1), if for any θ ∈ (0, 1), there is CHold = CHold (n, E, θ) so that for each T > 0 and each solution u to the equation (1.1),
1
|u(T, x)|2 dx ≤ eCHold (1+ T ) Rn
|u(T, x)|2 dx
θ
|u(0, x)|2 dx
1−θ .
(1.7)
Rn
E
Several remarks on the Hölder-type interpolation inequality (1.7) are given in order. (c1 ) The above Hölder-type interpolation inequality is equivalent to what follows: There is θ = θ(n, E) ∈ (0, 1) and CHold = CHold (n, E) so that (1.7) holds for all T > 0 and solutions u to (1.1). This can be verified by the similar way used in the proof of [37, Theorem 2.1]. (c2 ) The inequality (1.7) is a kind of quantitative unique continuation for the heat equation (1.1). It provides a Hölder-type propagation of smallness for solutions of the heat equation (1.1). In fact, if
|u(T, x)|2 dx = δ, then we derive from (1.7) that E
|u(T, x)|2 dx is bounded by Cδ θ for some conRn
stant C > 0. Consequently, u(T, ·) = 0 over Rn provided that it is zero over E. (c3 ) From perspective of control theory, the inequality (1.7) implies the approximate null controllability with cost for impulse controlled heat equations, i.e., given T > τ > 0, ε > 0, there is C = C(n, E, T, τ, ε) such that for each u0 ∈ L2 (Rn ), there is f ∈ L2 (Rn ) so that f L2 (Rn ) ≤ C u0 L2 (Rn ) and u(T, ·) L2 (Rn ) ≤ ε u0 L2 (Rn ) ,
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where u is the solution to the impulse controlled equation: ∂t u − Δu = δ{t=τ } χE f in (0, T ) × Rn , with the initial condition u(0, x) = u0 (x), x ∈ Rn . (See [37, Theorem 3.1].) (c4 ) The Hölder-type interpolation inequality (1.7) can imply the observability inequality (1.3). Moreover, it leads to the following stronger version of (1.3):
|u(T, x)| dx ≤ Cobs
|u(t, x)|2 dx dt, with Cobs = Cobs (n, T, E, F ),
2
Rn
(1.8)
F E
where F ⊂ (0, T ) is a subset of positive measure. This will be presented in Lemma 2.4. We derive (1.8) from (1.7), through using the telescoping series method developed in [36] (see also [38,2]) for heat equations in bounded domains. (c5 ) We can compare (1.7) with an interpolation inequality for the heat equation (1.4). A measurable set ω ⊂ Ω is said to satisfy the Hölder-type interpolation inequality for the equation (1.4), if for any θ ∈ (0, 1), there is C = C(n, Ω, ω, θ) so that for any T > 0 and any solution u to (1.4),
C(1+ T1 )
|u(T, x)| dx ≤ e 2
|u(T, x)| dx 2
θ
ω
Ω
|u(0, x)|2 dx
1−θ .
(1.9)
Ω
In [35], the authors proved that any open and nonempty subset ω ⊂ Ω satisfies the Hölder-type interpolation inequality (1.9) for heat equations with potentials in bounded and convex domains. The frequency function method used in [35] was partially borrowed from [10]. In [2], the authors proved that any subset ω of positive measure satisfies the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) where Ω is a bounded Lipschitz and locally star-shaped domain in Rn . More about this inequality for heat equations in bounded domains, we refer the reader to [36–38]. Finally, we will introduce a spectral inequality for some functions in L2 (Rn ). The spectral inequality. A measurable set E ⊂ Rn is said to satisfy the spectral inequality, if there is a positive constant Cspec = Cspec (n, E) so that for each N > 0,
|f (x)| dx ≤ e 2
Rn
Cspec (1+N )
|f (x)|2 dx for all f ∈ L2 (Rn ) with supp f ⊂ BN .
(1.10)
E
Several notes on the spectral inequality (1.10) are given in order. (d1 ) Recall the Lebeau-Robbiano spectral inequality (see [20,21]): Let Ω be a bounded smooth domain in Rn and let ω be a nonempty open subset of Ω. Write ΔΩ for the Laplacian operator on L2 (Ω) with Domain(ΔΩ ) = H01 (Ω) H 2 (Ω). Let {λj }j≥1 (with 0 < λ1 < λ2 ≤ · · · ) be the eigenvalues of −ΔΩ and let {φj }j≥1 be the corresponding eigenfunctions. Then there is a positive constant C(n, Ω, ω) so that for each λ > 0,
|f (x)|2 dx ≤ eC(n,Ω,ω)(1+ Ω
√ λ)
|f (x)|2 dx for all f ∈ span{φj : λj < λ}.
(1.11)
ω
This inequality was extended to the case where Ω is a bounded C 2 domain via a simpler way in [26]. Then it was extended to the case that Ω is a bounded Lipschitz and locally star-shaped domain; ω is a subset of positive measure so that ω ⊂ BR (x0 ) ⊂ B4R (x0 ) ⊂ Ω for some R > 0 and x0 ∈ Ω; and C(n, Ω, ω) = C(n, Ω, |ω|/|BR |) (see [2, Theorem 5 and Theorem 3]).
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By our understanding, the inequality (1.10) is comparable to (1.11) from two perspectives as follows: First, the inequality (1.10) is satisfied by functions in the subspace: EN f ∈ L2 (Rn ) : supp f ⊂ BN with N > 0, while the inequality (1.11) is satisfied by functions in the subspace: Fλ f = aj φj ∈ L2 (Ω) : {aj }j≥1 ⊂ R with λ > 0. λj <λ
From the definition of the spectral projection in the abstract setting given in [39] (see Pages 262-263 in [39]), we can define two spectral projections: χ[0,N 2 ) (−Δ) and χ[0,λ) (−ΔΩ ) on L2 (Rn ) and L2 (Ω), respectively. Then after some computations, we find that EN and Fλ are the ranges of χ[0,N 2 ) (−Δ) and χ[0,λ) (−ΔΩ ), respectively. Second, the square root of the integral of χ[0,N 2 ) over R is√N which corresponds to the √ N in (1.10), while the square root of the integral of χ[0,λ) over R is λ which corresponds to the λ in (1.11). (d2 ) Though the inequality (1.10) was first named as the spectral inequality in [23] (to our best knowledge), it has been extensively studied for long time. (See, for instance, [5,15,18,19,32,24,33,34,40].) In [19], the author announced that if E is γ-thick at scale L for some γ > 0 and L > 0, then E satisfies the spectral inequality (1.10), and further proved this announcement for the case when n = 1. Earlier, the authors of [24] (see also [15]) proved that E is γ-thick at scale L for some γ > 0 and L > 0 if and only if E satisfies the following inequality: For each N > 0, there is a positive constant C(n, E, N ) so that
|f (x)| dx ≤ C(n, E, N ) 2
Rn
|f (x)|2 dx for each f ∈ L2 (Rn ) with supp f ⊂ BN .
(1.12)
E
This result is often referred as the Logvinenko-Sereda theorem. Before [24], the above equivalence was proved by B.P. Paneyah for the case that n = 1 (see [34,33,15]). In [18], the author claimed (1.12), with C(n, E, N ) = eCspec (1+N ) , and proved this claim for the case when n = 1. In the proof of our main theorem of this paper, the expression C(n, E, N ) = eCspec (1+N ) will play an important role. From this point of view, (1.12) is weaker than the spectral inequality (1.10). (d3 ) The inequality (1.12) is also important. It is closely related to the uncertainty principle (which is an extensive research topic in the theory of harmonic analysis and says roughly that a nonzero function and its Fourier transform cannot be both sharply localized, see [13]). In fact, a measurable set E satisfies the inequality (1.12) if and only if it satisfies the following uncertainty principle:
Rn
⎛
⎜ |f (x)|2 dx ≤ C (n, E, N ) ⎝
E
|f (x)|2 dx +
⎞ ⎟ |f(ξ)|2 dξ ⎠ for all f ∈ L2 (Rn ).
c BN
We refer the interested readers to [5,15,17,32] for the proof of the above result, as well as more general c uncertainty principle, where E and BN are replaced by more general sets. It deserves mentioning what follows: The uncertainty principle can help us to get the exact controllability for the Schrödinger equation with controls located outside of two balls and at two time points. This was realized in [43]. (See [16] for more general cases.) (d4 ) By using a global Carleman estimate, the authors in [23] proved the spectral inequality (1.10) for such an open subset E that satisfies the property: there exists δ > 0 and r > 0 so that ∀y ∈ Rn , ∃ y ∈ E such that Br (y ) ⊂ E and |y − y | ≤ δ.
(1.13)
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It is clear that a set with the above property (1.13) is a set of γ-thick at scale L for some γ > 0 and L > 0.3 (d5 ) With the aid of the spectral inequality (1.10), one can use the same strategy given in [20] to derive the null controllability described in the note (b1 ). 1.2. Aim, motivation and main result Aim According to the note (d2 ) in the previous subsection, the characteristic of a measurable set holding the spectral inequality (1.10) is the γ-thickness at scale L for some γ > 0 and L > 0. Natural and interesting questions are as follows: What is the characteristic of observable sets for (1.1)? How to characterize a measurable set E satisfying the Hölder-type interpolation inequality (1.7)? What are the connections among inequalities (1.3), (1.7) and (1.10)? The aim of this paper is to answer the above questions. Motivation The motivations of our studies are given in order. (i) The first motivation arises from two papers [4] and [1]. In [4], the authors gave, for the wave equation in a bounded physical domain Ω ⊂ Rn , a sufficient and almost necessary condition to ensure an open subset Γ ⊂ ∂Ω to be observable (i.e., Γ satisfies the observability inequality for the wave equation with observations on Γ). This condition is exactly the well known Geometric Control Condition (GCC for short).4 Thus, we can say that the GCC condition is a characteristic of observable open sets on ∂Ω, though this condition is not strictly necessary (see [4] or [22]). The authors in [1] presented a sufficient and necessary condition to ensure a measurable subset ω ⊂ Ω satisfying (1.5). This condition is as: |ω| > 0. Hence, the characteristic of observable sets for the equation (1.4) is as: |ω| > 0. Analogically, it should be very important to characterize observable sets for the heat equation (1.1). However, it seems for us that there is no any such result in the past publications. These motivate us to find the characteristic of observable sets for the equation (1.1). (ii) For the heat equation (1.4), the observability inequality (1.5), the Hölder-type interpolation inequality (1.9) and the spectral inequality (1.11) are equivalent. More precisely, we have that if ω ⊂ Ω is a measurable set, then |ω| > 0 ⇐⇒ ω satisfies (1.11) ⇐⇒ ω satisfies (1.9) ⇐⇒ ω satisfies (1.5).
(1.14)
The proof of (1.14) was hidden in the paper [2]. (See Theorem 5, Theorem 6, as well as its proof, Theorem 1, as well as its proof, in [2].) However, for the heat equation (1.1), the equivalence among these three inequalities has not been touched upon. These motivate us to build up the equivalence among inequalities (1.3), (1.7) and (1.10). It deserves mentioning that for heat equations with lower terms in bounded physical domains, we do not know if (1.14) is true. Main Result The main result of the paper is the next Theorem 1.1. Theorem 1.1. Let E ⊂ Rn be a measurable subset. Then the following statements are equivalent: (i) The set E is γ-thick at scale L for some γ > 0 and L > 0. (ii) The set E satisfies the spectral inequality (1.10). In fact, one can choose L = 2(δ + r), γ = r n (2(δ + r))−n Vn . An open subset Γ ⊂ ∂Ω is said to satisfy the GCC if there exists T0 > 0 such that any geodesic with velocity one meets Γ within time T0 at a nondiffractive point. 3 4
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(iii) The set E satisfies the Hölder-type interpolation inequality (1.7). (iv) The set E satisfies the observability inequality (1.3). Several remarks about Theorem 1.1 are given in order. (e1 ) The equivalence of statements (i) and (iv) in Theorem 1.1 tells us: the characteristic of observable sets for the heat equation (1.1) is the γ-thickness at scale L for some γ > 0 and L > 0. This seems to be new for us. (e2 ) The equivalence among statements (ii), (iii) and (iv) in Theorem 1.1 presents closed connections of the three inequalities. This seems also to be new for us. (e3 ) We find the following way to prove Theorem 1.1: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). We prove (i) ⇒ (ii) by some ideas from [19]. Indeed, this result was announced in [19] and then proved for the case that n = 1 in the same reference. We prove (ii) ⇒ (iii) ⇒ (iv), though using some ideas and techniques from [2,36]. Finally, we show (iv) ⇒ (i) via the structure of a special solution to the equation (1.1). (e4 ) We noticed that four days after we put our current work in arXiv, the paper [9] appeared in arXiv. In [9], the authors independently got the equivalence (i) and (iv) in Theorem 1.1. 1.3. Extensions to bounded observable sets From Theorem 1.1, we see that in order to have (1.3) or (1.7), the set E has to be γ-thick at scale L for some γ > 0 and L > 0. Then a natural and interesting question arises: What are possible substitutions of (1.3) or (1.7), when E is replaced by a ball in Rn ? (It deserves to mention that any ball in Rn does not satisfy the thick condition (1.2).) We try to find the substitutes from two perspectives as follows: (i) We try to add weights on the left hand side and ask ourself if the following inequalities hold for all solutions of (1.1):
T
χBr (x)|u(T, x)| dx ≤ C(T, r , r, n) 2
|u(t, x)|2 dx dt
(1.15)
0 Br
Rn
and
T
ρ(x)|u(T, x)| dx ≤ C(T, ρ, r, n)
|u(t, x)|2 dx dt,
2
Rn
(1.16)
0 Br
where ρ(x) = x−ν or e−|x| . On one hand, we proved that (1.15) is true when r < r, while (1.15) is not true when r > r (see Theorem 3.2 in Subsection 3.2). Unfortunately, we do not know if (1.15) holds when r = r. On the other hand, we showed that (1.16) fails for all r > 0 (see Corollary 3.3 in Subsection 3.2). (ii) We try to find a class of initial data so that (1.3) (where E is replaced by Br ) holds for all solutions of (1.1) with initial data in this class. We have obtained some results on this issue (see Theorem 3.3 in Subsection 3.2). More interesting question is as: what is the biggest class of initial data so that (1.3) (where E is replaced by Br ) holds for all solutions of the heat equation (1.4) with initial data in this class? Unfortunately, we are not able to answer it. We now turn to possible substitutions of (1.7) where E is replaced by B1 . We expect to find b(ε) > 0 for each ε ∈ (0, 1) so that for any T > 0, there is C(n, T ) > 0 such that when u solves (1.1),
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⎛ |u(T, x)|2 dx ≤ C(n, T ) ⎝ε
Rn
⎞
|u(T, x)|2 dx⎠ .
|u(0, x)|2 dx + b(ε)
Rn
(1.17)
B1
Let us explain why (1.17) deserves to be expected. Reason One. Let θ ∈ (0, 1) and T > 0. Then the next two inequalities are equivalent. The first inequality is as: there is C(n, T, θ) so that when u solves (1.1),
θ
1−θ |u(T, x)|2 dx ≤ C(n, T, θ) |u(T, x)|2 dx |u(0, x)|2 dx , Rn
Rn
B1
while the second inequality is as: there is C(n, T, θ) > 0 so that for any ε ∈ (0, 1) and any solution u to (1.1),
⎛ |u(T, x)|2 dx ≤ C(n, T, θ) ⎝ε
Rn
|u(0, x)|2 dx + ε
Rn
− 1−θ θ
⎞ |u(T, x)|2 dx⎠ .
