Annali di Matematica DOI 10.1007/s10231-013-0381-3
A new result on backward uniqueness for parabolic operators Daniele Del Santo · Martino Prizzi
Received: 2 April 2013 / Accepted: 8 October 2013 © Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag Berlin Heidelberg 2014
Abstract Using Bony’s paramultiplication, we improve a previous backward uniqueness result for parabolic operators having coefficients non-Lipschitz continuous with respect to t but C 2 with respect to x (see Del Santo and Prizzi in J. Math. Pures Appl. 84:471–491, 2005), showing that a similar result is valid when C 2 regularity in x is replaced by Lipschitz regularity in x. Keywords
Parabolic equations · Backward uniqueness · Bony’s paramultiplication
Mathematics Subject Classification (2010)
35A02 · 35K10 · 35S50
1 Introduction We consider the operator L = ∂t +
n
∂x j (a jk (t, x)∂xk ) + c(t, x).
(1.1)
j,k=1
We suppose that the coefficients are defined in [0, T ] × Rnx , measurable and bounded; we suppose moreover that the coefficients a jk are Lipschitz continuous with respect to the variables x, uniformly with respect to t, and we suppose finally that the matrix (a jk (t, x)) jk is real, symmetric, and uniformly positive definite in [0, T ] × Rnx , i.e., there exists λ0 ∈ (0, 1] such that n a jk (t, x)ξ j ξk ≥ λ0 |ξ |2 , j,k=1
for all (t, x) ∈
[0, T ] × Rnx
and ξ ∈ Rnξ . L is a backward parabolic operator.
D. Del Santo (B) · M. Prizzi Dipartimento di Matematica e Geoscienze, Università di Trieste, Via A. Valerio 12/1, 34127 Trieste, Italy e-mail:
[email protected]
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Given a functional space H (in which it makes sense to look for the solutions of the equation Lu = 0), we say that L has the H-uniqueness property if, whenever u ∈ H, Lu = 0 in [0, T ] × Rnx and u(0, x) = 0 in Rnx , then u = 0 in [0, T ] × Rnx . We choose H to be the space of functions H = H 1 ((0, T ), L 2 (Rnx )) ∩ L 2 ((0, T ), H 2 (Rnx )).
(1.2)
This choice is natural, since it follows from elliptic regularity results (see, e.g., Theorem 8.8 in [10]) that the domain of the operator nj,k=1 ∂x j (a jk (t, x)∂xk ) in L 2 (Rn ) is H 2 (Rn ) for all t ∈ [0, T ]. The problem we are interested in is the following: find the minimal regularity on the coefficients a jk , with respect to the variable t, ensuring to L the H-uniqueness property. A classical result of Lions and Malgrange ([11], see also [1] and [9]) shows that a sufficient condition for backward uniqueness is given by the assumption that the map t → a jk (t, ·) is Lipschitz continuous from [0, T ] to L ∞ (Rnx ). On the other hand, a well-known nonuniqueness example due to Miller ([14]), concerning an operator having coefficients that are Hölder continuous of order 1/6 with respect to t and Cb∞ with respect to x, shows that a certain amount of regularity on the a jk ’s, with respect to the variable t, is necessary for the H-uniqueness. In our previous paper [7], we proved the H-uniqueness property for the operator (1.1) when the coefficients a jk were C 2 in the variables x and the regularity in t was given in terms of a modulus of continuity μ satisfying the so-called Osgood condition 1 0
1 ds = +∞. μ(s)
(1.3)
This uniqueness result was a consequence of a Carleman estimate in which the weight function depended on the modulus of continuity; such kind of weight functions in Carleman estimates was introduced by Tarama in [16] in the case of second-order elliptic operators. In obtaining our Carleman estimate, the integrations by parts, which could not be used since the coefficients were not Lipschitz continuous, was replaced by a microlocal approximation procedure. From the point of view of the regularity with respect to t, the result in [7] was satisfactory: in fact, the cited regularity, with the modulus of continuity satisfying (1.3), cannot be weakened, as shown by an improvement of the example of Miller given by Mandache in [12] (see also [7] for another similar non-uniqueness example inspired by the well-known result of Pli´s for elliptic equations in [15]). On the other hand, the requirement of the C 2 regularity with respect to x was due to a technical difficulty in the estimate of a commutator. In the present paper, we improve the result of [7] showing that C 2 regularity in x can be replaced by Lipschitz regularity in x, giving in this way a complete answer to our problem. In order to achieve the result, we replace the second-order part of the operator L with a paradifferential operator (actually, only a modified Bony’s paramultiplication is concerned). We deduce a Carleman estimate in a space H −s , with 0 < s < 1, instead of the usual estimate in L 2 . However, with such an estimate, we can repeat the arguments of [7] and regain the desired uniqueness property. The estimate of the commutator, in the present case, is made more effective by a theorem due to Coifman and Meyer [3, Th. 35] (see also, for a similar use of that theorem, [6, Prop. 3.7]).
