The unique continuation property for second order evolution PDEs

Abstract

We present a simple and self-contained approach to establish the unique continuation property for some classical evolution equations of second order in a cylindrical domain. We namely discuss this property for wave, parabolic and Schödinger operators with time-independent principal part. Our method is builds on two-parameter Carleman inequalities combined with unique continuation across a pseudo-convex hypersurface with respect to the space variable. The most results we demonstrate in this work are more or less classical. Some of them are not stated exactly as in their original form.

1 Introduction

Let D be a domain of ${\mathbb {R}}^d$, $d\ge 1$. We recall that $f\in C^\infty (D)$ is said real-analytic if its Taylor series around any arbitrary point of D converges in a ball centered at this point. It is known that a real analytic function possesses the unique continuation property which means that if f vanishes in a nonempty open subset $D_0$ of D then f must vanishes identically (see for instance [22, Theorem, page 65]). Functions satisfying this unique continuation property are also called quasi-analytic.

A classical result shows that a strong solution of an elliptic operator with smooth principal coefficients is quasi-analytic. In the present work we consider the analogue of this property for second order evolution PDEs. The right property in this context should be the following: if a solution of an evolution equation of second order in the cylindrical domain $D\times (t_1,t_2)$ vanishes in $D_0\times (t_1,t_2)$, for some nonempty open subset $D_0$ of D, then this solution must vanishes identically. Unfortunately this property holds only in the parabolic case. Less optimal result still holds for the Schödinger case. The worst case is that for wave equations for which we have only a weak version of unique continuation property, even for a large time interval.

We provide a simple and self contained approach to show the unique continuation property for wave, parabolic and Schrödinger equations. The approach we carry out is quite classical and it is based on two-parameter Carleman inequalities. The unique continuation property in each case is obtained as a consequence of the property of unique continuation across a non characteristic hypersurface satisfying in addition a pseudo-convexity condition in the case of wave and Schrödinger equations.

The core of our analysis consists in establishing two-parameter Carleman inequalities. We follow a classical scheme for obtaining these $L^2$-weighted energy estimates, essentially based on conjugating the original operator with a well chosen exponential function, splitting the resulting operator into its self-adjoint part and skew-adjoint part and finally making integrations by parts. The main assumption on the weight function is a pseudo-convexity condition with respect to the operator under consideration. A systematic approach was considered by Hörmander [19, Section 28.2, page 234] for a general operator P of an arbitrary order m where the pseudo-convexity condition is expressed in term of the principal symbol of P. The method we develop in this work is more simple and does not appeal to fine analysis of PDEs and our results have no pretension nor for generality neither for optimality.

It is worth remarking that splitting the conjugated operator into self-adjoint and skew-adjoint parts is not the best possible way to get two-parameter Carleman inequalities for elliptic and parabolic operators. There is a particular way to split the conjugated operator into two parts. However this particular decomposition is not applicable for wave and Schrödinger equations. But this is not really surprising since solutions of wave and Schrödinger equations do not enjoy the same regularity properties of solutions of elliptic and parabolic equations.

We choose to start with the more subtle case corresponding to the wave equation. Since most calculations to get Carleman inequalities are common for different type of equations, Carleman inequalities for parabolic and Schrödinger equations are obtained by making some modifications in the proof of the Carleman inequality for the wave equation. We also added a short section for the elliptic case whose analysis is almost similar to that of the parabolic case.

One can find in the literature two-parameter Carleman inequalities with degenerate weight function. We refer for instance to [8, 12, 13, 15, 17, 25] for parabolic operators and [3,4,5, 26, 27] for Schrödinger operators.

A Carleman inequality for wave equations on compact Riemannian manifold can be found in [7] and quite recently Huang [20] proved a Carleman inequality for a general wave operators with time-dependent principal part. The interested reader is referred to [16] for a unified approach establishing Carleman inequalities for second order PDEs and their applications to control theory and inverse problems.

We point out that it is possible to derive Carleman inequalities with single parameter in the weight function. We refer for instance to [18, 19, 23, 38, 39] and references therein for more details. Introducing a second parameter in the Carleman weight is a simple way to guarantee pseudo-convexity condition as it is remarked in [18]. It certainly contributed to the development of Carleman inequalities with two-parameter. Two-parameter Carleman inequalities appear to be more flexible then one-parameter Carleman inequalities since we have to our disposal two parameters. Of course, one-parameter Carleman inequalities can be deduced from two-parameter Carleman inequalities (see for instance Theorem 3.3 for the case of the wave equation).