(1.18)
B1
However, (1.18) is not true, for otherwise, we can use the same method developed in [36] (see also [2]) to derive (1.3) (where E is replaced by B1 ) which contradicts the equivalence of (i) and (iv) in Theorem 1.1. Thus, b(ε) in (1.17) cannot grow like a polynomial of ε. But it seems not to be hopeless for us to find some kind of b(ε) so that (1.17) holds. Reason Two. The space-like strong unique continuation of the heat equation (1.1) (see [10]) yields that if u(T, ·) = 0 on the ball B1 , then u(T, ·) = 0 over Rn . The inequality (1.17) is a quantitative version of the aforementioned unique continuation. Though we have not found any b(ε) so that (1.17) is true, we obtained some b(ε) so that (1.17) holds for all solutions to (1.1) with initial data having some slight decay (see Theorem 3.1 in Subsection 3.1). Finally, we would like to mention what follows: With the aid of an abstract lemma (i.e., Lemma 5.1 in [43]), each of extended inequalities mentioned above corresponds to a kind of controllability for controlled heat equations. In Subsection 3.3 of this paper, we will present some applications of some extended inequalities mentioned above to some kinds of controllability for some controlled heat equations. 1.4. Plan of the paper The paper is organized as follows: In Section 2, we prove Theorem 1.1. In Section 3, we present several weak observability inequalities and weak interpolation inequalities (where observations are made in a ball of Rn ) and some applications of these inequalities to controllability. 2. Proof of Theorem 1.1 We are going to prove Theorem 1.1 in the following way: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). The above steps are based on several lemmas: Lemmas 2.1, 2.3, 2.4 and 2.5. We begin with Lemma 2.1 connecting the spectral inequality with sets of γ-thickness at scale L. Lemma 2.1. Suppose that a measurable set E ⊂ Rn is γ-thick at scale L for some γ > 0 and L > 0. Then E satisfies the spectral inequality (1.10), with 1 for some C = C(n). Cspec (n, E) = C(1 + L) 1 + ln γ
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Remark 2.1. The manner that the√ constant eCspec (n,E)(1+N ) (in (1.10)) depends on N is comparable with the manner that the constant eC λ in (1.11) depends on λ. (This has been explained in the remark (d1 ) in Subsection 1.1.) The latter one played an important role in the proof of the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) (see [2]). Analogically, the previous one will play an important role in the proof of Theorem 1.1. Remark 2.2. In [19], the author announced the result in Lemma 2.1 and proved it for the case when n = 1. For the completeness of the paper, we give a detailed proof for Lemma 2.1, based on some ideas and techniques in [19]. To show Lemma 2.1, we need the following result on analytic functions: Lemma 2.2 ([19, Lemma 1]). Let Φ be an analytic function in D5 (0) (the disc in C, centered at origin and ˆ ⊂ I be a subset of positive measure. If of radius 5). Let I be an interval of length 1 such that 0 ∈ I. Let E |Φ(0)| ≥ 1 and M = max|z|≤4 |Φ(z)|, then there exists a generic constant C > 0 such that ln M ˆ ln 2 sup |Φ(x)|. sup |Φ(x)| ≤ C/|E| x∈I
ˆ x∈E
We are now in the position to prove Lemma 2.1. Proof of Lemma 2.1. We only need to prove this lemma for the case when L = 1. In fact, suppose that this is done. Let E be γ-thick at scale L > 0. Define a new set: L−1 E := L−1 x : x ∈ E . One can easily check that L−1 E is γ-thick at scale 1. Given N > 0 and f ∈ L2 (Rn ) with suppf ⊂ BN , let g(x) := f (Lx), x ∈ Rn . One can directly check that g(ξ) = L−n f(L−1 ξ), ξ ∈ Rn ;
supp g ⊂ BLN .
From these, we can apply Lemma 2.1 (with L = 1) to the set L−1 E and the function g to find C = C(n) so that
1 |g(x)|2 dx ≤ e2C(1+ln γ )(1+LN ) |g(x)|2 dx L−1 E
Rn 1
≤ e2C(1+L)(1+ln γ )(1+N )
|g(x)|2 dx.
(2.1)
L−1 E
Meanwhile, by changing variable x → Lx, we deduce that
Rn
|f (x)|2 dx = L−n
|g(x)|2 dx;
Rn
Hence, from (2.1) and (2.2), we find that
E
|f (x)|2 dx = L−n
L−1 E
|g(x)|2 dx.
(2.2)
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|f (x)| dx ≤ e 2
|f (x)|2 dx,
Cspec (n,E)(1+N )
Rn
E
with 1 Cspec (n, E) = 2C(1 + L) 1 + ln . γ This proves the lemma for the general case that L > 0. We now show Lemma 2.1 for the case when L = 1 by several steps. First of all, we arbitrarily fix N > 0 and f ∈ L2 (Rn ) with suppf ⊂ BN . Without loss of generality, we can assume that f = 0. Step 1. Bad and good cubes. For each multi-index j = (j1 , j2 , · · · , jn ) ∈ Zn , let Q(j) := {x = (x1 , . . . , xn ) ∈ Rn : |xi − ji | < 1/2 for all i = 1, 2, · · · , n} . It is clear that Q(j)
Q(k) = ∅ for all j = k ∈ Zn ; Rn =
Q(j),
j∈Zn
where Q(j) denotes the closure of Q(j). From these, we have that
|f (x)| dx = 2
|f (x)|2 dx.
(2.3)
j∈Zn
Rn
Q(j)
We will divide {Q(j) : j ∈ Zn } into two called “good cubes”
disjoint parts whose elements are respectively
|f |2 with
and “bad cubes”. And then we compare
|f |2 and
Rn
|f |2 , respectively. First,
Q(j) is bad
Q(j) is good
we define the function: h(s) := sn (s − 1)−n − 1, s ∈ [2, +∞). It is a continuous and strictly decreasing function satisfying that h(2) ≥ 1,
lim h(s) = 0.
s→+∞
Thus we can take A0 as the unique point in [2, +∞) so that h(A0 ) = 1/2. Clearly, A0 depends only on n, i.e., A0 = A0 (n). Given j ∈ Zn , Q(j) is said to be a good cube, if for each β ∈ N n ,
|∂xβ f (x)|2
dx ≤
|β| A0 N 2|β|
Q(j)
|f (x)|2 dx.
(2.4)
Q(j)
When Q(j) is not a good cube, it is called as a bad cube. Thus, when Q(j) is a bad cube, there is at least one β ∈ N n , with |β| > 0, so that
|β|
|∂xβ f (x)|2 dx > A0 N 2|β| Q(j)
Q(j)
|f (x)|2 dx.
(2.5)
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Using the Plancherel theorem and the assumption that suppf ⊂ BN (0), we obtain that for each β ∈ N n ,
|∂xβ f (x)|2
2 2 β β dx = ∂x f (ξ) dξ = (iξ) f (ξ) dξ =
Rn
Rn
≤ N 2|β|
Rn
|f(ξ)|2 dξ = N 2|β|
|ξ|≤N
β 2 ξ f (ξ) dξ
|ξ|≤N
|f (x)|2 .
(2.6)
Rn
Meanwhile, it follows by (2.5) that when Q(j) is a bad cube,
|f (x)|2 dx ≤
−|β|
A0
β∈N n ,|β|>0
Q(j)
N −2|β|
|∂xβ f (x)|2 dx.
(2.7)
Q(j)
Since Q(j), j ∈ Zn , are disjoint, by taking the sum in (2.7) for all bad cubes, we find that
|f (x)|2 dx ≤
−|β|
A0
β∈N n ,|β|>0
Q(j) is bad
≤
−|β|
A0
N −2|β|
N −2|β|
β∈N n ,|β|>0
|∂xβ f (x)|2 dx Q(j) is bad
|∂xβ f (x)|2 dx.
(2.8)
Rn
From (2.6) and (2.8), we have that
|f (x)|2 dx ≤
−|β|
β∈N n ,|β|>0
Q(j) is bad
A0
|f (x)|2 dx
Rn
= An0 (A0 − 1)−n − 1
|f (x)|2 dx.
(2.9)
Rn
Since h(A0 ) = 12 , it follows from (2.9) that
|f (x)|2 dx ≤
1 2
|f (x)|2 dx.
(2.10)
|f (x)|2 dx.
(2.11)
Rn
Q(j) is bad
By (2.3) and (2.10), we obtain that
1 |f (x)| dx ≥ 2
2
Rn
Q(j) is good
Step 2. Properties on good cubes. Arbitrarily fix a good cube Q(j). We will prove some properties related to Q(j). First of all, we claim that there is C0 (n) > 0 so that ∂xβ f
L∞ (Q(j))
≤ C0 (n)(1 + N )n
|β| A0 N f L2 (Q(j)) for all β ∈ N n .
In fact, according to (2.4), there is C1 (n) > 0 so that
(2.12)
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⎛
∂xβ f W n,2 (Q(j)) =
⎜ ⎝
μ∈N n ,|μ|≤n
≤
⎞1/2
⎟ |∂xβ+μ f (x)|2 dx⎠
Q(j) |β+μ| 2
N |β+μ| f L2 (Q(j))
A0
μ∈N n ,|μ|≤n
≤ C1 (n)(1 + N )n
|β| A0 N f L2 (Q(j)) for all β ∈ N n .
(2.13)
Meanwhile, because Q(j) satisfies the cone condition, we can apply the Sobolev embedding theorem W n,2 (Q(j)) → L∞ (Q(j)) to find C2 (n) > 0 so that ϕ L∞ (Q(j)) ≤ C2 (n) ϕ W n,2 (Q(j)) for all ϕ ∈ W n,2 (Q(j)). This, along with (2.13), leads to (2.12). Next, we let y ∈ Q(j) satisfy that f L∞ (Q(j)) = |f (y)|.
(2.14)
(Due to the continuity of |f | over Rn , such y exists.) Because the diameter of Q(j) is spherical coordinates centered at y to obtain that |E
χE Q(j) (x)dσ
dr |x−y|=r
0 √
n
=
χE Q(j) (x)dσ
dr |x−y|=r
0
=
n, we can use the
∞ Q(j)| =
√
√
1 n
χE Q(j) (x)dσ.
dr
(2.15)
√ |x−y|= nr
0
In (2.15), we change the variable: x=y+
√
nrw with w ∈ Sn−1 ,
and then obtain that |E
Q(j)| =
√
1 n
√ ( nr)n−1 dr
0
χE Q(j) (y +
√
nrw)dσ
Sn−1
1 ≤n
n/2
dr 0
χE Q(j) (y +
√
nrw)dσ.
(2.16)
Sn−1
For each w ∈ Sn−1 , let √ Iw r ∈ [0, 1] : y + nrw ∈ E Q(j) . Since
(2.17)
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n
|Sn−1 | =
2π 2 , where Γ(·) is the Gamma function, Γ( n2 )
it follows from (1.2) and (2.16) that Γ( n2 ) |E Q(j)| |Iω0 | ≥ n/2 n−1 ≥ γ for some w0 ∈ Sn−1 . n |S | 2(nπ)n/2
(2.18)
Then we define a function φ(·) over [0, 1] by √ f (y + ntw0 ) φ(t) = , t ∈ [0, 1]. f L2 (Q(j))
(2.19)
(Since f ∈ L2 (Rn ) satisfies that supp f ⊂ BN , we have that f is analytic over Rn . Consequently, f L2 (Q(j)) = 0 because we assumed that f = 0 over Rn .) We claim that φ(t) can be extended to an entire function in the complex plane. In fact, by (2.19), one can directly check that 3
(k) n 2 k max|β|=k Dβ f φ (0) ≤ f L2 (Q(j))
L∞ (Q(j))
for all k ≥ 0.
(2.20)
k (k) φ (0) ≤ C0 (n)(1 + N )n n 32 A0 N for all k ≥ 0.
(2.21)
By (2.20) and (2.12), we see that
From (2.21), we find that φ(t) = φ(0) + φ (0)t + · · · +
φ(k) (0) k t + ··· , k!
t ∈ [0, 1],
(2.22)
and that the series in (2.22), with t being replaced by any z ∈ C, is convergent. Thus, the above claim is true. From now on, we will use φ(z) to denote the extension of φ(t) over C. Step 3. Recovery of the L2 (Rn ) norm. We will finish our proof in this step. Applying Lemma 2.2, where ˆ = Iw (defined by (2.17) and (2.18)) and Φ = φ, I = [0, 1], E 0 and then using (2.18), we can find C3 = C3 (n) so that ln M
sup |φ(t)| ≤ (C/|Iw0 |) ln 2 sup |φ(t)| t∈Iw0
t∈[0,1]
! ≤
2C(nπ)n/2 γΓ( n2 )
≤M
C3 1+ln
1 γ
" lnlnM2 sup |φ(t)|
t∈Iw0
sup |φ(t)|,
(2.23)
t∈Iw0
where M = max |φ(z)|. |z|≤4
(2.24)
158
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Two facts are given in order. First, it follows from (2.14) and (2.19) that f L∞ (Q(j)) |f (y)| = = |φ(0)|. f L2 (Q(j)) f L2 (Q(j)) Second, it follows by the definition of Iw0 (see (2.17) and (2.18)) that sup |φ(t)| ≤
t∈Iw0
f L∞ (E Q(j)) . f L2 (Q(j))
The above two facts, along with (2.23), yield that f L∞ (Q(j)) ≤ M
C3 1+ln
1 γ
f L∞ (E Q(j)) .