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2 Definitions and result Definition 2.1 A function μ is said to be a modulus of continuity if μ is continuous, concave, and strictly increasing on [0, 1], with μ(0) = 0 and μ(1) = 1. Let I be an interval in R and let ϕ : I → B, where B is a Banach space. We say that ϕ ∈ C μ (I, B) if ϕ ∈ L ∞ (I, B) and ϕ(t) − ϕ(s) B < +∞. μ(|t − s|)
sup t,s∈I 0<|t−s|≤1
It is immediate to verify the following properties • • • • •
μ(s) ≥ s for all s ∈ [0, 1]; the function s → μ(s)/s is decreasing on ]0, 1]; there exists lims→0+ μ(s)/s; the function σ → σ μ(1/σ ) is increasing on [1, +∞[; the function σ → 1/(σ 2 μ(1/σ )) is decreasing on [1, +∞[.
Definition 2.2 A modulus of continuity is said to satisfy the Osgood condition if 1 0
1 ds = +∞. μ(s)
(2.1)
Example It is easily verified that a modulus of continuity μα (s), which coincides near 0 with s(log(1/s))α , with α ∈ [0, 1], is a modulus of continuity satisfying the Osgood condition. In particular, when α = 0, C μα is the set of Lipschitz continuous functions while, if α = 1, C μα is the set of log-Lipschitz continuous ones. An immediate computation gives that μα does not satisfy the Osgood condition if α > 1 and the same is valid for νβ (s) = s β , for β ∈ (0, 1). Theorem 2.3 Let L be the operator L = ∂t +
n
∂x j (a jk (t, x)∂xk ) + c(t, x),
(2.2)
j,k=1
where all the coefficients are supposed to be defined in [0, T ]× Rnx , measurable, and bounded; let the coefficient c be complex valued; let (a jk (t, x)) jk be a real symmetric matrix for all (t, x) ∈ [0, T ] × Rnx and suppose that there exists λ0 ∈ (0, 1) such that n
a jk (t, x)ξ j ξk ≥ λ0 |ξ |2 ,
(2.3)
j,k=1
for all (t, x) ∈ [0, T ] × Rnx and ξ ∈ Rnξ . Let H be the space of functions H = H 1 ((0, T ), L 2 (Rnx )) ∩ L 2 ((0, T ), H 2 (Rnx )).
(2.4)
Let μ be a modulus of continuity satisfying the Osgood condition. Suppose that a jk ∈ C μ ([0, T ], L ∞ (Rnx )) ∩ L ∞ ([0, T ], Lip (Rnx )),
(2.5)
for all j, k = 1 . . . , n. Then, L has the H-uniqueness property, i.e., if u ∈ H, Lu = 0 in [0, T ] × Rnx and u(0, x) = 0 in Rnx , then u = 0 in [0, T ] × Rnx .
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Remark 2.4 As it will be clear from the proof, the result above is valid also for the operator L˜ = ∂t +
n
∂x j (a jk (t, x)∂xk ) +
j,k=1
n
b j (t, x)∂x j + c(t, x).
(2.6)
j=1
Nevertheless, the coefficients b j cannot be simply L ∞ as in the case of our previous result [7], but since the Carleman estimate is performed in H −s , s ∈ (0, 1) (see Proposition 3.1), it is necessary that n b j (t, x)∂x j v ≤ Cb ∇x v H −s j=1 −s H
for some s > 0. This inequality holds only under a certain amount of regularity on the coefficients b j , e.g., if the b j s are Hölder continuous of index σ > 0 with respect to x, uniformly with respect to t. A similar requirement for the coefficients of the first-order terms, in the different contest of energy inequalities, can be found in [6, Assumption 1.1, p. 179] and [5, hypothesis (13), p. 4]; let us remark that from the result in [6], a local uniqueness result is deduced.
3 Proof 3.1 Modulus of continuity and Carleman estimate Theorem 2.3 will follow from a Carleman estimate in Sobolev spaces with negative index. The weight function in the Carleman estimate will be obtained from the modulus of continuity. The crucial idea of linking the weight function to the regularity of the coefficients goes back to the paper [16] in which a uniqueness result for elliptic operators with non-Lipschitz continuous coefficients was proved. We define 1 φ(t) = 1 t
1 ds. μ(s)
The function φ is a strictly increasing C 1 function. From (2.1), we have φ([1, +∞[) = [0, +∞[; moreover, φ (t) = 1/(t 2 μ(1/t)) > 0 for all t ∈ [1, +∞[. We set τ (τ ) =
φ −1 (s) ds.