The unique continuation property for elliptic and parabolic operators with unbounded lower order coefficients was obtained in [32,33,34,35]. The analysis in these references combine both classical tools used for establishing the property of unique continuation together with interpolation inequalities. Uniqueness and non-uniqueness for general operators were discussed in [1, 2, 40] (see also the references therein). We also mention [21, 28] as additional references on uniqueness of Cauchy problems.

We also discuss briefly observability inequalities which can be seen as the quantification of unique continuation for the Cauchy problem associated to IBVPs. We refer to [24] for general observability inequalities for wave and Schrödinger equations with arbitrary interior or boundary observation region. The reader can find in this work a detailed introduction to explain the main steps to get the property of unique continuation for an intermediate case between Holmgren’s analytic case and Hörmander’s general case for operators with partially analytic coefficients.

A more difficult problem consists in quantifying the property of unique continuation from an interior subdomain or the Cauchy data on a sub-boundary. The elliptic case is now almost completely solved with optimal results for $C^{1,\alpha }$-solutions and $C^{0,1}$-domains [11] or $H^2$-solutions and $C^{1,1}$-domains [9]. A non optimal result for $H^2$-solutions and $C^{0,1}$-domains was obtained in [10]. These kind of results can be obtained by a method based on three-sphere inequality which is deduced itself from a Carleman inequality. The case of parabolic and wave equations is extremely more difficult than the elliptic case. Concerning parabolic equations, a first result was obtained in [8] with Cauchy data in a particular subboundary. This result is based on a global Carleman inequality. The general case was tackled in [13] where a non optimal result was established using a three-cylinder inequality. A result for the wave equation was recently proved in [6]. This result was obtained via Fourier–Bros–Iagolnitzer transform allowing to transfer the quantification of unique continuation of an elliptic equation to that of the wave equation.

2 Preliminaries

2.1 Main notations and assumptions

Throughout $\Omega$ is bounded Lipschitz domain of ${\mathbb {R}}^n$, $n\ge 2$, with boundary $\Gamma$. The unit normal exterior vector field on $\Gamma$ is denoted by $\nu$.

We set $Q=\Omega \times (t_1,t_2)$ and $\Sigma =\Gamma \times (t_1,t_2)$, where $t_1,t_2\in {\mathbb {R}}$ are fixed so that $t_1<t_2$.

$A=(a_{k\ell })$ will denote a symmetric matrix with coefficients $a_{k\ell }\in C^{2,1}({\overline{\Omega }})$, $1\le k,\ell \le n$, and there exist two constants ${\mathfrak {m}}>0$ and $\varkappa \ge 1$ so that

$$\begin{aligned} \varkappa ^{-1}|\xi | ^2\le \sum _{k,\ell =1}^na_{k\ell }(x)\xi _\ell \xi _k \le \varkappa |\xi |^2,\quad x\in \Omega ,\quad \xi \in {\mathbb {R}}^n, \end{aligned}$$

and

$$\begin{aligned} \Vert A\Vert _{C^{2,1}({\overline{\Omega }} ;{\mathbb {R}}^{n\times n})}\le {\mathfrak {m}}. \end{aligned}$$

The set of such matrices will denoted in the rest of this text by ${\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$.

It is worth mentioning that, according to Rademacher’s theorem, $C^{2,1}({\overline{\Omega }})$ is continuously embedded in $W^{3,\infty }(\Omega )$.

For $\xi ,\eta \in {\mathbb {C}}^n$, $(\xi |\eta )$ and $\xi \otimes \eta$ are defined as usual respectively by

$$\begin{aligned} (\xi |\eta )=\sum _{k=1}^n \xi _k\overline{\eta _k} \quad \text{ and }\quad \xi \otimes \eta =(\xi _k\eta _\ell )_{1\le k,\ell \le n}. \end{aligned}$$

The Jacobian matrix and the Hessian matrix are denoted respectively by

$$\begin{aligned} U'&=(\partial _\ell U_k),\quad U\in H^1(\Omega ;{\mathbb {C}}^n), \\ \nabla ^2 u&=(\partial _{k\ell }^2u),\quad u\in H^2(\Omega ;{\mathbb {C}}). \end{aligned}$$

The norm (resp. the scalar product) of a Banach (resp. a Hilbert) space E is always denoted by $\Vert \cdot \Vert _E$ (resp. $\langle \cdot |\cdot \rangle _E$).

