On backward uniqueness for parabolic equations when Osgood continuity of the coefficients fails at one point

We prove the uniqueness for backward parabolic equations whose coefficients are Osgood continuous in time for t>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$t>0$$\end{document} but not at t=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$t=0$$\end{document}.


Introduction
Let us consider the following backward parabolic operator where all the coefficients are assumed to be defined in [0, T] × ℝ n , measurable and bounded; (a j,k (t, x)) j,k is a real symmetric matrix for all (t, x) ∈ [0, T] × ℝ n and there exists 0 ∈ (0, 1] such that for all (t, x) ∈ [0, T] × ℝ n and ∈ ℝ n . Given a functional space H , we say that the operator L has the H-uniqueness property if, whenever u ∈ H , Lu = 0 in [0, T] × ℝ n and u(0, x) = 0 in ℝ n , then u = 0 in [0, T] × ℝ n .
In the present paper, we are interested in the H-uniqueness property for the operator L defined in (1), when x j a j,k (t, x) x k + n ∑ j=1 b j (t, x) x j + c(t, x), (let us remark that this choice for H is, in some sense, natural, since, from elliptic regularity results, the domain of the operator − ∑ n

for all t ∈ [0, T]).
It is well known that, in dealing with the uniqueness property for partial differential operators, one of the main issues is the regularity of the coefficients. For example, in the case of elliptic operators, the uniqueness property in the case of Lipschitz continuous coefficients was proved by Hörmander in [14] (see [17] for a more refined result), while a famous non-uniqueness counterexample, for an elliptic operator having Hölder continuous coefficients, is due to Pliś (see [16]).
In [9,10], we investigated the problem of finding the minimal regularity assumptions on the coefficients a j,k ensuring the H-uniqueness property to (1). Namely, we proved the H -uniqueness property for (1) when the coefficients a j,k are Lipschitz continuous in x and the regularity in t is given in terms of a modulus of continuity , i.e., where satisfies the so-called Osgood condition A counterexample in [9], similar to that one of Pliś quoted here above, shows that, considering the regularity with respect to t for the a j,k , the Osgood condition is sharp: given any non-Osgood modulus of continuity , it is possible to construct a backward parabolic operator like (1), whose coefficients are C ∞ in x and -continuous in t, for which the H -uniqueness property does not hold.
It is interesting to remark that, in the recalled counterexample, the coefficients are in fact C ∞ in t for t ≠ 0 , and the Osgood continuity fails only at t = 0.
The loss of regularity for the coefficients at a single point is widely considered, e.g., in the case of well-posedness in the Cauchy problem for second-order hyperbolic operators of the type under the condition (2). For such class of operators, we have the well-posedness in Sobolev spaces when the coefficients are log-Lipschitz continuous with respect to t, there exist counterexamples to this property when the Lipschitz continuity fails only at t = 0 , and, finally, the well-posedness in Sobolev spaces can be recovered adding a control on the Lipschitz constant of the a j,k 's, for t going to 0 (the literature on such kind of problems is huge, see, e.g., [4-8, 13, 18]) In this paper, we show that if the loss of the Osgood continuity is properly controlled as t goes to 0, then the H-uniqueness property for (1) remains valid. Our hypothesis reads as follows: given a modulus of continuity satisfying the Osgood condition, we assume that the coefficients a j,k are Hölder continuous with respect to t on [0, T], and for all t ∈ (0, T] where 0 < < 1 . The coefficients a j,k are assumed to be globally Lipschitz continuous in x. Under such hypothesis, we prove that the H-uniqueness property holds for (1). As in [9,10], the uniqueness result is consequence of a Carleman estimate with a weight function shaped on the modulus of continuity . The weight function is obtained as solution of a specific second-order ordinary differential equation. In the previous results cited above, the corresponding o.d.e. is autonomous. Here, on the contrary, the time-dependent control (4) yields to a non-autonomous o.d.e. Also, the "Osgood singularity" of a j,k at t = 0 introduces a number of new technical difficulties which are not present in the fully Osgood-regular situation considered before. The result is sharp in the following sense: we exhibit a counterexample in which the coefficients a j,k are Hölder continuous with respect to t on [0, T], for all t ∈ (0, T] and for all > 0 and the operator (1) does not have the H-uniqueness property. The borderline case = 0 in (5) is considered in paper [11]. In such a situation, only a very particular uniqueness result holds and the problem remains essentially open.

Main result
We start with the definition of modulus of continuity. is decreasing.

Remark 4
The hypothesis (10), in particular the Hölder regularity with respect to t, is due to technical requirement for obtaining the Carleman estimate from which the main result is deduced. It does not seem easy to substitute it with different or weaker conditions.

Theorem 2 In the previous hypotheses, there exist
The way of obtaining the H-uniqueness from the inequality (17) is a standard procedure, the details of which can be found in [9,Par. 3.4].

Littlewood-Paley decomposition
We will use the so-called Littlewood-Paley theory. We refer to [2,3,15] and [1] for the details. Let ∈ C ∞ ([0, +∞), ℝ) such that is non-increasing and We set, for ∈ ℝ n , Given a tempered distribution u, the dyadic blocks are defined by 10 .
where we have denoted by F −1 the inverse of the Fourier transform. We introduce also the operator We recall some well-known facts on Littlewood-Paley deposition. where ‖a‖ Lip = ‖a‖ L ∞ + ‖∇a‖ L ∞.  Then, there exists C > 0 depending only on n and m such that for all u ∈ L 2 ,

Modified Bony's paraproduct
We end this subsection with a property which will needed in the proof of the Carleman estimate.
First of all, using Proposition 4, we fix a value for m in such a way that for all v ∈ C ∞ 0 ℝ n+1 such that Supp u ⊆ 0, T 2 × ℝ n . Next we use Proposition 3 and in particular from (22) we deduce that (26) will be a consequence of since the difference between (26) and (28) is absorbed by the right side part of (28) with possibly a different value of C and 0 . With a similar argument, using (19) and (25) , ∫ ℝ (s)ds = 1 , (s) ≥ 0 and (s) = 1 ( s ) . With a straightforward computation, from (10) and (11) where C depends only on n, m and ‖a j,k ‖ L ∞ and N > 0 can be chosen arbitrarily. Similarly where C depends only on n, m and ‖a j,k ‖ Lip and N > 0 can be chosen arbitrarily. As a conclusion, from (30), we finally obtain

End of the proof
We start considering (33) for h = 0 . We fix = 1 2 . Recalling (16) we have Choosing a suitable 0 , we have that, for all > 0 , Then, there exist 0 > 0 and C > 0 such that, for all > 0 and for all h ≥ 1,

Suppose finally that
From (15), the fact that 0 ≤ 1 and the properties of the modulus of continuity we obtain and Consequently Then, there exist 0 > 0 and C > 0 such that, for all > 0 and for all h ≥ 1, As a conclusion, from (34), (36) and (37), there exist 0 > 0 and C > 0 such that, for all > 0 and for all h ∈ ℕ, and (29) follows. The proof is complete.

A counterexample
Theorem 3 There exists with and there exist u, b 1 , b 2 , c ∈ C ∞ b (ℝ t × ℝ 2 x ) , with such that Remark 5 Actually, the function l will satisfy From (41) it is easy to obtain (40).
for some > 0 . Finally As a consequence (45), (46), (47) and (48) are satisfied for a suitable fixed j 0 . It remains to check (41). We have and consequently The conclusion of the theorem is reached simply exchanging t with −t . ◻