0
We obtain (τ ) = φ −1 (τ ) and consequently limτ →+∞ (τ ) = +∞. Moreover, 1 (τ ) = ( (τ ))2 μ , (τ )
(3.1)
for all τ ∈ [0, +∞[ and, as the function σ → σ μ(1/σ ) is increasing on [1, +∞[, we deduce that 1 = +∞. (3.2) lim (τ ) = lim ( (τ ))2 μ τ →+∞ τ →+∞ (τ )
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Now, we state the Carleman estimate. Proposition 3.1 For all s ∈ (0, 1), there exists γ0 , C > 0 such that T
2 e
2 γ (γ (T −t))
2 n ∂t u + ∂ (a (t, x)∂ u) x jk x j k j,k=1
dt
H −s
0 T
≥ Cγ
1 2
2
2
eγ
(γ (T −t))
1
( ∇x u 2H −s + γ 2 u 2H −s ) dt,
(3.3)
0
for all γ > γ0 and for all u ∈ C0∞ (Rn+1 ) such that supp u ⊆ [0, T /2] × Rnx (the symbol ∇x f denotes the gradient of f with respect to the x variables). The way of obtaining the H-uniqueness from the inequality (3.3) is a standard procedure, and the details, in the case of a Carleman estimate in L 2 , can be found in [7, Par. 3.4]. 3.2 Paraproducts 3.2.1 Littlewood–Paley decomposition We review some known results on Littlewood–Paley decomposition and related topics. More can be found in [2], [13, Ch. 4 and Ch. 5], and [6, Par. 3]. Let χ ∈ C0∞ (R), 0 ≤ χ ≤ 1, even and such that χ(s) = 1 for |s| ≤ 11/10 and χ(s) = 0 for |s| ≥ 19/10. For k ∈ N and ξ ∈ Rn , let us consider χk (ξ ) = χ(2−k |ξ |), let us denote χ˜ k (x) its inverse Fourier transform and let us define the operators S−1 u = 0, and Sk u = χ˜ k ∗ u = χk (Dx )u, 0 u = S0 u, and, for k ≥ 1, k u = Sk u − Sk−1 u. In the following propositions, we recall the characterization of Sobolev spaces and Lipschitz continuous functions via Littlewood–Paley decomposition (see [13, Prop. 4.1.11], [6, Prop. 3.1 and Prop. 3.2] and [8, Lemma 3.2]. Proposition 3.2 Let s ∈ R. A temperate distribution u is in H s if and only if the following two conditions hold i) for all k ≥ 0, k u ∈ L 2 ; ii) the sequence (δk )k , with δk = 2ks k u L 2 , is in l 2 . Moreover, there exists Cs ≥ 1 such that for all u ∈ H s , +∞ 1/2 1 u H s ≤ δk2 ≤ Cs u H s . Cs
(3.4)
k=0
Proposition 3.3 Let s ∈ R and R ≥ 2. Let (u k )k a sequence of functions in L 2 such that i) the support of the Fourier transform of u 0 is contained in {|ξ | ≤ R} and the support of Fourier transform of u k is contained in { R1 2k ≤ |ξ | ≤ R 2k }, for all k ≥ 1; ii) the sequence (δk )k , with δk = 2ks u k L 2 , is in l 2 .
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Then, the series k u k is converging, with sum u, in H s and the norm of u in H s is equivalent, in the sense of (3.4), to the norm of (δk )k in l 2 . When s > 0, it is sufficient to assume the Fourier transform of u k to be contained in {|ξ | ≤ R2k }, for all k ≥ 0. Proposition 3.4 A bounded function a is in Lip, and the space of bounded Lipschitz continuous functions defined on Rnx , if and only if sup ∇x (Sk a) L ∞ < +∞.
k∈N
Moreover, there exists C > 0 such that if a ∈ Lip, then k a L ∞ ≤ C a Lip 2−k and ∇x (Sk a) L ∞ ≤ C a Lip , where f Lip = f L ∞ + ∇x f L ∞ . 3.2.2 Bony’s modified paraproduct Let a ∈ L ∞ . The Bony’s paraproduct of a with u ∈ H s (see [2, Par. 2]) is defined as Ta u =
+∞
Sk−3 ak u.
k=3
We modify the definition of Bony’s paraproduct introducing the following operator +∞
Tam u = Sm−1 aSm+1 u +
Sk−3 ak u.
k=m+2
where m ∈ N (remark that Ta = Ta0 ). Useful properties of the (modified) paraproduct are contained in the following propositions (see also [13, Prop. 5.2.1], [6, Prop. 3.4]). Proposition 3.5 Let m ∈ N, s ∈ R and a ∈ L ∞ . Then, Tam maps H s into H s and Tam u H s ≤ Cm,s a L ∞ u H s .
(3.5)
Let m ∈ N, s ∈ (0, 1) and a ∈ Lip. Then, u → au − Tam u maps H −s into H 1−s and au − Tam u H 1−s ≤ Cm,s a Lip u H −s .