We recall that the anisotropic Sobolev space $H^{2,1}(Q)$ is defined as follows

$$\begin{aligned} H^{2,1}(Q)=L^2((t_1,t_2);H^2(\Omega ))\cap H^1((t_1,t_2);L^2(\Omega )). \end{aligned}$$

The following notations will be useful in the sequel, where $S=\Omega$ or $S=\Gamma$,

$$\begin{aligned} \nabla _Au(x)&=A(x)^{1/2}\nabla u(x), \quad u\in H^1(\Omega ;{\mathbb {C}}), \\ \text{ div}_AU(x)&=\text{ div }(A(x)^{1/2}U(x)),\quad U \in H^1(S ;{\mathbb {C}}^n), \\ \Delta _Au(x)&=\text{ div}_A\nabla _Au(x)=\text{ div }(A(x)\nabla u(x)),\quad u\in H^2(\Omega ;{\mathbb {C}}), \\ (U|V)_A(x)=(A(x)U(x)|V(x))&=(U(x)|A(x)V(x)),\quad U,V \in L^2(S ;{\mathbb {C}}^n), \\ |U|_A(x)&=[(U|U)_A(x)]^{1/2},\quad U\in L^2(S ;{\mathbb {C}}^n). \end{aligned}$$

We readily obtain from the above definitions the following identity

$$\begin{aligned} (\nabla _Au|\nabla _Av) =(\nabla u|\nabla v)_A,\quad u,v\in H^1(\Omega ;{\mathbb {C}}). \end{aligned}$$

Furthermore, the Green type formula

$$\begin{aligned} \int _\Omega \Delta _Au {\overline{v}}{ d}x&=-\int _\Omega (\nabla _Au|\nabla _Av){ d}x+\int _\Gamma (\nabla u|\nu )_Av{ d}\sigma \nonumber \\&=-\int _\Omega (\nabla u|\nabla v)_A{ d}x+\int _\Gamma (\nabla u|\nu )_Av{ d}\sigma , \end{aligned}$$

(2.1)

holds for any $u\in H^2(\Omega ;{\mathbb {C}} )$ and $v\in H^1(\Omega ;{\mathbb {C}})$.

We need also to introduce the following notations

$$\begin{aligned} {\mathcal {L}}_{A,0}^e&=\Delta _A \quad { (elliptic\, operator)}, \\ {\mathcal {L}}_{A,0}^p&=\Delta _A-\partial _t\quad \; { (parabolic\, operator)}, \\ {\mathcal {L}}_{A,0}^w&=\Delta _A-\partial _t^2\quad { (wave\, operator)}, \\ {\mathcal {L}}_{A,0}^s&=\Delta _A+i\partial _t\quad ({ Schr}\ddot{{ o}}{ dinger\,operator}). \end{aligned}$$

The $n\times n$ identity matrix will denoted by $\mathbf{I }$.

We shall use for notational convenience the following notation

$$\begin{aligned} {[}h]_{t=t_1}^{t_2}=h(\cdot ,t_2)-h(\cdot ,t_1),\quad h\in H^1((t_1,t_2);L^2(\Omega )). \end{aligned}$$

Finally, we equip $\partial Q$ with following measure

$$\begin{aligned} { d}\mu (x,t) =\mathbbm {1}_{\Gamma \times (t_1,t_2)}(x,t){ d}\sigma (x){ d}t +\mathbbm {1}_{\Omega \times \{t_1,t_2\}}(x,t){ d}x\delta _t, \end{aligned}$$

where ${ d}\sigma (x)$ is the Lebesgue measure on $\Gamma$ and $\delta _t$ is the Dirac measure at t.