(2.25)
We next define E x ∈ E Q(j) : |f (x)| ≤
2 |E Q(j)| E
|f (x)| dx .
Q(j)
By the Chebyshev inequality, we have that
|E | ≥
|E
γ Q(j)| ≥ . 2 2
By the same argument as that used in the proof of (2.25), one can obtain that f L∞ (Q(j)) ≤ M
C4 1+ln
1 γ
f L∞ (E Q(j)) for some C4 = C4 (n).
(2.26)
Meanwhile, it follows by the definition of E that f L∞ (E Q(j)) ≤
2 |E Q(j)| E
|f (x)| dx.
(2.27)
|f |2 dx.
(2.28)
Q(j)
From (2.26), (2.27) and the Hölder inequality, we find that
|f |2 dx ≤
4 2C4 M γ
1+ln
1 γ
Q(j)
E
Q(j)
The term M (given by (2.24)) can be estimated by (2.21) as follows: M = max |φ(z)| |z|≤4
≤ max
|z|≤4
∞ φ(k) (0) k=0
k!
|z|k
k 3√ ∞ 4n 2 A0 N ≤ C2 (n)(1 + N ) k! n
k=0
≤e
C5 (1+N )
for some C5 = C5 (n).
(2.29)
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159
Finally, combining (2.28) and (2.29) leads to that
C6 (1+N ) 1+ln
|f |2 dx ≤ e
1 γ
Q(j)
|f |2 dx for some C6 = C6 (n). E
(2.30)
Q(j)
Taking the sum in (2.30) for all good cubes, using (2.11), we see that
|f | dx ≤ 2 2
|f |2 dx
Q(j) is good Q(j)
Rn
≤2
e
C6 (1+N ) 1+ln
1 γ
Q(j) is good
≤ 2e
C6 (1+N ) 1+ln
1 γ
|f |2 dx E
Q(j)
|f |2 dx, E
which leads to (1.10), where 1 Cspec (n, E) = (C6 + 1)(1 + L) 1 + ln , with L = 1. γ This ends the proof of Lemma 2.1.
2
Lemma 2.3 and Lemma 2.4 deal with connections among the spectral inequality (1.10), the Hölder-type interpolation inequality (1.7) and the observability inequality (1.8). In their proofs, we borrowed some ideas and techniques from [2,36]. Lemma 2.3. Suppose that a measurable set E ⊂ Rn satisfies the spectral inequality (1.10). Then E satisfies the Hölder-type interpolation (1.7), with CHold =
1 (Cspec + 1)2 + ln 12. 1−θ
Proof. Let E ⊂ Rn satisfy the spectral inequality (1.10). Arbitrarily fix T > 0, θ ∈ (0, 1) and a solution u to (1.1). Write u0 (x) = u(0, x), x ∈ Rn . Then we have that u(T, x) = eT Δ u0 (x) for all x ∈ Rn . Given N > 0, write respectively χ≤N (D) and χ>N (D) for the multiplier operators with the symbols χ{|ξ|≤N } and χ{|ξ|>N } . Namely, for each g ∈ L2 (Rn ), χ≤N (D)g(ξ) = χ{|ξ|≤N } g(ξ) and χ>N (D)g(ξ) = χ{|ξ|>N } g(ξ) for a.e. ξ ∈ Rn . Then we can express u0 as: u0 = χ≤N (D)u0 + χ>N (D)u0 .
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160
From this and (1.10), we can easily check that
|u(T, x)|2 dx Rn
≤2
T Δ e χ≤N (D)u0 (x)2 dx + 2
Rn
≤ 2eCspec (1+N ) E
≤ 4eCspec (1+N )
T Δ e χ>N (D)u0 (x)2 dx
Rn
T Δ e χ≤N (D)u0 (x)2 dx + 2
T Δ e χ>N (D)u0 (x)2 dx
Rn
T Δ 2 e u0 (x) dx + 2 + 4eCspec (1+N )
T Δ e χ>N (D)u0 (x)2 dx.
(2.31)
Rn
E
Since
T Δ e χ>N (D)u0 (x)2 dx =
Rn
2 −T |ξ|2 e χ>N (ξ)# u0 (ξ) dξ
Rn
≤ e−T N
2
2 u #0 (ξ) dξ
Rn
= e−T N
2
|u0 (x)|2 dx, Rn
it follows from (2.31) that
|u(T, x)|2 dx ≤ 4eCspec (1+N )
Rn
⎛
2 |u(T, x)|2 dx + 2 + 4eCspec (1+N ) e−T N
|u0 (x)|2 dx Rn
E
≤ 6eCspec ⎝eCspec N
|u(T, x)|2 dx + eCspec N −T N
2
⎞
|u0 (x)|2 dx⎠ .
(2.32)
Rn
E
Given ε ∈ (0, 1), choose N = N (ε) so that $ % exp Cspec N − T N 2 = ε. (This can be done since the set: {Cspec s − T s2 : s > 0} contains (−∞, 0].) With the above choice of N , we have that Cspec + N=
& 2 Cspec + 4T ln 1ε 2T
1 ≤ T
' Cspec +
(
1 T ln ε
) .
Thus, with θ ∈ (0, 1) fixed before, we see that + * + ( 2 Cspec Cspec 1 exp √ exp [Cspec N ] ≤ exp ln T ε T + * + * 2 2 Cspec 1−θ 1 θ Cspec exp ln + ≤ exp T θ ε 1−θ T *
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
*
2 Cspec = exp (1 − θ)T
161
+ ε−
1−θ θ
.
From this and (2.32), we find that for every ε ∈ (0, 1), ⎛
2 Cspec
|u(T, x)|2 dx ≤ 6eCspec ⎝e (1−θ)T ε
− 1−θ θ
Rn
⎞
|u0 (x)|2 dx⎠ .
|u(T, x)|2 dx + ε Rn
E
Choosing in the above "θ ! |u(T, x)|2 dx E ε= , |u0 (x)|2 dx Rn we obtain that
⎛
⎞θ ⎛
|u(T, x)|2 dx ≤ 12eCspec e (1−θ)T ⎝
|u(T, x)|2 dx⎠ ⎝
2 Cspec
Rn
E
≤e
2 1 1−θ (Cspec +1) +ln
12 1+ T1
⎞1−θ |u0 (x)|2 dx⎠
Rn
⎛
⎝
⎞θ ⎛
|u(T, x)|2 dx⎠ ⎝
⎞1−θ |u0 (x)|2 dx⎠
,
Rn
E
which leads to (1.7) with CHold = This ends the proof of Lemma 2.3.
1 2 (Cspec + 1) + ln 12. 1−θ
2
Lemma 2.4. Suppose that a measurable set E ⊂ Rn has the property: there is a positive constant CHold = CHold (n, E) so that for any T > 0,
|u(T, x)|2 dx ≤ eCHold
1+ T1
Rn
|u(T, x)|2 dx
1/2
|u(0, x)|2 dx
1/2 ,
(2.33)
Rn
E
when u solves the equation (1.1). Then for each T > 0 and each subset F ⊂ (0, T ) of positive measure, there is a positive constant Cobs = Cobs (n, T, F, CHold ) so that when u solves (1.1),
|u(T, x)| dx ≤ Cobs
|u(s, x)|2 dxds.
2
Rn
(2.34)
F E
In particular, if F = (0, T ) then the constant Cobs in (2.34) can be expressed as: Cobs = exp [36(1 + 3CHold )(1 + 1/T )] . Proof. Suppose that E ⊂ Rn satisfies (2.33). Arbitrarily fix T > 0 and F ⊂ (0, T ) of positive measure. Applying Cauchy’s inequality to (2.33), we find that for all t > 0 and ε > 0,
1 1 |u(t, x)| dx ≤ e2CHold 1+ t ε
Rn
|u(t, x)| dx + ε
2
2
E
Rn
|u0 (x)|2 dx.
(2.35)
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By a translation in time, we find from (2.35) that for all 0 < t1 < t2 and ε > 0,
1 2CHold e ε
|u(t2 , x)|2 dx ≤
1+ t
1 2 −t1
|u(t2 , x)|2 dx + ε
Rn
|u(t1 , x)|2 dx.
(2.36)
Rn
E
Let l be a Lebesgue density point of F . Then according to [36, Proposition 2.1], for each λ ∈ (0, 1), there + is a sequence {lm }∞ l=1 ⊂ (l, T ) so that for each m ∈ N , lm+1 − l = λm (l1 − l)
(2.37)
1 F (lm+1 , lm ) ≥ (lm − lm+1 ). 3
(2.38)
and
Arbitrarily fix m ∈ N + . Take s so that 0 < lm+2 < lm+1 ≤ s < lm < T. Using (2.36) (with t1 = lm+2 and t2 = s) and noting that
2
|u (lm , x)| dx ≤ Rn
|u(s, x)|2 dx and lm+1 − lm+2 ≤ s − lm+2 , Rn
we see that
1 2C |u (lm , x)| dx ≤ e Hold ε 2
1 m+1 −lm+2
1+ l
2
|u(s, x)| dx + ε
|u (lm+2 , x)| dx.
2
Rn
(2.39)
Rn
E
Integrating with s over F (lm+1 , lm ) in (2.39) implies that
2
2
|u (lm , x)| dx ≤ ε Rn
1 1 2C e Hold + ε |F (lm+1 , lm )|
|u (lm+2 , x)| dx
Rn
1 m+1 −lm+2
1+ l
|u(s, x)|2 dxds.
F
(2.40)
(lm+1 ,lm ) E
Since it follows by (2.38) that 1 1 1 − (lm+1 , lm ) ≥ (lm − lm+1 ) ≥ e lm −lm+1 , F 3 3 we obtain from (2.40) that
2
|u (lm , x)| dx ≤ ε Rn
2
|u (lm+2 , x)| dx
Rn
+
3 lm −l1m+1 +2CHold e ε
(2.41) 1 m+1 −lm+2
|u(s, x)|2 dxds.
F
Meanwhile, it follows by (2.37) that
1+ l
(lm+1 ,lm ) E
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lm − lm+1 =
163
1 (lm − lm+2 ) 1+λ
(2.42)
λ (lm − lm+2 ) . 1+λ
(2.43)
and lm+1 − lm+2 = Inserting (2.42) and (2.43) into (2.41), we find that
2
2
|u (lm , x)| dx ≤ ε Rn
|u (lm+2 , x)| dx
Rn
C 1 + 3e2CHold e lm −lm+2 ε
|u(s, x)|2 dxds
F
(2.44)
(lm+1 ,lm ) E
with C = 1 + λ +
2CHold (1 + λ) . λ
(2.45)
Rewrite (2.44) as εe
C − lm −l
2
Rn
C 2 − lm −lm+2
|u (lm , x)| dx − ε e
m+2
2
|u (lm+2 , x)| dx Rn
|u(s, x)|2 dxds.
≤ 3e2CHold F
(2.46)
(lm+1 ,lm ) E
√ Next, we fix λ ∈ (1/ 2, 1). Let μ :=
1 2−λ−2 .
Then μ > 1. Setting, in (2.46),
, (μ − 1)C , ε = exp − lm − lm+2 we have that e
μC − lm −l
2
|u (lm , x)| dx − e
m+2
Rn
− l(2μ−1)C m −l
2
|u (lm+2 , x)| dx
m+2
Rn
|u(s, x)|2 dxds.
≤ 3e2CHold F
(2.47)
(lm+1 ,lm ) E
Meanwhile, one can easily check that , , μC (2μ − 1)C = exp − 2 . exp − lm − lm+2 λ (lm − lm+2 ) Because lm+2 − lm+4 = λ2 (lm − lm+2 ) , we deduce from (2.47) and (2.48) that
(2.48)
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
164
e
μC − lm −l
2
|u (lm , x)| dx − e
m+2
Rn
2
|u (lm+2 , x)| dx Rn
≤ 3e2CHold
μC m+2 −lm+4
−l
|u(s, x)|2 dxds. F
(lm+1 ,lm ) E
Summing the above inequality for all odd m derives that
e
− l μC −l 1
2
|u (l1 , x)| dx ≤ 3e
3
2CHold
m=1 F
Rn
∞
|u(s, x)|2 dxds
(lm+1 ,lm ) E
|u(s, x)|2 dxds
≤ 3e2CHold F
(l,l1 ) E
≤ 3e2CHold
|u(s, x)|2 dxds. F E
Thus, we have that
|u(T, x)| dx ≤
|u(l1 , x)| dx ≤ 3e
2
Rn
2
2CHold
e
μC l1 −l3
Rn
|u(s, x)|2 dxds,
(2.49)
F E
which leads to (2.34) with Cobs
, μC . = 3 exp 2CHold + l 1 − l3
Finally, in the case when F = (0, T ), we set T 2T , l= l1 = 3 3
( and λ =
2 . 3
Then we have that (see (2.45)) l 1 − l3 =
T , μ = 2 and C ≤ 2 + 6CHold . 9
Now, we derive from (2.49) that
|u(T, x)| dx ≤ 3e 2
2CHold
e
36 1+3CHold T
T
|u(s, x)|2 dxds 0 E
Rn 36(1+3CHold ) 1+ T1
T
≤e
|u(s, x)|2 dxds. 0 E
This completes the proof of Lemma 2.4. 2 The next lemma seems to be new to our best knowledge. The key of its proof is the structure of a special solution to the equation (1.1). This structure is based on the heat kernel.
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Lemma 2.5. Suppose that a measurable set E ⊂ Rn satisfies the observability inequality (1.3). Then the set E is γ-thick at scale L for some γ > 0 and L > 0. Proof. Let E ⊂ Rn be a measurable set satisfying the observability inequality (1.3). Recall that the heat kernel is as: K(t, x) = (4πt)−n/2 e−|x|
2
/4t
, t > 0, x ∈ Rn .
Given u0 ∈ L2 (Rn ), the function defined by
(t, x) −→
K(t, x − y)u0 (y) dy, (t, x) ∈ (0, ∞) × Rn ,
(2.50)
Rn
is a solution to the equation (1.1) with the initial condition u(0, x) = u0 (x), x ∈ Rn . Arbitrarily fix x0 ∈ Rn . By taking u0 (x) = (4π)−n/2 e−|x−x0 |
2
/4
, x ∈ Rn ,
in (2.50), we get the following solution to the equation (1.1): v(t, x) = (4π(t + 1))− 2 e− n
|x−x0 |2 4(t+1)
,
t ≥ 0, x ∈ Rn .