(3.6)
Proof We prove only the second part of the statement. We have ⎛ au − Tam u =
+∞ k=max{3,m}
k aSk−3 u +
+∞ k=m
⎞
⎜ ⎟ ⎜ k a j u ⎟ ⎝ ⎠. j≥0
| j−k|≤2
We remark that the support of the Fourier transform of k aSk−3 u is contained in {2k−2 ≤ |ξ | ≤ 2k+2 }. Moreover, by Proposition 3.4, we have that k aSk−3 u L 2 ≤ k a L ∞ Sk−3 u L 2 ≤ C a Lip 2−k
k−3 j=0
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2 js δ j ,
A new result on backward uniqueness
where δ j = 2− js j u L 2 . From Proposition 3.2, we have that (δ j ) j ∈ l 2 and its l 2 norm is equivalent to the H −s norm of u. On the other hand, setting δ˜k =
k
2−(k− j)s δ j ,
j=0
we have that (δ˜k )k ∈ l 2 and (δ˜k )k l 2 ≤ Cs (δk )k l 2 . Consequently, k aSk−3 u L 2 ≤ Cs a Lip 2−k(1−s) δ˜k−3 , 1−s with and then, by Proposition 3.3, we have that +∞ k=max{3,m} k aSk−3 u ∈ H +∞ aS u ≤ Cm,s a Lip u H −s . k k−3 k=max{3,m} 1−s
(3.7)
H
Next, we see that, for m ≥ 2, ⎞ ⎛ +∞ +∞ +∞ ⎟ ⎜ ⎟= ⎜ a u a u + · · · a + k ak+2 u, k j k k−2 ⎠ ⎝
k=m
j≥0
k=m
| j−k|≤2
k=m
with a slight modification in the case m = 0, 1. We have that the support of the Fourier transform of k ak−2 u is contained in {|ξ | ≤ 2k+2 } and similarly for the other four terms, e.g., the support of the Fourier transform of k ak+2 u is contained in {|ξ | ≤ 2k+4 }. Moreover, k ak−2 u L 2 ≤ k a L ∞ k−2 u L 2 ≤ Cs a Lip 2−k(1−s) δk−2 . 1−s and Again from Proposition 3.3, we have that +∞ k=m k ak−2 u ∈ H +∞ k−2 ak u ≤ Cm,s a Lip u H −s . k=m
H 1−s
Arguing similarly for the other terms, we have that ⎛ ⎞ +∞ ⎜ ⎟ 1−s ⎜ k a j u ⎟ ⎝ ⎠∈H
k=m
and
j≥0
| j−k|≤2
⎛ ⎞ +∞ ⎜ ⎟ ⎜ ⎟ k a j u ⎠ ⎝ k=m j≥0 | j−k|≤2
≤ Cm.s a Lip u H −s .
(3.8)
H 1−s
The conclusion of the proof of the proposition is reached putting together (3.7) and (3.8). As pointed out in [6, Par. 3], the positivity of the function a does not imply, for all m ≥ 0, the positivity of Tam . Nevertheless, the following proposition holds (see [6, Cor. 3.12]).
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Proposition 3.6 Let a ∈ Lip and suppose that a(x) ≥ λ0 > 0 for all x ∈ Rn . Then, there exists m depending on λ0 and a Lip such that, for all u ∈ L 2 , Re Tam u, u L 2 ≥
λ0 u L 2 , 2
(3.9)
where ·, · L 2 denotes the scalar product in L 2 . A similar result is valid for vector-valued functions when a is replaced by a positive symmetric matrix (a jk ) jk . We state now a property of commutation, which will be crucial in the proof of the Carleman estimate. Proposition 3.7 Let m ∈ N, a ∈ Li p, s ∈ R and u ∈ H 1−s . Then, +∞ 1/2 −2νs m 2 2 ∂x j ([ν , Ta ]∂xh u) L 2 ≤ Cm,s a Lip u H 1−s ,
(3.10)
ν=0
where [A, B] denotes the commutator between the operators A and B, i.e., [A, B]w = A(Bw) − B(Aw). Proof We start remarking that +∞
[ν , Tam ]w = [ν , Sm−1 a]Sm+1 w +
[ν , Sk−3 a]k w,
k=m+2
and consequently, ∂x j ([ν , Tam ]∂xh u) = ∂x j ([ν , Sm−1 a]Sm+1 (∂xh u)) +∞ [ν , Sk−3 a]k (∂xh u) . +∂x j k=m+2