We used above $\mathbbm {1}_X$ to denote the characteristic function of the measurable set X:

$$\begin{aligned} \mathbbm {1}_X(x)=\left\{ \begin{array}{ll} 1\quad &{}\text{ if }\; x\in X,\\ 0\quad &{}\text{ if }\; x\not \in X.\end{array} \right. \end{aligned}$$

2.2 Pseudo-convexity condition

Define, for $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$,

$$\begin{aligned} \Lambda _{k\ell }^m(A)(x)=-\sum _{p=1}^n\partial _pa_{k\ell }(x)a_{pm}(x)+2\sum _{p=1}^na_{kp}(x)\partial _pa_{\ell m}(x), \end{aligned}$$

where $x\in {\overline{\Omega }}$ and $1\le k,\ell ,m\le n$.

We associate to $h\in C^1({\overline{\Omega }})$ the matrix $\Upsilon _A (h)$ given by

$$\begin{aligned} (\Upsilon _A (h))_{k\ell }(x)=\sum _{m=1}^n\Lambda _{k\ell }^m(A)(x)\partial _m h (x),\quad x\in {\overline{\Omega }} . \end{aligned}$$

Note that $\Upsilon _A (h)$ is not necessarily symmetric.

Inspired by the definition introduced in [19, Section 28.2, page 234] we consider the following one:

Definition 2.1

We say that $h\in C^2({\overline{\Omega }})$ is A-pseudo-convex with constant $\kappa >0$ in $\Omega$ if $\nabla h(x)\ne 0$ for any $x\in {\overline{\Omega }}$ and if

$$\begin{aligned} (\Theta _A(h)(x)\xi |\xi )\ge \kappa |\xi |^2,\; x\in {\overline{\Omega }},\; \xi \in {\mathbb {R}}^n, \end{aligned}$$

where

$$\begin{aligned} \Theta _A(h)=2A\nabla ^2hA+\Upsilon _A(h). \end{aligned}$$

It worth noticing that $A\rightarrow \Theta _A$ is positively homogenous of degree two:

$$\begin{aligned} \Theta _{\lambda A}=\lambda ^2\Theta _A,\quad \lambda >0. \end{aligned}$$

Since $\Theta _{\mathbf{I }}(h)=2\nabla ^2h$, h is $\mathbf{I }$-pseudo-convex in $\Omega$ if $\nabla h(x)\ne 0$ and $\nabla ^2h(x)$ is positive definite for any $x\in {\overline{\Omega }}$. In other words, when $A=\mathbf{I }$ pseudo-convexity is reduced to local strict convexity.

2.3 Carleman weights

It will be convenient to define the notion of Carleman weight for different kind of operators we are interested in. In the rest of this paper $\psi =\psi (x,t)$ is a function of the form

$$\begin{aligned} \psi (x,t)=\psi _0(x)+\psi _1(t),\quad (x,t)\in Q, \end{aligned}$$

and $\phi =e^{\lambda \psi }$, $\lambda >0$.

Definition 2.2

• (a) Let $0\le \psi _0 \in C^4({\overline{\Omega }})$. We say that $\phi _0=e^{\lambda \psi _0}$, $\lambda >0$, is a weight function for the elliptic operator ${\mathcal {L}}_{A,0}^e$ if $\nabla \psi _0(x)\ne 0$ for any $x\in {\overline{\Omega }}$.

• (b) If $0\le \psi \in C^4({\overline{Q}})$ and $\nabla \psi _0(x)\ne 0$, for any $x\in {\overline{\Omega }}$, we say that $\phi$ is a weight function for the parabolic operator ${\mathcal {L}}_{A,0}^p$.

• (c) Assume that $0\le \psi \in C^4({\overline{Q}})$. Then $\phi$ is said a weight function for the Schrödinger operator ${\mathcal {L}}_{A,0}^s$ if $\psi _0$ is A-pseudo-convex in $\Omega$.

• (d) We say that $\phi$, with $0\le \psi \in C^4({\overline{Q}})$, is a weight function for the wave operator ${\mathcal {L}}_{A,0}^w$ if $\psi _0$ is A-pseudo-convex with constant $\kappa >0$ in $\Omega$ and if, in addition, the following two conditions hold:

  $$\begin{aligned}&\min _{{\overline{Q}}}\left[ |\nabla \psi _0|_A^2-(\partial _t\psi _1)^2\right] ^2>0, \end{aligned}$$

  (2.2)
  $$\begin{aligned}&|\partial _t^2\psi _1|\le \varkappa ^{-1}\kappa /4. \end{aligned}$$

  (2.3)