(2.51)
From (2.51), we obtain by direct computations that
|v(1, x)|2 dx = 4−n π − 2 . n
(2.52)
Rn
From (2.51), we also find that for an arbitrarily fixed L > 0, L2
v(t, x) ≤ (4π)− 2 e− 16 e− n
|x−x0 |2 16
for all 0 ≤ t ≤ 1 and x ∈ Rn , with |x − x0 | ≥ L.
(2.53)
By (2.53), the above solution v satisfies that
1
|v(t, x)|2 dx dt ≤ (2π)− 2 e− n
L2 8
.
(2.54)
0 |x−x0 |≥L
Meanwhile, by taking T = 1 and u = v in the observability inequality (1.3), we see that
1
|v(1, x)| dx ≤ C
|v(t, x)|2 dx dt.
2
(2.55)
0 E
Rn
Here and in what follows, C stands for the constant Cobs (n, 1, E) in (1.3). Now, it follows from (2.52), (2.55) and (2.54) that
4
−n − n 2
π
1
≤C 0 E
BL (x0 )
|v(t, x)|2 dx dt + C(2π)− 2 e− n
L2 8
.
(2.56)
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166
Choose L > 0 in such a way that C(2π)− 2 e− n
L2 8
≤
1 −n − n 4 π 2. 2
Then by (2.56) and (2.51), we obtain that 1 −n − n 4 π 2 ≤C 2
1
|v(t, x)|2 dx dt
0 E
1
BL (x0 )
(4π(t + 1))−n e−
=C 0 E
1
(4π)−n dx dt ≤ C(4π)−n |E
≤C
dx dt
BL (x0 )
0 E
|x−x0 |2 2(t+1)
BL (x0 )|,
BL (x0 )
from which, it follows that |E
BL (x0 )| ≥ (2C)−1 π 2 . n
(2.57)
Since BL (x0 ) ⊂ (x0 + 2LQ), we see from (2.57) that n E (x0 + 2LQ) ≥ (2C)−1 π 2 . From this, as well as the choice of L, we can find L > 0 and γ > 0, which are independent of the choice of x0 , so that E (x0 + L Q) ≥ γ(L )n .
(2.58)
Notice that x0 in (2.58) was arbitrarily taken from Rn . Hence, the set E is γ-thick at scale L . This ends the proof of Lemma 2.5. 2 We now on the position to prove Theorem 1.1. Proof of Theorem 1.1. We can prove it in the following way: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). Indeed, the conclusions (i) ⇒ (ii), (ii) ⇒ (iii), (iii) ⇒ (iv) and (iv) ⇒ (i) follow respectively from Lemma 2.1, Lemma 2.3, Lemma 2.4 and Lemma 2.5. This ends the proof of Theorem 1.1. 2 Tracking the constants in Lemma 2.1, Lemma 2.3 and Lemma 2.4, we can easily get the following consequences of Theorem 1.1: Corollary 2.1. Let E ⊂ Rn be a set of γ-thick at scale L for some γ > 0 and L > 0. Then the following conclusions are true for a constant C = C(n) > 0:
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
167
(a) The set E satisfies the Hölder-type interpolation (1.7) with CHold (n, E, θ) =
C 1 2 (1 + L)2 1 + ln . 1−θ γ
(b) The set E satisfies the observability inequality (1.3) with 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
2
1 γ
1+ T1
.
Remark 2.3. Let E be γ-thick at scale L. From (b) of Corollary 2.1, we can take the observable constant as: 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
1 γ
2
1+ T1
,
which depends only on γ, when L and T and n are given. Meanwhile, we define the average measure of E as: γ0 γ0 (E) infn
|E
x∈R
(x + LQ)| . Ln
(2.59)
One can easily check that E is also γ0 -thick at scale L. Then by making use of (b) of Corollary 2.1 again, we can take the observable constant as: 300(1+C)(1+L)2 1+ln
Cobs (n, E, T ) = e
1 γ0
2
1+ T1
.
(2.60)
From (2.60) and (2.59), we see that when L, T and n are given, Cobs (n, E, T ) can be taken as a constant depending only on the average measure of E, but independent of the shape and the position of E. In particular, if E1 , E2 ⊂ Rn satisfy that |Ei
(x + LQ)| = γLn for all x ∈ Rn with γ ∈ (0, 1], i = 1, 2,
then Cobs (n, E1 , T ) and Cobs (n, E2 , T ) can be taken as the same. 3. Weak interpolation and observability inequalities In this section, we introduce several weak observability inequalities and interpolation inequalities, where observations are made over a ball in Rn . On the one hand, these inequalities can be viewed as extensions of (1.3) and (1.7) in some senses, while on the other hand, they are independently interesting. 3.1. Weak interpolation inequalities with observation on the unit ball We begin with introducing two spaces. Given a > 0 and ν > 0, we set ν
L2 (ea|x| dx) := {f : Rn → R : f is measurable and f L2 (ea|x|ν
dx)
< +∞},
equipped with the norm: ⎛ f L2 (ea|x|ν
dx)
:= ⎝
Rn
⎞1/2 |f (x)|2 ea|x| dx⎠ ν
ν
, f ∈ L2 (ea|x| dx).
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168
Given ν > 0, we set L2 (xν dx) := {f : Rn → R : f is measurable and f L2 ( x ν dx) < +∞}, equipped with the norm: ⎛ f L2 ( x ν dx) := ⎝
⎞1/2
|f (x)|2 xν dx⎠
, f ∈ L2 (xν dx).
Rn
Notice that any function in one of the above spaces decays along the radical direction. In this subsection, we will build up some interpolation inequalities for solutions to (1.1), with initial data ν in L2 (ea|x| dx) (or L2 (xν dx)). In these inequalities, observations are made over the unit ball in Rn and at one time point. The purpose to study such observability has been explained in Subsection 1.3. Our main results about this subject are included in the following theorem: Theorem 3.1. (i) There is θ = θ(n) ∈ (0, 1) and C = C (n) such that for any ε > 0, T > 0 and a > 0,
4| ln θ| − |u(T, x)|2 dx ≤ C1 (a, T ) ε |u0 (x)|2 ea|x| dx + ε−1 eε a |u(T, x)|2 dx ,
Rn
Rn
B1
when u solves (1.1) with the initial condition u(0, ·) = u0 (·) ∈ L2 (ea|x| dx). Here, C1 (a, T ) = eC
1+ T1 +a+a2 T
(
1 + a−n Γ
a . 2| ln θ|
(ii) There is θ = θ(n) ∈ (0, 1) and C = C (n) so that for any ε ∈ (0, 1), T > 0 and ν ∈ (0, 1],
1
(3| ln θ|+1) 1 ν ε 2 ν e |u(T, x)| dx ≤ C2 (ν, T ) ε |u0 (x)| x dx + e |u(T, x)|2 dx ,
2
Rn
Rn
B1
when u solves (1.1) with the initial condition u(0, ·) = u0 (·) ∈ L2 (xν dx). Here, ν
C2 (ν, T ) = (1 + T 2 )eC
1+ T1
.
Remark 3.1. (a) The condition that ν ≤ 1 in (ii) of Theorem 3.1 is not necessary. We make this assumption only for the brevity of the statement of the theorem. Indeed, from the definition of L2 (xν dx), we see that L2 (xν dx) → L2 (x dx) for any ν ≥ 1. From this and (ii) of Theorem 3.1, one can easily check that when ν > 1, any solution of (1.1) satisfies that
1
(3| ln θ|+1) 1 1 ε |u(T, x)|2 dx ≤ 1 + T 2 eC (n) 1+ T ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx .
Rn
Rn
B1
(b) [11, Theorem 1] contains the following result: There is a universal constant C > 0 so that for each T > 0 and R > 0, sup 0≤t≤T
e
2 t|x| 4 t2 +R2
u(t)
L2 (Rn )
≤ C u(0) L2 (Rn ) + e
2 T |x| 4 T 2 +R2
u(T )
L2 (Rn )
,
(3.1)
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when u solves (1.1). The first inequality in Theorem 3.1 is comparable to the above inequality (3.1). By our understanding, these two inequalities can be viewed as different versions of Hardy uncertainty principle. On one hand, the inequality (3.1) can be understood as follows: From some information on a solution to (1.1) at infinity in Rn at two time points 0 and T , one can know the behaviour of this solution at infinity in Rn at each time t ∈ [0, T ]. On the other hand, the first inequality in Theorem 3.1 can be explained in the following way: From some information on a solution to (1.1) at infinity in Rn at time 0, and in the ball B1 in Rn at time T , one can know the behaviour of this solution at infinity in Rn at time T . Similarly, we can compare the second inequality in Theorem 3.1 with (3.1). It deserves to mention that we can only prove inequalities in Theorem 3.1 for the pure heat equation (1.1), while [11, Theorem 1] gave the inequality (3.1) for heat equations with general potentials. (c) The first inequality in Theorem 3.1 can also be understood as follows: If we know in advance that the initial datum of a solution to (1.1) is in the unit ball of L2 (ea|x| dx), then by observing this solution in the unit ball of Rn at time T , we can approximately recover this solution over Rn at the same time T , with the error C1 (a, T )ε. The second inequality in Theorem 3.1 can be explained in a very similar way. Before proving Theorem 3.1, we give a consequence of its conclusion (ii). This consequence can be viewed as an extension of the spectral inequality (1.10) when the γ-thickness set E is replaced by a unit ball. Corollary 3.1. There are two constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for each ε ∈ (0, 1) and each f ∈ L2 ((1 + |x|2 ) dx), with supp f ⊂ BN ,
⎛ |f (x)|2 dx ≤ C ⎝ε(2 + 8N 2 )e2N
2
Rn
where cε = ee
|f (x)|2 (1 + |x|2 ) dx + cε
Rn
(3| ln θ|+1) 1 ε
⎞ |f (x)|2 dx⎠ ,
(3.2)
B1
.
Proof. First of all, according to (ii) of Theorem 3.1 and (a) of Remark 3.1 (with T = 1 and ν = 2), there exists θ = θ(n) ∈ (0, 1) and C = C(n) so that for each ε ∈ (0, 1) and each u0 ∈ L2 ((1 + |x|2 ) dx), ⎛
|eΔ u0 (x)|2 dx ≤ C ⎝ε
Rn
|u0 (x)|2 (1 + |x|2 ) dx + cε
Rn
⎞ |eΔ u0 (x)|2 dx⎠ ,
(3.3)
B1
(3| ln θ|+1) 1
ε where cε = ee . Next we arbitrarily fix ε ∈ (0, 1) and f ∈ L2 ((1 + |x|2 ) dx), with supp f ⊂ BN for some N > 0. Define a function h in the following manner:
2 h(ξ) = e|ξ| f(ξ) for a.e. ξ ∈ Rn .
(3.4)
Then we have two facts: First, f = eΔ h; Second, by the Plancherel theorem and (3.4), using the fact that supp f ⊂ BN , after some computations, we obtain that
|h(x)|2 (1 + |x|2 ) dx = Rn
| h(ξ)|2 + |∇ξ h(ξ)|2 dξ
Rn
≤ (2 + 8N 2 )e2N
2
Rn
|f(ξ)|2 + |∇ξ f(ξ)|2 dξ
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170
2N 2
2
|f (x)|2 (1 + |x|2 ) dx < ∞.
= (2 + 8N )e
Rn
From which, it follows that h ∈ L2 ((1 + |x|2 ) dx). Finally, by the above two facts, we can use (3.3), with u0 = h, to get (3.2). This ends the proof of Corollary 3.1. 2 To show Theorem 3.1, we need some preliminaries. We begin with some auxiliary lemmas on the persisν tence of the heat semigroup in the spaces L2 (ea|x| dx) and L2 (xν dx). ν
Lemma 3.1. Let a > 0 and 0 < ν ≤ 1. Then when u0 ∈ L2 (ea|x| dx), etΔ u0
n
L2 (ea|x|ν
dx)
≤ 2 2 ea
2 2−ν
ν
u0 L2 (ea|x|ν
t 2−ν
for all t > 0.
dx)
ν
Proof. Arbitrarily fix a > 0, 0 < ν ≤ 1 and u0 ∈ L2 (ea|x| dx). Using the fundamental solution of (1.1) and ν the definition of L2 (ea|x| dx), we have that etΔ u0
L2 (ea|x|ν
=
dx)
a|x|ν 2
e
(4πt)−n/2
Rn
e−
|x−y|2 4t
2 u0 (y) dy
dx
12
.
(3.5)
Rn
Since |x|ν ≤ (|x − y| + |y|)ν ≤ |x − y|ν + |y|ν for all x, y ∈ Rn , (Here, we used the elementary inequality: (τ + s)ν ≤ τ ν + sν , τ, s > 0.) it follows from (3.5) that e
tΔ
u0
L2 (ea|x|ν
dx)
≤
−n/2
e−
(4πt) Rn
≤
|x−y|2 4t
Rn
(4πt)−n/2 e−
|x|2 4t
ν
+ a|x| 2
dx ·
Rn
=
ν
+ a|x−y| 2
e
e
a|y|ν 2
a|y|ν 2
|u0 (y)| dy
2
2 u0 (y)
dy
dx
12
12
Rn
(4πt)−n/2 e 4t (−|x| 1
2
ν
+2ta|x| )
dx · u0 L2 (ea|x|ν
dx) .
(3.6)
Rn
Meanwhile, by the Young inequality: 2ta|x| ≤ ν
|x|2 2 ν
2
+
(2ta) 2−ν 2 2−ν
≤
2 1 2 |x| + (2ta) 2−ν , 2
we obtain that
−n/2
(4πt) Rn
e
2 ν 1 4t (−|x| +2ta|x| )
dx ≤
e
2 (2ta) 2−ν 4t
(4πt)−n/2 e−
|x|2 8t
dx
Rn n 2
=2 e
2 (2ta) 2−ν 4t
n
≤ 2 2 ea
2 2−ν
ν
t 2−ν
.
Now, the desired inequality follows from (3.6) and (3.7). This ends the proof of Lemma 3.1.
(3.7) 2
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Remark 3.2. The inequality in Lemma 3.1 does not hold for the case when ν > 1. Indeed, given ν > 1, ν ν 1 let u0 (x) = e− 2 |x| x−n , x ∈ Rn . It is clear that u0 ∈ L2 (e|x| dx). However, we have that for any t > 0, ν etΔ u0 ∈ / L2 (e|x| dx). This can be proved as follows: Arbitrarily fix t > 0. By some direct calculations, we find that when |x| ≥ 2,
1 1 1 ν etΔ u0 (x) ≥ C(4πt)−n/2 e− 4t e− 2 |x|− 2 |x| − 1/2−n for some C = C(n).