In fact, ν and k commute, so that ν (Sm−1 aSm+1 w) − Sm−1 aSm+1 (ν w) = ν (Sm−1 aSm+1 w) − Sm−1 aν (Sm+1 w), similarly for the other term. Let us consider ∂x j ([ν , Sm−1 a]Sm+1 (∂xh u)) = ∂x j ([ν , Sm−1 a]∂xh (Sm+1 u). Looking at the support of the Fourier transform, it is easy to see that this term is identically equal to 0 if ν ≥ m + 4. Moreover, the support of the Fourier transform is contained in {|ξ | ≤ 2m+3 }. From Bernstein’s inequality, we have ∂x j ([ν , Sm−1 a]∂xh (Sm+1 u)) L 2 ≤ 2m+3 [ν , Sm−1 a]∂xh (Sm+1 u) L 2 . We remark that k , considered as a Fourier multiplier, is an operator of order 0 with symbol bounded independently of ν, so that, using the result of [3, Th. 35] (see also [17, Par. 3.6] and [6, Prop. 3.7]), we deduce that [ν , Sm−1 a]∂xh (Sm+1 u) L 2 ≤ C a Lip Sm+1 u L 2 . But Sm+1 u L 2 ≤ Cm,s u H 1−s ,
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A new result on backward uniqueness
so that ∂x j ([ν , Sm−1 a]Sm+1 (∂xh u)) L 2 ≤ Cm,s 2m+3 a Lip u H 1−s , and +∞ ν=0
2−2νs ∂x j ([ν , Sm−1 a]Sm+1 (∂xh u)) 2L 2 =
m+3
2−2νs ∂x j ([ν , Sm−1 a]Sm+1 (∂xh u)) 2L 2 ν=0 ≤ Cm,s a 2Lip u 2H 1−s . (3.11)
Let us consider +∞ +∞ ∂x j [ν , Sk−3 a]k (∂xh u) = ∂x j [ν , Sk−3 a]∂xh (k u) . k=m+2
k=m+2
Again looking at the support of the Fourier transform, it is possible to see that [ν , Sk−3 a]∂xh (k u) is identically 0 if |k − ν| ≥ 4. Consequently, the sum is on at most 7 terms: ∂x j ([ν , Sν−6 a]∂xh (ν−3 u))+· · ·+∂x j ([ν , Sν a]∂xh (ν+3 u)), each of them having the support of the Fourier transform contained in {|ξ | ≤ C2ν }. Let us consider one of these terms, e.g., ∂x j ([ν , Sν−3 a]∂xh (ν u)). We have, from Bernstein’s inequality ∂x j ([ν , Sν−3 a]∂xh (ν u)) L 2 ≤ C2ν [ν , Sν−3 a]∂xh (ν u) L 2 . and consequently, using again [3, Th. 35], ∂x j ([ν , Sν−3 a]∂xh (ν u)) L 2 ≤ C2ν a Lip ν u L 2 . Since u ∈ H 1−s and, consequently, the sequence 2ν(1−s) ν u L 2 ν is in l 2 , the same is −νs valid for 2 ∂x j ([ν , Sν−3 a]∂xh (ν u)) L 2 ν and +∞ ν=0
2−2νs ∂x j ([ν , Sν−3 a]∂xh (ν u)) 2L 2 ≤ Cm,s a 2Lip u 2H 1−s .
The computation of the other terms being similar, we obtain +∞ 2 +∞ −2νs 2 [ν , Sk−3 a]∂xh (k u) ≤ Cm,s a 2Lip u 2H 1−s . (3.12) ∂ x j ν=0
k=m+2
L2
The estimate (3.10) follows from (3.11) and (3.12). We end this subsection with a result on the adjoint of Tam (see [6, Prop. 3.8 and Prop. 3.11]). Proposition 3.8 Let m ∈ N, a ∈ Lip and u ∈ L 2 . Then, (Tam − (Tam )∗ )∂x j u L 2 ≤ Cm a Lip u L 2 .
(3.13)
Proof We remark that (Tam − (Tam )∗ )∂x j u = [Sm−1 a, Sm+1 ]∂x j u +
+∞
[Sk−3 a, k ]∂x j u.
k=m+2
From [3, Th. 35], we deduce that [Sm−1 a, Sm+1 ]∂x j u L 2 ≤ C ∇x (Sm−1 a) L ∞ u L 2 ,
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and hence, from Proposition 3.4, we obtain [Sm−1 a, Sm+1 ]∂x j u L 2 ≤ C a Lip u L 2 .
(3.14)
On the other hand, we have that the support of Fourier transform of [Sk−3 a, k ]∂x j u is contained in {2k−2 ≤ |ξ | ≤ 2k+2 }. Moreover, it is easy to see that [Sk−3 a, k ]∂x j u = [Sk−3 a, k ]∂x j ((k−2 + k−1 + k + k+1 + k+2 )u). Again from [3, Th. 35] and Proposition 3.4, we have [Sk−3 a, k ]∂x j u L 2 = [Sk−3 a, k ]∂x j ((k−2 + · · · + k+2 )u) L 2 ≤ C a Lip ( k−2 u L 2 + · · · + k+2 u L 2 ). From Proposition 3.3, we finally obtain that
+∞
+∞ [Sk−3 a, k ]∂x j u k=m+2
k=m+2 [Sk−3 a, k ]∂x j u
∈ L 2 and
≤ Cm a Lip u L 2 .