Example 2.1

Fix $t_0\in {\mathbb {R}}$, $x_0\in {\mathbb {R}}^n{{\setminus }} {\overline{\Omega }}$ and set, for $\gamma \in {\mathbb {R}}$,

$$\begin{aligned} \psi (x,t)=\left[ |x-x_0|^2+\gamma (t+t_0)^2\right] /2+C, \quad (x,t)\in {\overline{Q}}, \end{aligned}$$

where the constant C is chosen sufficiently large in order to guarantee that $\psi \ge 0$. In that case

$$\begin{aligned} (\Upsilon _A(\psi _0))_{\alpha \beta }=-\sum _{k,\ell =1}^n\partial _ka_{\alpha \beta }a_{k\ell }(x_\ell -x_{0,\ell })+2\sum _{k,\ell =1}^na_{\alpha k}\partial _ka_{\beta \ell }(x_\ell -x_{0,\ell }) . \end{aligned}$$

Let us first discuss A-pseudo-convexity condition of $\psi _0$ in different cases.

• (i) Assume that $\Omega =B(0,r)$ and $x_0\in B(0,2r){\setminus } {\overline{B}}(0,r)$. We can then choose r sufficiently small in such a way that

  $$\begin{aligned} (\Upsilon _A(\psi _0)\xi |\xi )\ge -\varkappa ^2 |\xi |^2, \end{aligned}$$

  from which we deduce that

  $$\begin{aligned} (\Theta _A(\psi _0)\xi |\xi )\ge \varkappa ^2|\xi |^2. \end{aligned}$$

• (ii) As the mapping

  $$\begin{aligned} A \in C^{2,1}({\overline{\Omega }} ,{\mathbb {R}}^{n\times n})\mapsto \Upsilon _A(\psi _0)\in C^{1,1}({\overline{\Omega }} ,{\mathbb {R}}^{n\times n}) \end{aligned}$$

  is continuous in a neighborhood of $\mathbf{I }$ and $\Upsilon _{\mathbf{I }}(\psi _0)=0$, we conclude that there exists ${\mathcal {N}}$, a neighborhood of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }} ,{\mathbb {R}}^{n\times n})$, so that, for any $A\in {\mathcal {N}}$, we have

  $$\begin{aligned} (A(x)\xi |\xi )&\ge |\xi |^2/2,\quad x\in {\overline{\Omega }} ,\; \xi \in {\mathbb {R}}^n, \\ (\Upsilon _A(\psi _0)\xi |\xi )&\ge -|\xi |^2/4,\quad x\in {\overline{\Omega }} ,\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

  Whence

  $$\begin{aligned} (\Theta _A(\psi _0)\xi |\xi )\ge |\xi |^2/4,\quad x\in {\overline{\Omega }} ,\; \xi \in {\mathbb {R}}^n, \end{aligned}$$

  provided that $A\in {\mathcal {N}}$.

• (iii) Consider the particular case in which $A=a\mathbf{I }$ with $a\in C^{2,1}({\overline{\Omega }})$ satisfying $a\ge \varkappa$. Simple computations then yield

  $$\begin{aligned} \Upsilon _A(\psi _0)= -a\left( \nabla a |x-x_0\right) \mathbf{I }+2a\nabla a\otimes (x-x_0). \end{aligned}$$

  In consequence

  $$\begin{aligned} (\Theta _A(\psi _0)\xi |\xi )\ge \varkappa \left( 2\varkappa -3\left| \nabla a\right| |x-x_0|\right) |\xi |^2. \end{aligned}$$

  Hence a condition guaranteeing that $\psi _0$ is $a\mathbf{I }$-pseudo-convex is

  $$\begin{aligned} \left| \nabla a\right| |x-x_0|<2\varkappa /3. \end{aligned}$$

  This condition is achieved for instance if $\Omega$ has sufficiently small diameter and $x_0$ is close to $\Omega$ or else $|\nabla a|$ is small enough.