This leads to that etΔ u0
≥ C(4πt)−n/2 e− 4t 1
L2 (e|x|ν dx)
e|x|
ν
− |x|− 12 ν
|x| − 1/2−2n dx
1/2 .
(3.8)
|x|≥2
Meanwhile, one can easily find a constant M > 2 so that ν
|x|ν − (|x| − 1/2) ≥
ν ν−1 |x| , when |x| ≥ M. 4
From this and (3.8), we obtain that etΔ u0
≥ C(4πt)−n/2 e− 4t 1
L2 (e|x|ν dx)
e 4 |x| ν
ν−1
|x| − 1/2−2n dx
1/2
= ∞.
|x|≥M
Lemma 3.2. Let ν ≥ 0. Then for any u0 ∈ L2 (xν dx), ν etΔ u0 L2 ( x ν dx) ≤ 4ν+2 Γ(ν/2 + n) 1 + t 4 u0 L2 ( x ν dx) for all t > 0. Proof. The proof is similar to that of Lemma 3.1. (See also [41, Lemma B.6.1].)
2
s Lemma 3.3. Given s > 0, there is C = C(n, s) so that when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ) for some a > 0,
Dα f L∞ (Rn ) ≤ C |α|+1 a−
2|α|+3n 2s
1 (α!) s f(ξ)
L2 (ea|ξ|s dξ)
for each α ∈ N n .
(Here, we adopt the convention that α! = α1 !α2 ! · · · αn !.) s Remark 3.3. From Lemma 3.3, we see that if f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ), with s > 0 and a > 0, then f is analytic, when s = 1, while f is ultra-analytic, when s > 1. s Proof of Lemma 3.3. Arbitrarily fix s > 0, a > 0 and f ∈ L2 (Rn ), with f ∈ L2 (ea|ξ| dξ). Then arbitrarily fix α = (α1 , . . . , αn ) ∈ N n , β = (β1 , . . . , βn ) ∈ N n and γ = (γ1 , . . . , γn ) ∈ N n , with |γ| ≤ n. Several facts are given in order.
Fact One: By direct computations, we see that
Rn
2β −a|ξ|s ξ e dξ ≤
n . 2β −a(Σn |ξ |s /n) ξ e i=1 i dξ =
|ξi |
2βi
e−a|ξi |
s
/n
dξi
i=1 Rξi
Rn
∞ n n n 2βis+1 ∞ 2βi +1 . . 2βi −ar s /n = 2 r e dr = 2 t s −1 e−t dt a i=1 i=1 0
n 2β + 1 n 2|β|+n . s i Γ = 2n . a s i=1
0
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172
From this, we obtain that ⎛ ⎝
⎞1/2
|ξ
2(α+γ)
|e
−a|ξ|s
dξ ⎠
≤2
n/2
Rn
n ( n 2|α|+3n . 2αi + 2γi + 1 2s . Γ a s i=1
(3.9)
Fact Two: By the Sobolev embedding H n (Rn ) → L∞ (Rn ), we can find C1 (n) > 0 so that Dα f L∞ (Rn ) ≤ C1 (n)
Dα+γ f
L2 (Rn )
.
(3.10)
γ∈N n ,|γ|≤n
Fact Three: By the Plancheral theorem and the Hölder inequality, we obtain that ⎛ Dα+γ f
L2 (Rn )
= ξ α+γ f(ξ)
L2 (Rn )
≤ f(ξ)ea|ξ|
s
⎝
/2
L2 (Rn )
⎞1/2 |ξ 2(α+γ) |e−a|ξ| dξ ⎠ s
.
(3.11)
Rn
Fact Four: There exists C2 = C2 (n, s) so that ( 1 2αi + 2γi + 1 Γ ≤ C2αi +1 Γ1/s (αi ) = C2αi +1 (αi !) s . s
(3.12)
The proof of (3.12) is as follows: From the Stirling formula, we have that lim √
x→+∞
Γ(x) 1 = 1. 2πe−x xx+ 2
(3.13)
From (3.13), we can find constants M1 = M1 (s) and C3 = C3 (n, s) so that for all αi > M1 , i +1 + 1 2α + 2γ + 1 2α + 2γ + 1 2αi +2γ √ 2αi +2γi +1 s 2 i i i i s ≤ 2 2πe− s s 2α + 2γ + 1 2γis+1 + 12 2α + 2γ + 1 2αs i √ 2αi +2γi +1 i i i i s = 2 2π · e− · s s 2α + 2γ + 1 2αs i √ 2γi +1 i i −x + 12 s ≤ 2 2π sup e x · s x>0 2γ +1 2α √ 2γ +1 1 is + 12 2αi + 2n + 1 s i −( is + 12 ) 2γi + 1 + = 2 2π · e · s 2 s
Γ
2αi
≤ C3αi αi s .
(3.14)
From (3.13), we can also find an absolute constant M2 ≥ 1 so that for all αi > M2 , √ 1 Γ(αi ) ≥ 2−1 2πe−αi αi αi + 2 .
(3.15)
According to (3.14) and (3.15), there is a constant C4 (n, s) so that & i +1 Γ 2αi +2γ s Γ1/s (αi )
≤ [C4 (n, s)]
αi
for all αi > M := max{M1 , M2 }.
Meanwhile, it is clear that there is a constant C5 (n, s) so that
(3.16)
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& i +1 Γ 2αi +2γ s Γ1/s (αi )
173
≤ C5 (n, s) for all αi ≤ M.
(3.17)
(Here we agree that Γ(0) = ∞.) Combining (3.16) and (3.17) leads to (3.12). Inserting (3.12) into (3.9), noticing that |γ| ≤ n, we find that for some C6 = C6 (n, s), ⎞1/2
n n 2|α|+3n . 1 s 2s ⎝ ξ 2(α+γ) e−a|ξ| dξ ⎠ ≤ 2n/2 C2αi +1 (αi !) s a i=1 ⎛
Rn
|α|+1 − 2|α|+3n 2s
≤ C6
a
1
(α!) s .
(3.18)
Finally, it follows from (3.10), (3.11) and (3.18) that for some C = C(n, s),
Dα f L∞ (Rn ) ≤ C1 (n)
s f(ξ)ea|ξ| /2
γ∈N n ,|γ|≤n
≤ C |α|+1 a−
2|α|+3n 2s
(α!) s f(ξ)
|α|+1
L2 (Rn )
C6
1
L2 (ea|ξ|s dξ)
1
(α!) s
.
2
This ends the proof of Lemma 3.3.
The next corollary is a consequence of Lemma 3.3. 2 Corollary 3.2. There is C = C(n) > 0 so that when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ) for some a > 0,
2
Dα f L∞ (Rn ) ≤ eC(1+b
1 |α|! ) 1+ a b|α|
f(ξ)
L2 (ea|ξ|2 dξ)
for all b > 0 and α ∈ N n .
Remark 3.4. Let u0 ∈ L2 (Rn ) be arbitrarily given. Set u(t, x) = etΔ u0 (x), (t, x) ∈ (0, ∞) × Rn . Then u is the solution of (1.1) with u(0, ·) = u0 (·). Arbitrarily fix t > 0. By applying Corollary 3.2 (where f (·) = u(t, ·) and a = 2t), we see that the radius of analyticity of u(t, ·) (which is treated as a function of x) is independent of t. It is an analogy result for solutions of the heat equation in a bounded domain with an analytic boundary (see [2,12]). This property plays a very important role in the proof of the observability estimates from measurable sets when using the telescoping series method developed in [2,12]. Proof of Corollary 3.2. Arbitrarily fix a > 0 and f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ). Then arbitrarily fix b > 0 and α ∈ N n . According to Lemma 3.3 (with s = 2), there is C = C (n) so that 2
Dα f L∞ (Rn ) ≤ C
|α|+1 − 2|α|+3n 4
≤ g(|α|)
a
|α|! f (ξ) b|α|
1 (α!) 2 f(ξ)
L2 (ea|ξ|2 dξ)
L2 (ea|ξ|2 dξ)
,
(3.19)
where r −1/2 g(r) = C a−3n/4 bC a−1/2 (r!) , r ∈ N. To estimate g(r) pointwisely, we use (3.15) to find that when r > M2 ≥ 1 (where M2 is given by (3.15)),
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r −1/2 √ 1 g(r) ≤ C a−3n/4 bC a−1/2 2−1 2πe−r rr+ 2 r ≤ 21/4 π −1/4 C a−3n/4 be1/2 C a−1/2 r−r/2 ≤ 21/4 π −1/4 C a−3n/4 sup (be1/2 C a−1/2 )r r−r/2 r>0
= 21/4 π −1/4 C a−3n/4 e
b2 C 2 2a
.
(3.20)
Meanwhile, it is clear that when r ≤ M2 , M2 g(r) ≤ C a−3n/4 bC a−1/2 + 1 .
(3.21)
From (3.20) and (3.21), it follows that g(r) ≤ eC
1 1+b2 1+ a
for all r = 0, 1, 2, . . . ,
which, together with (3.19), yields the desired inequality. This ends the proof of Corollary 3.2. 2 To prove Theorem 3.1, we also need the decomposition: Rn =
Ωj , with Ωj := {x ∈ Rn : j − 1 ≤ |x| < j}.
(3.22)
j≥1
The next lemma concerns with the propagation of smallness for some real-analytic functions with respect to the above decomposition of Rn . Lemma 3.4. There are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0, j ≥ 1 and 2 f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ),
1
|f |2 dx ≤ j (n−1)(1−θ) eC(1+ a ) Ωj+1
|f |2 dx
θ
|f|2 ea|ξ| dξ 2
1−θ .
Rn
Ωj
The proof of Lemma 3.4 needs Corollary 3.2 and the next lemma which is quoted from [1] (see also [2, Theorem 4]), but is originally from [42]. Lemma 3.5 ([1, Theorem 1.3]). Let R > 0 and let f : B2R ⊂ Rn → R be real analytic in B2R verifying |Dα f (x)| ≤ M (ρR)−|α| |α|!, when x ∈ B2R and α ∈ N n for some positive numbers M and ρ ∈ (0, 1]. Let ω ⊂ BR be a subset of positive measure. Then there are two constants C = C(ρ, |ω|/|BR |) > 0 and θ = θ(ρ, |ω|/|BR |) ∈ (0, 1) so that f L∞ (BR ) ≤ CM 1−θ
θ 1
|f (x)| dx . |ω| ω
We are now in the position to show Lemma 3.4. Proof of Lemma 3.4. Let a > 0 and j ≥ 1 be arbitrarily given. Arbitrarily fix f ∈ L2 (Rn ) with f ∈ 2 L2 (ea|ξ| dξ). The rest proof is divided into the following several steps.
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Step 1. The decompositions of Ωj and Ωj+1 in the polar coordinates. In the polar coordinate system, we have that /
Ωj = (r, ϑ1 , · · · , ϑn−1 ) ∈ [j − 1, j) × [0, 2π)n−1 , Ωj+1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ [j, j + 1) × [0, 2π)n−1 .
(3.23)
When j is large, the distance between two points in Ωj can be very large. This makes our studies on the propagation from Ωj to Ωj+1 harder. To pass this barrier, we need to build up a suitable refinement for each Ωj . We set [0, 2π] =
,
Δl , with Δl :=
1≤l≤j
" l l−1 2π, 2π . j j
(3.24)
Given (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, · · · , n − 1), we set /
Ωj;k1 ,··· ,kn−1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ Ωj : ϑ1 ∈ Δk1 , · · · , ϑn−1 ∈ Δkn−1 , Ωj+1;k1 ,··· ,kn−1 = (r, ϑ1 , · · · , ϑn−1 ) ∈ Ωj+1 : ϑ1 ∈ Δk1 , · · · , ϑn−1 ∈ Δkn−1 .
(3.25)
Then one can easily check that for each ˆj ∈ {j, j + 1}, Ωˆj;k1 ,··· ,kn−1
Ωˆj;k1 ,··· ,k
n−1
= ∅, if (k1 , · · · , kn−1 ) = (k1 , · · · , kn−1 ),
(3.26)
and Ωˆj =
Ωˆj;k1 ,··· ,kn−1 ,
(3.27)
where the union is taken over all different (k1 , . . . , kn−1 ) ∈ N n−1 , with 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. In what follows, we write d Ωj;k1 ,··· ,kn−1 for the diameter of Ωj;k1 ,··· ,kn−1 . Step 2. To prove the following three properties: (O1) There are constants c1 = c1 (n) and c2 = c2 (n) so that for any (k1 , . . . , kn−1 ) ∈ N n−1 , with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), c1 Vn ≤ Ωj;k1 ,··· ,kn−1 ≤ c2 Vn .
(3.28)
(O2) We have that for any (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), d Ωj;k1 ,··· ,kn−1 = d (Ωj;1,··· ,1 ) :=
sup
x,x ∈Ωj;1,··· ,1
|x − x |.
(3.29)
(O3) We have that for any fix (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j (i = 1, . . . , n − 1), d Ωj;k1 ,··· ,kn−1 ≤ 2π
0
3
i2 ≤ 2πn 2 ;
(3.30)
1≤i≤n
j+1 d Ωj+1;k1 ,··· ,kn−1 ≤ 2π j
0 1≤i≤n
3
i2 ≤ 4πn 2 .