(3.15)
L2
The estimate (3.13) follows from (3.14) and (3.15). 3.3 Approximation and Carleman estimate 1
(γ (T −t))
We set v(t, x) = e γ exists γ0 , C > 0 such that
u(t, x). The inequality (3.3) becomes: for all s ∈ (0, 1), there
T 2 2 n ∂t v + ∂ (a (t, x)∂ v) + (γ (T − t))v xj jk xk j,k=1
dt
H −s
0
T
1
2
≥ Cγ 2
1
( ∇x v 2H −s + γ 2 v 2H −s ) dt,
(3.16)
0
for all γ > γ0 and for all v ∈ C0∞ (Rn+1 ) such that supp v ⊆ [0, T /2] × Rn . Using the Proposition 3.6, we fix the parameter m in such a way that the modified paraproduct associated with the matrix (a jk ) jk satisfies (3.9). From the second part of the Proposition 3.5 (estimate (3.6)), the inequality (3.16) will be deduced from the following T 2 2 n m ∂t v + ∂x j (Ta jk ∂xk v) + (γ (T − t))v j,k=1
dt
H −s
0
T
≥ Cγ
1 2
2
1
( ∇x v 2H −s + γ 2 v 2H −s ) dt,
(3.17)
0
T /2 as the extra term from 0 nj,k=1 ∂x j ((a jk − Tamjk )∂xk v) 2H −s dt can be absorbed by the right-hand side part of (3.16), possibly taking different C and γ0 .
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A new result on backward uniqueness
Let us go back to the Littlewood–Paley decomposition. Denoting k u by u k a, the estimate (3.4) reads now 1 −2νs 2 u ν 2L 2 ≤ u 2H −s ≤ C−s 2−2νs u ν 2L 2 C−s ν ν for all u ∈ H −s . We have T 2 2 n m ∂t v + ∂x j (Ta jk ∂xk v) + (γ (T − t))v j,k=1
dt
H −s
0
T
2
1 ≥ C−s
ν
0
2 n −2νs m 2 ∂x j (Ta jk ∂xk v) + (γ (T − t))v) dt ν (∂t v + 2 j,k=1
n 1 −2νs ∂t vν + 2 ∂x j (Tamjk ∂xk vν ) + (γ (T − t))vν ≥ C−s ν j,k=1 0 2 n m + ∂x j ([ν , Ta jk ]∂xk v) dt 2 j,k=1
L
T 2
L
T
1 2C−s
≥
2 ν
0
2 n −2νs m 2 ∂x j (Ta jk ∂xk vν ) + (γ (T − t))vν ∂t vν + dt 2 j,k=1 L
2 n 1 −2νs m 2 ∂x j ([ν , Ta jk ]∂xk v) − dt. C−s 2 j,k=1 ν T 2
0
L
From the result of Proposition 3.7 it is immediate that (3.17) will be deduced from the same estimate from below for T 2 2 n m dt, ∂ 2−2νs v + ∂ (T ∂ v ) + (γ (T − t))v x j a jk xk ν ν t ν 2 ν j,k=1 0
L
possibly taking different C and γ0 . We obtain T
2 0
ν
2
−2νs
2 m ∂t vν + dt ∂ (T ∂ v ) + (γ (T − t))v x j a jk xk ν ν 2 jk L
2 −2νs 2 m 2 ( ∂t vν L 2 + ∂ (T ∂ v ) + (γ (T − t))v = x j a jk xk ν ν jk 2 ν 0 L 2 m + γ (γ (T − t)) vν L 2 + 2 Re ∂t vν , ∂x j (Ta jk ∂xk vν ) L 2 ) dt. T 2
jk
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D. Del Santo, M. Prizzi