Next, we discuss a bound on $\gamma$ for which (2.2) and (2.3) hold simultaneously. If $d=\text{ dist }(x_0,{\overline{\Omega }})$ $(>0)$ and $\sigma =\Vert t+t_0\Vert _{L^\infty ((t_1,t_2))}$, then (2.2) is satisfied whenever

$$\begin{aligned} |\nabla \psi |^2-(\partial _t \psi )^2\ge d^2- \sigma ^2 \gamma ^2 >0. \end{aligned}$$

As $\partial _t^2\psi =\gamma$, we see that both (2.2) and (2.3) are satisfied when

$$\begin{aligned} 0<|\gamma | <\min \left[ d/\sigma ,\kappa /(4\varkappa )\right] . \end{aligned}$$

2.4 Pseudo-convex hypersurface

We begin by a lemma concerning the action of an orthogonal transformation on $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$. If ${\mathcal {O}}$ is an orthogonal transformation and $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$, we set $A_{{\mathcal {O}}}(y)={\mathcal {O}}A({\mathcal {O}}^ty){\mathcal {O}}^t$. Here ${\mathcal {O}}^t$ denotes the transposed matrix of ${\mathcal {O}}$.

The proof of the following lemma is straightforward.

Lemma 2.1

Let $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$ and ${\mathcal {O}}$ an orthogonal transformation. Then $A_{{\mathcal {O}}}\in {\mathscr {M}}({\mathcal {O}}\Omega ,\varkappa ,{\mathfrak {m}}')$, where ${\mathfrak {m}}'={\mathfrak {m}}'(n,{\mathfrak {m}})>0$ is a constant.

The gradient with respect to the variable $x'\in {\mathbb {R}}^{n-1}$ or $y'\in {\mathbb {R}}^{n-1}$ is denoted hereafter by $\nabla '$.

Let $\theta$ be a $C^{3,1}$-function defined in a neighborhood ${\mathcal {U}}$ of ${\tilde{x}}$ in $\Omega$ with $\nabla \theta ({\tilde{x}})\ne 0$. Consider then the hypersurface

$$\begin{aligned} H=\{x\in {\mathcal {U}};\; \theta (x)=\theta ({\tilde{x}})\}. \end{aligned}$$

Making a translation and change of coordinates we may assume that ${\tilde{x}}=0$, $\theta ({\tilde{x}})=0$, $\nabla '\theta (0)=0$ and $\partial _n\theta (0)\ne 0$. With the help of the implicit function theorem $\theta (x)=0$ near 0 may rewritten as $x_n=\vartheta (x')$ with $\vartheta (0)=0$ and $\nabla '\vartheta (0) =0$.

Let ${\hat{A}}$ be the matrix obtained after this transformations. According to Lemma 2.1, ${\hat{A}}\in {\mathscr {M}}({\mathcal {O}}({\mathcal {U}}+{\tilde{x}}),\varkappa ,{\mathfrak {m}}')$, ${\mathfrak {m}}'={\mathfrak {m}}'(n,{\mathfrak {m}})>0$ is a constant, where ${\mathcal {O}}$ is the orthogonal transformation corresponding to the above change of coordinates. Also, note that $\hat{\mathbf{I }}=\mathbf{I }$.

Consider, in some neighborhood of 0, the mapping

$$\begin{aligned} \varphi =\varphi _H :(x',x_n)\in \omega \mapsto (y',y_n)=(x',x_n-\vartheta (x')+|x'|^2). \end{aligned}$$

(2.4)

Elementary calculations yield

$$\begin{aligned} \varphi '(x',x_n)=\left( \begin{array} {cccc} 1 &{}\ldots &{}0 &{}0 \\ \vdots &{}\ddots &{}0 &{}\vdots \\ 0 &{}\ldots &{}1 &{}0 \\ g_1(x') &{}\ldots &{}g_{n-1}(x') &{}1 \end{array} \right) \end{aligned}$$

with $g_k(x')=-\partial _k\vartheta (x')+2x_k'$, $0\le k\le n-1$. Whence

$$\begin{aligned} (\varphi '(x',x_n)\xi |\xi )= |\xi '|^2+(-\nabla '\vartheta (x')+2x'|\xi ')\xi _n +\xi _n^2,\quad \xi =(\xi ',\xi _n) \in {\mathbb {R}}^n. \end{aligned}$$