(3.31)
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To see (O1), we use the definitions of Ωj;k1 ,··· ,kn−1 and Ωj to find that Ωj;k
1 ,···
,kn−1
=
1 j n−1
|Ωj | = Vn
j n − (j − 1)n , j n−1
which leads to (3.28). The conclusion (O2) follows immediately from the definitions of Ωj;k1 ,··· ,kn−1 and d Ωj;k1 ,··· ,kn−1 . To show (3.30) in (O3), we let (r, ϑ1 , · · · , ϑn−1 ) and (r, ϑ1 , · · · , ϑn−1 ) be the polar coordinates of x = (x1 , . . . , xn ) and x = (x1 , . . . , xn ), respectively. Then we have that x, x ∈ Ωj;1,··· ,1 ⇐⇒ j − 1 ≤ r < j, 0 ≤ ϑl , ϑl <
2π for all l = 1, · · · , n − 1. j
(3.32)
Notice that the connection between (x1 , . . . , xn ) and (r, ϑ1 , . . . , ϑn−1 ) is as: ⎧ x1 = r cos ϑ1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x = r sin ϑ1 cos ϑ2 , ⎪ ⎪ ⎨ 2 ··· ⎪ ⎪ ⎪ ⎪ = r sin ϑ1 sin ϑ2 · · · sin ϑn−2 cos ϑn−1 , x ⎪ ⎪ ⎪ n−1 ⎪ ⎩ xn = r sin ϑ1 sin ϑ2 · · · sin ϑn−2 sin ϑn−1 . Then, by the mean value theorem, we have that for some ζ ∈ (0, 2π/j), |x1 − x1 | = r| cos ϑ1 − cos ϑ1 | = r| sin ζ · (ϑ1 − ϑ1 )| ≤ j| sin ζ|
2π ≤ 2π. j
By inserting suitable terms and using the mean value theorem, we have that |x2 − x2 | ≤ r | sin ϑ1 cos ϑ2 − sin ϑ1 cos ϑ2 | + | sin ϑ1 cos ϑ2 − sin ϑ1 cos ϑ2 | ≤ 2π · 2. Similarly, we can verify that |xi − xi | ≤ 2π · i for all i = 3, . . . , n. These, along with (3.29), lead to (3.30). The inequality (3.31) in (O3) can be proved in the same way. The reason that the factor j+1 j appears in (3.31) is as follows: Since 1 ≤ ki ≤ j (i = 1, . . . , n − 1), we see from the definition of Ωj+1;1,··· ,1 that x, x ∈ Ωj+1;1,··· ,1 ⇐⇒ j ≤ r < j + 1, 0 ≤ ϑl , ϑl <
2π for all l = 1, · · · , n − 1. j
(The above is comparable with (3.32).) Step 3. To prove that there are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) (both independent of a, j and f ) so that
1 C(1+ a )
|f | dx ≤ e 2
Ωj+1;k1 ,··· ,kn−1
Since Ωj;k1 ,··· ,kn−1
Ωj;k1 ,··· ,kn−1
|f | dx 2
θ
|f|2 ea|ξ| dξ 2
1−θ .
Rn
Ωj+1;k1 ,··· ,kn−1 is connected (see (3.25)), it follows from (3.30) and (3.31) that
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3 Ωj+1;k1 ,··· ,kn−1 ≤ d Ωj;k1 ,··· ,kn−1 + d Ωj+1;k1 ,··· ,kn−1 ≤ 6πn 2 . d Ωj;k1 ,··· ,kn−1 Thus, there exists x 1 ∈ Rn such that Ωj;k1 ,··· ,kn−1
3
Ωj+1;k1 ,··· ,kn−1 ⊂ BR0 (1 x), with R0 = 7πn 2 .
(3.33)
= C(n) According to Corollary 3.2 where b = R0 , there is C > 0 such that for all α ∈ N n , 1+ 1 a
Dα f L∞ (Rn ) ≤ eC
|α|! |α|
f(ξ)
R0
L2 (ea|ξ|2 dξ)
.
(3.34)
By (3.34), as well as (3.33), we can apply Lemma 3.5, where 1+ 1 a
ρ = 1, R = R0 , BR = BR0 (1 x), B2R = B2R0 (1 x), ω = Ωj;k1 ,··· ,kn−1 , M = eC
f(ξ)
L2 (ea|ξ|2 dξ)
,
to find constants C0 = C0 (n) > 0 and θ = θ(n) ∈ (0, 1) so that ! f L2 (BR0 (1x)) ≤ C0 f θ 2 L
Ωj;k1 ,··· ,kn−1
1+ 1 a
≤ C0 eC
1+ 1 a
eC
f θ 2 L
Ωj;k1 ,··· ,kn−1
"1−θ
f(ξ)
L2 (ea|ξ|2 dξ)
1−θ L2 ea|ξ|2 dξ
f(ξ)
.
(3.35)
(Here, we used (3.28) and a coordinate translation.) Finally, the desired inequality of this step follows from (3.35) and (3.33). Step 4. To prove the inequality of this lemma. From (3.26) and (3.27), we see that Ωj is the disjoint union of all Ωj;k1 ,··· ,kn−1 with different (k1 , . . . , kn−1 ) ∈ N n−1 satisfying 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. Meanwhile, by (3.23), (3.24) and (3.25), one can also check that Ωj+1 is the disjoint union of all Ωj+1;k1 ,··· ,kn−1 with different (k1 , . . . , kn−1 ) ∈ N n−1 satisfying 1 ≤ ki ≤ j for all i = 1, . . . , n −1. These, along with Lemma 3.4, yield that for some C = C(n) > 0 and θ = θ(n),
|f |2 dx = Ωj+1
≤
|f |2 dx
Ωj+1;k1 ,··· ,kn−1
eC
1
1+ a
|f |2 dx
≤e
|f | dx 2
Ωj;k1 ,··· ,kn−1
= j (n−1)(1−θ) eC
1 1+ a
Ωj
|f|2 ea|ξ| dξ 2
1−θ
Rn
Ωj;k1 ,··· ,kn−1 1 C 1+ a
θ
|f |2 dx
θ
θ
|f|2 ea|ξ| dξ 2
1−θ
Rn
|f|2 ea|ξ| dξ 2
1−θ ,
(3.36)
Rn
where the sums are taken over all different (k1 , . . . , kn−1 ) ∈ N n−1 with 1 ≤ ki ≤ j for all i = 1, . . . , n − 1. (Notice that there are j n−1 such (k1 , . . . , kn−1 ).). Hence, the desired conclusion follows from (3.36). This ends the proof of Lemma 3.4. 2 Based on Lemma 3.4, we can have the next propagation result which will be used later.
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Lemma 3.6. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0 and j ≥ 1,
|f |2 dx ≤ j n−1 eC
1 1+ a
|f |2 dx
θj
2
1−θj ,
Rn
B1
Ωj+1
|f|2 ea|ξ| dξ
2 when f ∈ L2 (Rn ) satisfies that f ∈ L2 (ea|ξ| dξ). 2 Proof. Arbitrarily fix a > 0 and j ≥ 1. And then arbitrarily fix f ∈ L2 (Rn ) with f ∈ L2 (ea|ξ| dξ). From Lemma 3.4, we can use the induction method to verify that
2 j−2 j−1 (n−1)(1−θ) |f |2 dx ≤ j(j − 1)θ (j − 2)θ · · · 2θ 1θ
Ωj+1 1
× eC(1+ a )(1+θ+···+θ C
1
≤ j n−1 e 1−θ (1+ a )
j−1
)
|f |2 dx
θj
2
1−θj
Rn
Ω1
|f |2 dx
|f|2 ea|ξ| dξ
θj
|f|2 ea|ξ| dξ 2
1−θj .
(3.37)
Rn
Ω1
Since Ω1 = B1 and θ = θ(n) ∈ (0, 1), the desired conclusion in the lemma follows from (3.37). This ends the proof of Lemma 3.6. 2 The next proposition plays a very important role in the proof of Theorem 3.1. Proposition 3.1. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0, t > 0 and ε > 0,
e−a|x| |f |2 dx ≤ eC
1+ 1t +a
1 + a−n Γ
Rn
a ε 2| ln θ|
|f|2 et|ξ| dξ + eε 2
Rn
−
2| ln θ| a
|f |2 dx ,
B1
2 when f ∈ L2 (Rn ) satisfies f ∈ L2 (et|ξ| dξ).
To prove Proposition 3.1, we need the following result quoted from [43]: Lemma 3.7 ([43, Lemma 3.1]). Let a > 0, b ∈ (0, 1) and θ ∈ (0, 1). Then ∞
bθ e−ak ≤ k
k=1
a ea a Γ | ln b|− | ln θ| . | ln θ| | ln θ|
Proof of Proposition 3.1. Arbitrarily fix a > 0 and t > 0. And then arbitrarily fix f ∈ L2 (Rn ) satisfies 2 f ∈ L2 (et|ξ| dξ). It suffices to show the inequality in Proposition 3.1 for the above fixed a, t, f and any ε > 0. Without loss of generality, we can assume that
A := Rn
|f|2 et|ξ| dξ = 0 and B := 2
|f |2 dx = 0.
(3.38)
B1
For otherwise, when A = 0, we have that f = 0 over Rn , thus the desired inequality is trivial; while when B = 0, we can use the analyticity of f (which follows from Corollary 3.2) to see that f = 0 over Rn and then the desired inequality is trivial again.
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By (3.22), we have that
e−a|x| |f |2 dx =
Rn
e−a|x| |f |2 dx +
e−a|x| |f |2 dx
j≥1Ω j+1
B1
|f |2 dx +
≤
e−aj |f |2 dx.
(3.39)
j≥1Ω j+1
B1
We now estimate the last term of (3.39). According to Lemma 3.6, there is C1 = C1 (n) > 0 and θ = θ(n) ∈ (0, 1) so that
e−aj |f |2 dx ≤ e
1
C1 1+ 1t
j≥1Ω j+1
⎛
⎞θj ⎛
j n−1 e−aj ⎝
|f |2 dx⎠ ⎝
j≥1
≤ eC1
1+ t
B1
a n θj n! (2/a) e− 2 j A (B/A) ,
⎞1−θj |f|2 et|ξ| dξ ⎠ 2
Rn
(3.40)
j≥1
where A and B are given by (3.38). In the proof of (3.40), we used the inequality: j n e− 2 j ≤ n! (2/a) a
n
for all j ≥ 1.
Meanwhile, by Lemma 3.7 (with b = B/A), we have that
a e2 a Γ A| ln(B/A)|− 2| ln θ| . | ln θ| 2| ln θ| a
e− 2 j A(B/A)θ ≤ j
a
j≥1
(3.41)
About A/B, there are only two possibilities: either A/B > e or A/B ≤ e. In the first case when A/B > e, we claim that −
A| ln(B/A)|− 2| ln θ| ≤ εA + eε a
2| ln θ| a
B for all ε > 0.
In fact, when ε satisfies that A| ln(B/A)|− 2| ln θ| ≤ εA, a
(3.42) is trivial. On the other hand, when ε > 0 satisfies that A| ln(B/A)|− 2| ln θ| > εA, a
we have that A/B < eε
−
2| ln θ| a
.
This, along with the fact that A/B > e, yields that −
A| ln(B/A)|− 2| ln θ| ≤ A ≤ B · A/B ≤ eε a
Thus, we have proved (3.42).
2| ln θ| a
B for all ε > 0.
(3.42)
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180
Inserting (3.42) into (3.41) leads to
2| ln θ| − e 2 a Γ εA + eε a B for all ε > 0. | ln θ| 2| ln θ| a
e− 2 j A(B/A)θ ≤ j
a
j≥1
(3.43)
Now it follows from (3.39),(3.40) and (3.43) that for some C2 = C2 (n) > 0,
e
−a|x|
C1 (1+ 1t )
|f | dx ≤ e 2
Rn
a a 2| ln θ| 2 − n e Γ n! 1 + (2/a) εA + eε a B | ln θ| 2| ln θ|
1 ≤ eC2 (1+ t +a) 1 + a−n Γ
2| ln θ| − a εA + eε a B for any ε > 0. 2| ln θ|
This proves the desired inequality for the first case that A/B > e. In the second case where A/B ≤ e, we derive directly that
e
−a|x|
|f | dx ≤
|f | dx ≤
2
Rn
2
Rn
Rn
⎛
≤ e ⎝ε
2 |f|2 et|ξ| dξ ≤ e
|f |2 dx
B1 −
|f|2 et|ξ| dξ + eε 2
2| ln θ| a
Rn
⎞ |f |2 dx⎠ for any ε > 0.
B1
This proves the desired inequality for the second case that A/B ≤ e. Hence, we end the proof of Proposition 3.1. 2 We now are on the position to show Theorem 3.1. Proof of Theorem 3.1. (i). Arbitrarily fix u0 ∈ L2 (ea|x| dx). Let u(T, x) = (eT Δ u0 )(x), x ∈ Rn . By the Hölder inequality, we have that
|u(T, x)| dx ≤ 2
Rn
|u(T, x)| e
2 a|x|
1/2
dx
Rn
|u(T, x)|2 e−a|x| dx
1/2 .
(3.44)
Rn
We will estimate the two terms on right side of (3.44) one by one. For the first term, we apply Lemma 3.1 (with ν = 1) to obtain that
|u(T, x)|2 ea|x| dx ≤ 2n e2a Rn
2
|u0 (x)|2 ea|x| dx.
T
(3.45)
Rn
T Δ u ∈ L2 (eT |ξ| dξ), since To estimate the second term (on right side of (3.44)), we first notice that e 0 2 a|x| 2 n u0 ∈ L (e dx) ⊂ L (R ). Thus, we can apply Proposition 3.1 (with f = eT Δ u0 and t = 2T ) to find C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for each ε > 0, 2
e Rn
−a|x|
2| ln θ| 2 ε− a |u(T, x)| dx ≤ C(T, a, n) ε |u0 (x)| dx + e |u(T, x)|2 dx , 2
Rn
B1
where 1 C(T, a, n) = eC(1+ T +a) 1 + a−n Γ
a . 2| ln θ|
(3.46)
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Inserting (3.45) and (3.46) into (3.44), we get that for each ε > 0
|u(T, x)|2 dx ≤ C
Rn
ε1/2 ≤C
|u0 (x)|2 ea|x| dx
|u0 (x)| e
2 a|x|
Rn
ε1/2 ≤ 2−1 C
ε1/2 ≤C
1/2
1/2 2| ln θ| − |u(T, x)|2 dx ε |u0 (x)|2 dx + eε a
Rn
1/2 dx
Rn
B1
|u0 (x)| dx + ε
1/2
2
ε
−1/2
e
Rn
|u0 (x)|2 ea|x| dx + ε1/2
Rn
|u(T, x)|2 dx
1/2
B1 −
|u0 (x)|2 dx + ε−1/2 eε
Rn
|u0 (x)|2 ea|x| dx + ε−1/2 e
2| ln θ| ε− a
2| ln θ| ε− a
Rn
2| ln θ| a
|u(T, x)|2 dx
B1
|u(T, x)|2 dx ,
(3.47)
B1
where 2 = C(T, C a, n) = 2n/2 ea T
C(T, a, n).
Since ε > 0 can be arbitrary taken, we replace ε by ε2 in (3.47) to get the desired conclusion in (i) of Theorem 3.1. (ii). Arbitrarily fix u0 ∈ L2 (xν dx). Let u(T, x) = eT Δ u0 (x), x ∈ Rn . Three facts are given in order. Fact One. Using the inequality: 1
1 ≤ ε|x|ν + e(1/ε) ν e−|x| for all ε > 0 and x ∈ Rn , we find that
1
|u(T, x)|2 dx ≤ ε Rn
|u(T, x)|2 |x|ν dx + e(1/ε) ν
Rn
|u(T, x)|2 e−|x| dx.