We approximate the last term in the above equality using awell-known technique that goes back to [4]. Let ρ ∈ C0∞ (R) with supp ρ ⊆ [−1/2, 1/2], R ρ(s) ds = 1 and ρ(s) ≥ 0 for all s ∈ R; we set t −s 1 ds a jk, ε (t, x) = a jk (s, x) ρ ε ε R
for ε ∈ ]0, 1/2]. We obtain from (2.5) that there exists C such that |a jk, ε (t, x) − a jk (t, x)| ≤ Cμ(ε)
(3.18)
and |∂t a jk, ε (t, x)| ≤ C
μ(ε) , ε
(3.19)
for all j, k = 1 . . . , n and for all (t, x) ∈ [0, T ] × Rnx . We have
T
2
2 Re ∂t vν ,
∂x j (Tamjk ∂xk vν )
jk
0
dt L2
T
2 ∂x j ∂t vν , Tamjk ∂xk vν L 2 dt = −2 Re jk
0 T 2
∂x j ∂t vν , (Tamjk − Tamjk, ε )∂xk vν L 2 dt = −2 Re jk
0 T 2
−2 Re ∂x j ∂t vν , Tamjk, ε ∂xk vν L 2 dt. jk
0
We remark that Tamjk − Tamjk, ε = Tamjk −a jk, ε and consequently, from (3.5) and (3.18), we have that (Tamjk − Tamjk, ε )∂xk vν L 2 = Tamjk −a jk, ε ∂xk vν L 2 ≤ Cμ(ε) ∂xk vν L 2 . Moreover, ∂x j vν L 2 ≤ 2ν+1 vν L 2 and ∂x j ∂t vν L 2 ≤ 2ν+1 ∂t vν L 2 for all ν ∈ N. Hence, T 2 m m ∂x j ∂t vν , (Ta jk − Ta jk, ε )∂xk vν L 2 dt 2 Re 0 jk T
≤ 2Cμ(ε)
2 0
C ≤ N
T 2
T
2 ∂t vν 2L 2
0
123
∂x j ∂t vν L 2 ∂xk vν L 2 dt
jk
dt + C N 2
4(ν+1)
μ(ε)
vν 2L 2 dt 0
A new result on backward uniqueness
for all N > 0 (note that μ(ε)2 ≤ μ(ε)). On the other hand, ∂t (Tamjk, ε w) = T∂t a jk, ε w + Ta jk, ε ∂t w, then, using also the fact that the matrix (a jk ) jk is real and symmetric, T
T
2 2 m −2 Re ∂x j ∂t vν , Ta jk, ε ∂xk vν L 2 dt = (∂x j vν , T∂mt a jk, ε ∂xk vν L 2 jk
0
0
jk
+ ∂x j vν , (Tamjk, ε − (Tamjk, ε )∗ )∂xk ∂t vν L 2 )dt. From (3.5) and (3.19), we deduce T T 2 2 m 2(ν+1) μ(ε) ∂x j vν , T∂t a jk, ε ∂xk vν L 2 dt ≤ C 2 vν 2L 2 dt, ε 0 jk 0 and, from (3.13) and (3.18), T 2 m m ∗ ∂x j vν , (Ta jk, ε − (Ta jk, ε ) )∂xk ∂t vν L 2 )dt 0 jk T
2 ≤ Cμ(ε)
∇x vν L 2 ∂t vν L 2 dt 0
C ≤ N
T 2
T
2 ∂t vν 2L 2 dt + C N 22(ν+1)
0
vν 2L 2 dt
(3.20)
0
for all N > 0. Choosing suitably N, we finally obtain T
2 ∂t vν +
∂x j (Tamjk ∂xk vν ) + (γ (T − t))vν 2L 2 dt
jk
0 T 2
≥ 0
⎛ ⎝
∂x j (Tamjk ∂xk vν ) + (γ (T − t))vν 2L 2 + γ (γ (T − t)) vν 2L 2
jk
− C(24(ν+1) μ(ε) + 22(ν+1)
⎞ μ(ε) ) vν 2L 2 ⎠ dt, ε
where the last term in (3.20) is absorbed by C 22(ν+1)
(3.21)
μ(ε) 2 0 ε T
vν 2L 2 dt since
μ(ε) ε
≥ 1.
3.4 End of the proof of the Carleman estimate From now on, the proof is exactly the same as in [7, Par. 3.2]. We detail it for reader’s convenience.