Since

$$\begin{aligned} |(-\nabla '\vartheta (x')+2x'|\xi ')\xi _n|&\le |(-\nabla '\vartheta (x')+2x'|\xi ')|^2/2+\xi _n^2/2 \\&\le |-\nabla '\vartheta (x')+2x'|^2|\xi '|^2/2+2\xi _n^2/2 \\&\le \left( |\nabla '\vartheta (x')|^2+4|x'|^2\right) |\xi '|^2+\xi _n^2/2, \end{aligned}$$

there exists a neighborhood $\omega$ of 0, only depending of $\vartheta$, so that

$$\begin{aligned} |(-\nabla '\vartheta (x')+2x'|\xi ')\xi _n|\le |\xi '|^2/2+\xi _n^2/2=|\xi |^2/2. \end{aligned}$$

In consequence

$$\begin{aligned} (\varphi '(x',x_n)\xi |\xi )\ge |\xi |^2/2,\quad x=(x',x_n)\in \omega ,\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

(2.5)

Whence (2.5) together with Cauchy-Schwarz’s inequality yield

$$\begin{aligned} |(\varphi ')^t(x',x_n)\xi |^2\ge |\xi |^2/2,\quad x=(x',x_n)\in \omega ,\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

(2.6)

Let ${\tilde{\omega }}=\varphi (\omega )$ (hence $\varphi$ is a diffeomorphism from $\omega$ onto ${\tilde{\omega }}$) and define

$$\begin{aligned} {\tilde{A}} (y)=\varphi '\left( \varphi ^{-1}(y)\right) {\hat{A}}\left( \varphi ^{-1}(y)\right) (\varphi ')^t\left( \varphi ^{-1}(y)\right) ,\quad y\in {\tilde{\omega }}. \end{aligned}$$

(2.7)

In light of (2.6) we obtain

$$\begin{aligned} ({\tilde{A}}(y)\xi |\xi )\ge \varkappa |(\varphi ')^t\left( \varphi ^{-1}(y)\right) \xi |^2\ge (\varkappa /2)|\xi |^2,\quad y\in {\tilde{\omega }}. \end{aligned}$$

Also, by straightforward computations we get

$$\begin{aligned} \Vert {\tilde{A}}\Vert _{C^{2,1}({\overline{\Omega }} ;{\mathbb {R}}^{n\times n})}\le \tilde{{\mathfrak {m}}}, \end{aligned}$$

with $\tilde{{\mathfrak {m}}}$ only depending of n, ${\mathfrak {m}}$ and $\vartheta$.

We observe that the role of $\varphi$ is to transform the hypersurface $\{x_n=\vartheta (x')\}$ in a neighborhood of the origin into the convex hypersurface $\{y_n=|y'|^2\}$ in another neighborhood of the origin.

Define ${\tilde{\psi }}_0$ as follows

$$\begin{aligned} {\tilde{\psi }}_0(y)=(y_n-1)^2+|y'|^2. \end{aligned}$$

The matrix ${\tilde{A}}$ appearing in (2.7) is denoted hereafter by $A_H$. The following definition is motivated by the classical procedure used to establish the unique continuation property of an elliptic operator across the convex hypersurface $\{y_n=|y'|^2\}$.

Definition 2.3

We say that the hypersurface H is A-pseudo-convex if ${\tilde{\psi }}_0$ is $A_H$-pseudo-convex in ${\tilde{\omega }}$.

Lemma 2.2

• (a) There exists ${\mathcal {N}}$, a neighborhood of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }} ;{\mathbb {R}}^{n\times n})$, so that, for any $A\in {\mathcal {N}}$, H is A-pseudo-convex.

• (b) There exists ${\mathcal {N}}_0$, a neighborhood of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }} ;{\mathbb {R}}^{n\times n})$, so that, for any $A\in {\mathcal {N}}_0$ and any orthogonal transformation ${\mathcal {O}}$, we have $A_{\mathcal {O}}\in {\mathcal {N}}$.

Proof

• (a) Let us first discuss the case where $A=\mathbf{I }$. Note that it is not hard to check that

  $$\begin{aligned} \varphi ^{-1}(y',y_n)=\left( y',y_n+\vartheta (y')-|y'|^2\right) \end{aligned}$$

  and

  $$\begin{aligned} \mathbf{I }_H(y)=\tilde{\mathbf{I }} (y)=({\tilde{a}}_{ij}(y'))=\left( \begin{array} {cccc} 1 &{}\ldots &{}0 &{}{\tilde{g}}_1(y') \\ \vdots &{}\ddots &{}0 &{}\vdots \\ 0 &{}\ldots &{}1 &{}{\tilde{g}}_{n-1}(y') \\ {\tilde{g}}_1(y') &{}\ldots &{}{\tilde{g}}_{n-1}(y') &{}{\tilde{g}}_n(y') \end{array} \right) , \end{aligned}$$

  with ${\tilde{g}}_k(y')=\partial _k\vartheta (y')-2y_k'$, $0\le k\le n-1$ and ${\tilde{g}}_n=|\nabla \vartheta (y')-2y'|^2+1$.