(3.48)
Rn
Fact Two. Since 0 < ν ≤ 1, we can apply Lemma 3.2 to find C1 = C1 (n) so that
2 ν |u(T, x)|2 |x|ν dx ≤ 4ν+2 Γ(ν/2 + n) 1 + T 4 u0 L2 ( x ν dx)
Rn
ν ≤ C1 1 + T 2
|u0 (x)|2 xν dx.
(3.49)
Rn
Fact Three. We can use (3.46) (with a = 1 and ε = μ) to find C2 = C2 (n) and θ = θ(n) ∈ (0, 1) so that for all μ > 0,
e−|x| |u(T, x)|2 dx ≤ eC2
1+ T1
μ
Rn
−2| ln θ|
|u0 (x)|2 dx + eμ
Rn
2 |u(T, x)| dx .
(3.50)
B1
To continue the proof, we arbitrarily fix ε ∈ (0, 1). We will first use Fact Three, and then use Fact One 1
and Fact Two. By taking μ = εe−( ε ) ν in (3.50), we obtain that 1
1
e(1/ε) ν
Rn
e−|x| |u(T, x)|2 dx
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182
1
≤e
(1/ε) ν
= eC2
e
C2 1+ T1
1
1+ T
μ
|u0 (x)| dx + e 2
Rn
|u0 (x)|2 dx + bε
ε Rn
μ−2| ln θ|
|u(T, x)|2 dx
B1
|u(T, x)|2 dx ,
(3.51)
B1
where 2 3 1 1 bε = exp (1/ε) ν + (ε−2 e2(1/ε) ν )| ln θ| . Meanwhile, one can directly check the following two inequalities: s2 ≤ es for all s > 0;
(3.52)
s + e3| ln θ|s ≤ e(3| ln θ|+1)s for all s > 0.
(3.53)
Choosing s = ε−1 and s = (1/ε) ν in (3.52) and (3.53) respectively, using 0 < ν ≤ 1, we find that 1
2 3 1 1 bε ≤ exp (1/ε) ν + (e1/ε e2(1/ε) ν )| ln θ| 2 3 1 1 ≤ exp (1/ε) ν + (e3(1/ε) ν )| ln θ| 2 1 3 ≤ exp e(3| ln θ|+1)(1/ε) ν .
(3.54)
Combining (3.51) and (3.54) leads to that 1
e
e−|x| |u(T, x)|2 dx
(1/ε) ν
Rn
≤ eC2 (n)
1+ T1
ε
|u0 (x)|2 dx + ee
1 (3| ln θ|+1)(1/ε) ν
Rn
|u(T, x)|2 dx .
(3.55)
B1
Finally, inserting (3.49) and (3.55) into (3.48), we obtain that for some C3 = C3 (n),
|u(T, x)|2 dx Rn
1
(3| ln θ|+1)(1/ε) ν ν 1 ≤ C1 (1 + T 2 ) + eC2 (1+ T ) ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx
Rn
B1
1
(3| ln θ|+1)(1/ε) ν ν 1 ≤ (1 + T 2 )eC3 (1+ T ) ε |u0 (x)|2 xν dx + ee |u(T, x)|2 dx ,
Rn
B1
which leads to the desired conclusion in (ii) of Theorem 3.1. Hence, we end the proof of Theorem 3.1. 2 3.2. Weak observability inequalities with observations on balls According to Theorem 1.1, it is impossible to recover a solution of (1.1) by observing it over a ball. Thus, two interesting questions arise. First, can we recover a solution of (1.1) over a ball by observing it
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
183
on another ball? Second, can we have observability inequalities with observations over balls for solutions of (1.1) with some kind of initial values? The answer to the first question is almost negative, while we give partially positive answer for the second question. The first main result of this subsection is stated as follows: Theorem 3.2. (i) There is an absolute positive constant C so that for all T > 0 and 0 < r < r, !
u (T, x) dx ≤ 2
Br
Cn 1 + T (r − r )2
" T
u2 (t, x) dx dt, when u solves (1.1). 0 Br
(ii) Given T > 0 and r > r > 0, there is no constant C = C(T, r , r, n) so that
T
u (T, x) dx ≤ C 2
u2 (t, x) dx dt for any solution u to (1.1). 0 Br
Br
Proof. (i) Arbitrarily fix T > 0 and 0 < r < r. Arbitrarily fix a solution u to (1.1). Let u(0, x) = u0 (x), x ∈ Rn . Choose a C 2 function ϕ on Rn so that for some absolute constant C > 0, −|α|
0 ≤ ϕ(x) ≤ 1 over Rn ; |Dα ϕ(x)| ≤ C (r − r )
for all α ∈ N n , with |α| ≤ 2,
(3.56)
and so that / ϕ(x) =
1, x ∈ Br , 0, x ∈ Brc .
Set v = ϕu. Then v satisfies vt − Δv = −2∇ϕ · ∇u − Δϕu
in R+ × Rn ,
v(0, ·) = ϕ(·)u0 (·) in Rn .
(3.57)
Multiplying (3.57) by tv leads to 1 1 2 tv t − v 2 − tvΔv = −2tu∇ϕ · ∇v + t 2|∇ϕ|2 − ϕΔϕ u2 . 2 2
(3.58)
Integrating (3.58) over (0, T ) × Rn , we have that
1 2
1 T v (T, x) dx − 2
T
2
T
v (t, x) dx dt +
0 Rn
Rn
2
t |∇v(t, x)| dx dt
2
0 Rn
T
−2tu∇ϕ(x) · ∇v(t, x) dx dt
= 0 Rn
T
+
2 t 2 |∇ϕ(x)| − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
0 Rn
Meanwhile, by the Hölder inequality, we find that
(3.59)
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184
T
−2tu(t, x)∇ϕ(x) · ∇v(t, x) dx dt 0 Rn
T
T
2
≤
2
t |∇v(t, x)| dx dt + 0
t |∇ϕ(x)| u2 (t, x) dx dt. 0
Rn
(3.60)
Rn
Inserting (3.60) into (3.59) leads to that
v 2 (T, x) dx
T Rn
T
T
≤
2
v (t, x) dx dt + 2 0 Rn
t 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
(3.61)
0 Rn
Since v = u on Br and v = 0 on Brc , we can use (3.61) and (3.56) to get that
u2 (T, x) dx Br
1 ≤ T
T
2 u (t, x) dx dt + T
T
0 Br
! ≤
2
8Cn 1 + T (r − r )2
t 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt
0 Br \Br
" T
u2 (t, x) dx dt, 0 Br
which leads to the desired conclusion in (i) of Theorem 3.2. (ii) By contradiction, we suppose that there were T > 0, r > r > 0 and C = C(T, r , r, n) > 0 so that
T
u (T, x) dx ≤ C 2
u2 (t, x) dx dt for any solution u to (1.1).
(3.62)
0 Br
Br
We would use a constructive method to derive a contradiction with (3.62). For this purpose, we define, for each k ≥ 1, uk (t, x) =
|x1 −k|2 +|x |2 1 e− 4(t+1) , n/2 (4π(t + 1))
(t, x) = (t, x1 , x ) ∈ [0, ∞) × R × Rn−1 .
One can easily check that uk is the solution of (1.1) with initial value: uk (0, x) =
|x −k|2 +|x |2 1 − 1 4 e , (4π)n/2
x = (x1 , x ) ∈ R × Rn−1 .
It is clear that {uk (0, ·)}k≥1 is uniformly bounded in L2 (Rn ). We next show that when k is large enough, uk does not satisfy (3.62) (which leads to a contradiction). To this end, we need two estimates:
G. Wang et al. / J. Math. Pures Appl. 126 (2019) 144–194
T
u2k (t, x) dx dt ≤
185
(k−r)2 r n T Vn − 2(T +1) , when k > r + e 2n(T + 1); (4π(T + 1))n
(3.63)
0 Br
u2k (T, x) dx ≥
(k−r− σ )2 3 1 − 2(T +1) e (σ/3)n Vn , with σ = r − r, when k > r . (4π(T + 1))n
(3.64)
Br
To show (3.63), we obtain from a direct computation that ∂t (ln uk (t, x)) = −
|x1 − k|2 + |x |2 n 1 + for all t ≥ 0, x = (x1 , x ) ∈ Rn , 2 t+1 4(t + 1)2
from which, it follows that when t ≥ 0 and x = (x1 , x ) ∈ Rn , ∂t (ln uk (t, x)) > 0 ⇐⇒ |x1 − k|2 + |x |2 > 2n(t + 1). This implies that k>r+
2n(T + 1) =⇒ ∂t (ln uk (t, x)) > 0 for all (t, x) × (0, T ) × Br .
From the above, we see that when k > r + 2n(T +1), we have that for each x ∈ Br , uk (t, x) is an increasing function of t on [0, T ]. Hence, when k > r + 2n(T + 1),
T
T
u2k (t, x) dx dt
≤
0 Br
T
u2k (T, x) dx dt
0 Br
=
T (4π(T + 1))n
=
|x −k|2 +|x |2 1 − 1 2(T +1) e dx dt (4π(T + 1))n
0 Br
e
|x −k|2 +|x |2 − 1 2(T +1)
dx ≤
T (4π(T + 1))n
Br
≤
n
r T Vn e (4π(T + 1))n
|k−r|2
e− 2(T +1) dx
Br
(k−r)2 − 2(T +1)
,
which leads to (3.63). We next show (3.64). Given b > 0, use Br (b, 0 ) to denote the ball centered at (b, 0, 0, · · · , 0) and of radius r, namely Br (b, 0 ) := x ∈ Rn : |x1 − b|2 + |x |2 ≤ r2 . Let k > r and σ = r − r. Then
u2k (T, x) dx ≥ Br
Br
u2k (T, x) dx Bk−r− σ (k,0 )
3
|x1 −k|2 +|x |2 1 e− 2(T +1) dx n (4π(T + 1))
= Br
Bk−r− σ (k,0 ) 3
1 e− ≥ (4π(T + 1))n Meanwhile, it is clear that
2 k−r− σ 3 2(T +1)
Bk−r− σ3 (k, 0 ) . Br
(3.65)
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186
B σ3 (r + 2σ/3, 0 ) ⊂ Br
Bk−r− σ3 (k, 0 ),
which leads to that Bk−r− σ3 (k, 0 ) ≥ B σ3 (r + 2σ/3, 0 ) = (σ/3)n Vn . Br
(3.66)
Now (3.64) follows from (3.65) and (3.66) at once. From (3.63) and (3.64), we find that when k > max{r + 2n(T + 1), r },
T
u2k (t, x) dx dt 0 Br
u2k (T, x) dx
(k−r)2
− 2(T +1) r n T Vn (4π(T +1))n e
≤
− 1 (4π(T +1))n e
2 k−r− σ 3 2(T +1)
σ n 3
Vn
Br
=T =T
3r n σ 3r n
1
k−r− σ 3
e 2(T +1) 1
e 2(T +1)
2
−(k−r)2
2 2σ (σ 3 ) − 3 (k−r)
σ 3r n r −r 2 2(r −r) 1 =T e 2(T +1) 3 − 3 (k−r) . r −r
(3.67)
Since (
r − r 2 2(r − r) ) − (k − r) → −∞, as k → ∞, 3 3
we see from (3.67) that
T
u2k (t, x) dx dt lim
0 Br
k→+∞
= 0, u2k (T, x) dx
Br
from which, it follows that uk does not satisfy (3.62), when k is large enough. This shows the conclusion in (ii) of Theorem 3.2. Hence, we end the proof of Theorem 3.2. 2 The next corollary is a direct consequence of Theorem 3.2. Corollary 3.3. Given ν > 0, T > 0 and r > 0, there is no constant C = C(T, r, ν, n) > 0 so that
T
u (T, x)ρ(x) dx ≤ C 2
Rn
u2 (t, x) dx dt for any u solves (1.1), 0 Br
where either ρ(x) = x−ν , x ∈ Rn , or ρ(x) = e−|x| , x ∈ Rn .
(3.68)
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Remark 3.5. It was announced in [7, p. 384] (without proof) that given a bounded interval E, there is no positive weight function ρ such that
∞
T
|u(T, x)| ρ(x) dx ≤ C
|u(t, x)|2 dx dt
2
0
0 E
for all solutions of the heat equation in the physical space (0, ∞). The above Corollary 3.2 presents a similar result for the heat equation in the physical space Rn . The second main result of this subsection is stated as follows: Theorem 3.3. (i) There is a generic constant C so that for any T > 0, M > r > 0 and u0 ∈ L2 (Rn ) with supp u0 ⊂ Br , !
|u(T, x)|2 dx ≤ Rn
Cn 1 + T (M − r)2
" T
|u(t, x)|2 dx dt, 0 BM
where u is the solution to (1.1) with u(0, ·) = u0 (·). (ii) Assume that 0 ≤ u0 ∈ L1 (Rn ) so that
u0 (x) dx ≥ μ
u0 (x) dx for some r > 0 and μ ∈ (0, 1).
Rn
Br
Then for any T > 0, M > 0 and any solution u to (1.1) with u(0, ·) = u0 (·),
2 2 +1 π 2 T 2 −1 4r2 |u(T, x)| dx ≤ e T Vn (r ∧ M )n μ2 n
n
n
T
|u(t, x)|2 dx dt.
2
Rn
0 BM
Here, r ∧ M := min{r, M }. Proof. (i) The proof is similar to that of (i) of Theorem 3.2. Arbitrarily fix T > 0, M > r > 0 and u0 ∈ L2 (Rn ) with supp u0 ⊂ Br . Write u for the solution to (1.1) with u(0, ·) = u0 (·). Choose a C 2 function ϕ over Rn so that for some absolute constant C > 0, 0 ≤ ϕ(x) ≤ 1 over Rn ; |Dα ϕ(x)| ≤ C(M − r)−|α| for all α ∈ N n , with |α| ≤ 2
(3.69)
and so that / ϕ(x) =
0, x ∈ Br , c . 1, x ∈ BM
Set v = ϕu. Multiplying (3.57) by v, we find that 1 2 v t − vΔv = −2u∇ϕ · ∇v + 2|∇ϕ|2 − ϕΔϕ u2 . 2 Integrating (3.70) over (0, T ) × Rn , we obtain that
(3.70)
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188
1 2
1 v (T, x) dx − 2
T
2
Rn
|∇v(t, x)|2 dx dt
2
v (0, x) dx +
T
T
−2u(t, x)∇ϕ(x) · ∇v(t, x) dx dt +
=
(3.71)
0 Rn
Rn
0 Rn
2|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dt.