123
D. Del Santo, M. Prizzi
Let ν = 0. From (3.2), we can choose γ0 > 0 such that (γ (T − t)) ≥ 1 for all γ > γ0 and for all t ∈ [0, T /2]. Taking now ε = 1/2, we obtain from (3.21) that T 2 2 m ∂t v0 + dt ∂ (T ∂ v ) + (γ (T − t))v x x 0 0 a jk k j 2 jk
0
L
T 2
γ − 24Cμ
≥
1 v0 2L 2 dt, 2
0
for all γ > γ0 . Possibly choosing a larger γ0 , we have, again for all γ > γ0 , T T 2 2 2 γ m ∂t v0 + ∂x j (Ta jk ∂xk v0 ) + (γ (T − t))v0 v0 2L 2 dt. ≥ 2 2 jk
0
L
(3.22)
0
Let now ν ≥ 1. We take ε = 2−2ν . We obtain from (3.21) that T 2 2 m ∂t vν + dt ∂ (T ∂ v ) + (γ (T − t))v x j a jk xk ν ν 2 jk 0 L T ⎛ 2 2 ⎜ m ∂ (T ∂ v ) + (γ (T − t))v ≥ ⎝ x j a jk xk ν ν 2 jk 0 L 2 4ν −2ν +γ (γ (T − t)) vν L 2 − K 2 μ(2 ) vν 2L 2 dt T ⎛⎛ ⎞2 2 ⎜⎝ ⎠ ∂x j (Tamjk ∂xk vν ) ≥ ⎝ − (γ (T − t)) vν L 2 2 jk 0 L 2 + γ (γ (T − t)) vν L 2 − K 24ν μ(2−2ν ) vν 2L 2 dt,
where K = 24C. On the other hand, from (3.9), recalling that in this case ∇vν ≥ 2ν−1 vν , we have ∂x j (Tamjk ∂xk vν ) ∂x j (Tamjk ∂xk vν ), vν vν L 2 ≥ jk 2 jk 2 L L λ m λ0 2ν 0 ≥ Ta jk ∂xk vν , ∂x j vν 2 ≥ ∇vν 2L 2 ≥ 2 vν 2L 2 . (3.23) L 2 8 jk
Suppose first that (γ (T − t)) ≤ λ160 22ν . Then, from (3.23), we deduce that λ0 2ν m ∂x j (Ta jk ∂xk vν ) − (γ (T − t)) vν L 2 ≥ 16 2 vν L 2 jk 2 L
123
A new result on backward uniqueness
and then, using also the fact that (γ (T − t)) ≥ 1, we obtain that T
2 0
⎛⎛ ⎜⎝ ∂x j (Tamjk ∂xk vν ) ⎝ jk
⎞2 − (γ (T − t)) vν L 2 ⎠ L2
⎞
+γ (γ (T − t)) vν 2L 2 − K 24ν μ(2−2ν ) vν 2L 2 ⎠ dt 2 T
≥
λ0 2ν 2 16
2 +γ −K 2
4ν
μ(2
−2ν
) vν 2L 2
dt
0 T 2 2 1 λ0 γ −2ν 4ν ≥ − K μ(2 ) 2 + vν 2L 2 dt 2 16 3
0
2 2 1 λ0 2 4ν 2 + γ vν 2L 2 dt. + 2 16 3 T
0
Since limν→+∞ μ(2−2ν ) = 0, there exists γ0 > 0 such that 1 λ0 2 γ −2ν − K μ(2 ) 24ν + ≥ 0, 2 16 3 for all γ ≥ γ0 and for all ν ≥ 1. Consequently, there exists γ0 and c > 0 not depending on ν such that T
2 0
⎛⎛ ⎜⎝ m ∂ (T ∂ v ) ⎝ x j a jk xk ν jk
+γ (γ (T
⎞2 − (γ (T − t)) vν L 2 ⎠
L2
− t)) vν 2L 2
− K 24ν μ(2−2ν ) vν 2L 2
dt
2 1 1 λ0 γ 2 4ν 2 + γ vν 2L 2 dt ≥ ≥ + cγ 2 22ν vν 2L 2 dt, 2 16 3 2 T 2
T 2
0
(3.24)
0
for all γ ≥ γ0 . If on the contrary (γ (T − t)) ≥ properties of μ,
λ0 16
22ν , then, using (3.1), the fact that λ0 ≤ 1 and the
1 (γ (T − t)) 2 2 λ0 16 −2ν λ0 ≥ 24ν μ 2 24ν μ(2−2ν ). ≥ 16 λ0 16
(γ (T − t)) = ( (γ (T − t))2 μ
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D. Del Santo, M. Prizzi
Hence, also in this case, there exists γ0 and c > 0 such that T ⎛⎛ ⎞2 2 ⎜⎝ ⎠ ∂x j (Tamjk ∂xk vν ) ⎝ − (γ (T − t)) vν L 2 2 jk 0 L +γ (γ (T − t)) vν 2L 2 − K 24ν μ(2−2ν ) vν 2L 2 dt 2 T
≥
γ + 2
γ 2
λ0 16
2 −K
2 μ(2 4ν
−2ν
) vν 2L 2 dt
0 T
≥
2
γ + cγ 22ν vν 2L 2 dt, 2
(3.25)
0
for all γ ≥ γ0 and for all ν ≥ 1. Putting together (3.24) and (3.25), we have that there exists γ0 and c > 0 such that T 2 2 m ∂t vν + dt ∂ (T ∂ v ) + (γ (T − t))v x x ν ν a jk k j 2 jk
0
L
T 2
≥
1 γ + cγ 2 22ν vν 2L 2 dt, 2
(3.26)
0
for all ν ≥ 1 and for all γ ≥ γ0 . Form (3.22) and (3.26), we get that there exists γ0 and c > 0 such that T
2 0
2−2νs ∂t vν +
ν
∂x j (Tamjk ∂xk vν ) + (γ (T − t))vν 2L 2 dt
jk T 2
≥ cγ
1 2
0
ν
1
2−2νs ( ∇vν 2L 2 + γ 2 vν 2L 2 ) dt,
(3.27)
for all γ ≥ γ0 . Acknowledgments The authors would like to thank Prof. Marius Paicu for useful and interesting discussions on topics related to this paper.
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