0 Rn
Since the support of u0 is contained in Br , we have that v(0, ·) = 0 over Rn . Then by the Hölder inequality, we deduce from (3.71) that
T
2 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (s, x) dx ds.
v (T, x) dx ≤ 2
0
Rn
(3.72)
Rn
Note that (3.72) is still true if we replace T by any t ∈ (0, T ). This implies that
T
T t
v (t, x) dx dt ≤ 2
0
0
Rn
0
2 3|∇ϕ(x)|2 − ϕ(x)Δϕ(x) u2 (t, x) dx dτ dt.
(3.73)
Rn
c Since v = u on BM , it follows from (3.69) and (3.73) that
T
CnT u (t, x) dx ≤ (M − r)2
T
2
0 |x|≥M
u2 (t, x) dx dt.
(3.74)
0 r≤|x|≤M
Meanwhile, it is clear that
1 u (T, x) dx ≤ T
T
2
u2 (t, x) dx dt.
(3.75)
0 Rn
Rn
Combining (3.74) and (3.75), we obtain that
1 u (T, x) dx ≤ T
T
2
Rn
1 ≤ T ! ≤
1 u (t, x) dx dt + T
T
2
0 |x|≤M
T
u2 (t, x) dx dt 0 |x|≥M
1 CnT u (t, x) dx dt + T (M − r)2
T
2
0 |x|≤M
Cn 1 + T (M − r)2
u2 (t, x) dx dt 0 r≤|x|≤M
" T
u2 (t, x) dx dt, 0 |x|≤M
which leads to the conclusion (i) of Theorem 3.3. (ii) Let T > 0 and M > 0 be arbitrarily given. Arbitrarily fix u0 so that
0 ≤ u0 ∈ L1 (Rn );
u0 (x) dx ≥ μ
Br
Rn
u0 (x) dx for some r > 0 and μ ∈ (0, 1).
(3.76)
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189
Write u for the solution to (1.1) with u(0, ·) = u0 (·). We first prove that when 0 < M ≤ r,
2 2 +1 π 2 T 2 −1 4r2 |u(T, x)| dx ≤ e T Vn M n μ2 n
n
n
T
|u(t, x)|2 dx dt.
2
(3.77)
0 BM
Rn
For this purpose, we need the following two estimates: ⎛
T
T 2
u2 (t, x) dx dt ≥ 2−1 (4π)−n Vn M n μ2 T −(n−1) e
2 − 4r T
|x|≤M
⎝
⎞2 u0 (x) dx⎠ ;
(3.78)
Rn
⎛
u2 (T, x) dx ≤ 2
− 3n 2
(πT )−n/2 ⎝
Rn
⎞2
u0 (x) dx⎠ .
(3.79)
Rn
To show (3.78), we observe that
u(t, x) =
(4πt)−n/2 e−
|x−y|2 4t
u0 (y) dy,
(t, x) ∈ (0, ∞) × Rn .
(3.80)
Rn
By (3.80) and (3.76), we find that when t > 0, |x| ≤ r,
|x−y|2 r2 u(t, x) ≥ (4πt)−n/2 e− 4t u0 (y) dy ≥ (4πt)−n/2 e− t u0 (y) dy |y|≤r
≥ μ(4πt)−n/2 e
|y|≤r
2 − rt
u0 (x) dx.
(3.81)
Rn
Since M ≤ r, it follows from (3.81) that
T
⎛ u (t, x) dx dt ≥ Vn M μ ⎝ 2
T 2
n 2
|x|≤M
⎞2 u0 (x) dx⎠
Rn
T
(4πt)−n e−
T 2
≥ 2−1 (4π)−n Vn M n μ2 T −(n−1) e
⎛ 2 − 4r T
⎝
2r 2 t
dt ⎞2
u0 (x) dx⎠ ,
Rn
which leads to (3.78). We now show (3.79). By (3.80) and the Young inequality, we have that
|x|2 − 3n 4 (πT )−n/4 u(T, x) L2 (Rn ) ≤ (4πT )−n/2 e− 4T u (x) dx = 2 u0 (x) dx, 0 2 n L (R ) Rn
Rn
which leads to (3.79). Next, by (3.78) and (3.79), we see that
Rn
2 2 +1 π 2 T 2 −1 4r2 e T Vn M n μ2 n
u2 (T, x) dx ≤
n
n
T
u2 (t, x) dx dt T 2
|x|≤M
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190
2 2 +1 π 2 T 2 −1 4r2 ≤ e T Vn M n μ2 n
n
n
T
u2 (t, x) dx dt, 0 |x|≤M
which leads to (3.77) for the case when 0 < M ≤ r. Finally, when M > r, we apply (3.77) (with M = r) to obtain that
2 2 +1 π 2 T 2 −1 4r2 u (T, x) dx ≤ e T Vn rn μ2 n
n
n
T
2
Rn
2 2 +1 π 2 T 2 −1 4r2 e T ≤ Vn rn μ2 n
n
n
u2 (t, x) dx dt 0 |x|≤r
T
u2 (t, x) dx dt, 0 |x|≤M
which leads to (3.77) for the case that M > r. So the conclusion (ii) in Theorem 3.3 is true. Hence, we end the proof of Theorem 3.3. 2 3.3. Applications to controllability In this subsection, we will use Theorem 3.1 and Theorem 3.2 to derive some kinds of controllability for some kinds of controlled heat equations. We start with introducing the controlled equation for the applications of Theorem 3.1. Let 0 < τ < T . Consider the impulsively controlled heat equation: ⎧ ⎪ in (0, T ) \ {τ } × Rn , ⎨ ∂t y − Δy = 0 (3.82) Rn , y|t=τ = y|t=τ − + χB1 h in ⎪ ⎩ y| n in R , t=0 = y0 where y0 ∈ L2 (Rn ) is an initial state, h ∈ L2 (Rn ) is a control and y|t=τ − denotes the left limit of the solution y (which is treated as a function from R+ to L2 (Rn )) at time τ . Write y(·, ·; y0 , h) for the solution to (3.82). Theorem 3.4. The following two conclusions are true: (i) Let a > 0. Then for any y0 ∈ L2 (Rn ) and ε > 0, there is a control hy0 ,ε ∈ L2 (Rn ) so that
C1 (a, T − τ ) ≥
1 ε
|y0 (x)|2 dx
Rn
e−a|x| |y(T, x; y0 , hy0 ,ε )|2 dx + εe−ε
Rn
−
4| ln θ| a
|hy0 ,ε (x)|2 dx, Rn
where C1 (a, T − τ ) and θ are given by (i) of Theorem 3.1. ˜ y ,ε ∈ L2 (Rn ) so that (ii) Let ν ∈ (0, 1]. Then for any y0 ∈ L2 (Rn ) and ε > 0, there is a control h 0
C2 (ν, T − τ ) 1 ≥ ε
|y0 (x)|2 dx
Rn
x
−ν
˜ y ,ε )|2 dx + e−e |y(T, x; y0 , h 0
Rn
where C2 (ν, T − τ ) and θ are given by (ii) of Theorem 3.1.
1 (3| ln θ|+1) 1 ν ε
˜ y ,ε (x)|2 dx, |h 0 Rn
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Proof. (i) We will use [43, Lemma 5.1] and (i) of Theorem 3.1 to prove the conclusion (i). For this purpose, we set up our control problem in the framework of [43, Lemma 5.1]. Arbitrarily fix a > 0. Define the Banach space: ⎧ ⎨ Xa :=
f ∈ L2 (Rn ) :
⎩
ea|x| |f (x)|2 dx < ∞ Rn
⎫ ⎬ ⎭
,
(3.83)
endowed with the norm ⎛ f Xa := ⎝
⎞1/2
ea|x| |f (x)|2 dx⎠
, f ∈ Xa .
Rn
One can directly check that the dual space of Xa is as:
· Xa∗
Xa∗ = C0∞ (Rn )
,
(3.84)
with the norm: ⎛ g Xa∗ := ⎝
⎞1/2 e−a|x| |g(x)|2 dx⎠
, g ∈ Xa∗ .
(3.85)
Rn
We let X := L2 (Rn ) = X ∗ , Y := L2 (Rn ; C) = Y ∗ and Z := Xa ,
(3.86)
where the space Xa is given by (3.83). Define two operators R : Z → X and O : Z → Y by Rz := eΔ(T −τ ) z for each z ∈ Xa ;
Oz := χB1 eΔ(T −τ ) z for each z ∈ Xa .
(3.87)
One can directly check that R∗ f = eΔ(T −τ ) f, ∀ f ∈ L2 (Rn );
O∗ h = eΔ(T −τ ) (χB1 h) = y(T, ·; 0, h), ∀ h ∈ L2 (Rn ).
(3.88)
By (i) of Theorem 3.1, (3.87), (3.86) and (3.83), we can verify that for any ε > 0, 4| ln θ| − Rz 2L2 (Rn ) ≤ C1 (a, T − τ ) ε Oz 2L2 (Rn ) + ε−1 eε a z 2Xa for each z ∈ Xa ,
(3.89)
where C1 (·, ·) and θ are given by (i) of Theorem 3.1. Next, we arbitrarily fix ε > 0 and y0 ∈ L2 (Rn ). According to [43, Lemma 5.1] and (3.89), there exists hy0 ,ε ∈ L2 (Rn ) so that 1 1 hy0 ,ε 2Y ∗ + R∗ (eΔτ y0 ) − O∗ hy0 ,ε 2Z ∗ ≤ eΔτ y0 2X ∗ . 4| ln θ| − C1 (a, T − τ )ε a −1 ε C1 (a, T − τ )ε e Since {eΔt }t≥0 is contractive in L2 (Rn ), from the above, (3.88), (3.86) and (3.85), we obtain the conclusion (i) of this theorem.
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(ii) We will use [43, Lemma 5.1] and (ii) of Theorem 3.1 to prove the conclusion (ii). To this end, we arbitrarily fix ν ∈ (0, 1]. Define the Banach space: Xν :=
⎧ ⎨ ⎩
f ∈ L2 (Rn ) :
xν |f (x)|2 dx < ∞ Rn
⎫ ⎬ ⎭
,
(3.90)
endowed with the norm ⎛ f Xν := ⎝
⎞1/2
xν |f (x)|2 dx⎠
, f ∈ Xν .
Rn
One can directly check that the dual space of Xν is as:
· Xν∗
Xν∗ = C0∞ (Rn )
,
with the norm ⎛ g Xν∗ := ⎝
⎞1/2 x−ν |g(x)|2 dx⎠
, g ∈ Xν∗ .
(3.91)
Rn
We let X := L2 (Rn ) = X ∗ , Y := L2 (Rn ) = Y ∗ and Z := Xν ,
(3.92)
where the space Xν is given by (3.90). Define two operators R : Z → X and O : Z → Y by Rz := eΔ(T −τ ) z for each z ∈ Xν ;
Oz := χB1 eΔ(T −τ ) z for each z ∈ Xν .
(3.93)
One can directly check that R∗ f = eΔ(T −τ ) f, ∀ f ∈ L2 (Rn );
O∗ h = eΔ(T −τ ) (χB1 h) = y(T, ·; 0, h), ∀ h ∈ L2 (Rn ).
(3.94)
By (ii) of Theorem 3.1, (3.93), (3.92) and (3.90), we can verify that for any ε > 0, Rz 2L2 (Rn )
1 (3| ln θ|+1) 1 ν ε 2 e ≤ C2 (ν, T − τ ) ε Oz L2 (Rn ) + e z 2Xν for each z ∈ Xν ,
(3.95)
where C2 (·, ·) and θ are given by (ii) of Theorem 3.1. Next, we arbitrarily fix ε > 0 and y0 ∈ L2 (Rn ). According to [43, Lemma 5.1] and (3.95), there exists ˜ y ,ε ∈ L2 (Rn ) so that h 0 1 ˜ y ,ε 2 ∗ + h Y 0 C2 (a, T − τ )ε
1 1 (3| ln θ|+1) 1 ν ε e
˜ y ,ε 2 ∗ ≤ eΔτ y0 2 ∗ . R∗ (eΔτ y0 ) − O∗ h Z X 0
C2 (a, T − τ )e
Since {eΔt }t≥0 is contractive in L2 (Rn ), from the above, (3.94), (3.92) and (3.91), we obtain the conclusion (ii) of this theorem. Hence, we end the proof of Theorem 3.4. 2
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Remark 3.6. Theorem 3.4 presents some approximate null controllability with a cost for the impulsively controlled heat equation (3.82). The conclusion (i) in this theorem says that given y0 ∈ L2 (Rn ) and ε > 0, there is a control hy0 ,ε ∈ L2 (Rn ), with the cost: C (a, T − τ ) − 4| ln θ| 12 1 eε a y0 L2 (Rn ) , ε X∗
so that the corresponding solution to (3.82) derives y0 at time 0 into Br a , with r = (C1 (a, T − 1 X∗ τ )ε) 2 y0 L2 (Rn ) , at time T . Here, Br a is the ball in Xa∗ (which is defined by (3.84)), centered at the origin and of radius r. In the similar manner, we can explain the conclusion (ii) in Theorem 3.4. The approximate null controllability with a cost for impulsively controlled heat equations in a bounded domain was studied in [37]. Similar controllability was studied for impulsively controlled Schrödinger equations in [43]. We end this subsection with introducing an application of Theorem 3.2. Let T > 0 and r > 0. Consider the controlled equation: /
∂t z − Δz = χBr v z(0, ·) = z0 (·)
in in
0, T ) × Rn , Rn ,
(3.96)
where z0 ∈ L2 (Rn ) is an initial state and v ∈ L2 ((0, T ) × Rn ) is a control. Write z(·, ·; z0 , v) for the solution to (3.96). By a standard duality method, we can easily obtain the next controllability result, with the aid of Theorem 3.2. (We will omit the detailed proof.) Theorem 3.5. Let T > 0 and r > r > 0. Then for each z0 ∈ L2 (Rn ), with supp z0 ⊂ Br , there is a control vz0 ∈ L2 ((0, T ) × Rn ), with
T
0 Rn
1
Cn + |vz0 (t, x)| dxdt ≤ T (r − r )2 2
|z0 (x)|2 dx,
Rn
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