  We have clearly ${\tilde{A}} (0)=\mathbf{I }$ and, for $1\le p\le n-1$,

  $$\begin{aligned} \partial _p{\tilde{a}}_{k\ell }(y')= \left\{ \begin{array}{ll} 0,\quad 1\le k,\ell \le n-1, \\ \\ \partial _{pk}^2\vartheta (y')-2\delta _{pk}, \quad 1\le k\le n-1,\; \ell =n, \\ \\ 2\sum _{\alpha =1}^{n-1}(\partial _\alpha \vartheta (y')-2y_\alpha )(\partial _{p\alpha }^2 \vartheta (y')-2\delta _{p\alpha }) ,\quad k=n,\; \ell =n. \end{array} \right. \end{aligned}$$

  Therefore

  $$\begin{aligned} \partial _p{\tilde{a}}_{k\ell }(0)= \left\{ \begin{array}{ll} 0, &{}\quad 1\le k,\ell \le n-1, \\ \\ \partial _{pk}^2\vartheta (0)-2\delta _{pk} &{}1\le k\le n-1,\; \ell =n, \\ \\ 0 &{} k=n,\; \ell =n. \end{array} \right. \end{aligned}$$

  Let ${\tilde{\Lambda }}_{k\ell }^m$ given by

  $$\begin{aligned} {\tilde{\Lambda }}_{k,\ell }^m(y)=-\sum _{p=1}^n\partial _p{\tilde{a}}_{k\ell }(y){\tilde{a}}_{pm}(y)+2\sum _{p=1}^n{\tilde{a}}_{kp}(y)\partial _p{\tilde{a}}_{\ell m}(y) \end{aligned}$$

  and define ${\tilde{\Upsilon }}(y)=({\tilde{\Upsilon }}_{k\ell }(y))$ as follows

  $$\begin{aligned} {\tilde{\Upsilon }}_{k\ell }(y)=\sum _{m=1}^n{\tilde{\Lambda }}_{k\ell }^m(y)\partial _m {\tilde{\psi }}_0(y) . \end{aligned}$$

  It is then straightforward to check that

  $$\begin{aligned} {\tilde{\Upsilon }}_{k\ell }(0)=\sum _{m=1}^n\Lambda _{k,\ell }^m(0)\partial _m {\tilde{\psi }}_0(0)=-2\Lambda _{k,\ell }^n(0)=0,\quad 0\le k,\ell \le n. \end{aligned}$$

  Since

  $$\begin{aligned} {\tilde{\Theta }}(y)=\Theta _{\mathbf{I }}({\tilde{\psi }}_0)(y)= 2\nabla ^2{\tilde{\psi }}_0(y)+{\tilde{\Upsilon }}(y), \end{aligned}$$

  we get

  $$\begin{aligned} {\tilde{\Theta }}(0)= 4\mathbf{I }, \end{aligned}$$

  and hence

  $$\begin{aligned} ({\tilde{\Theta }}(0)\xi |\xi )\ge 4|\xi |^2,\quad \xi \in {\mathbb {R}}^n. \end{aligned}$$

  Continuity argument, first with respect to y and then with respect to A, shows that, by reducing ${\tilde{\omega }}$ if necessary,

  $$\begin{aligned} (\Theta _{{\tilde{A}}}(\psi _0)(y)\xi |\xi )\ge 2|\xi |^2,\quad y\in {\tilde{\omega }},\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

• (b) Immediate from Lemma 2.1. $\square$

3 The wave equation

3.1 Carleman inequality

In this subsection $\psi (x,t)=\psi _0(x)+\psi _1(t)$ is a weight function for the wave operator ${\mathcal {L}}_{A,0}^w$ with A-pseudo-convexity constant $\kappa >0$. We set

$$\begin{aligned} \delta = \min _{{\overline{Q}}}\left[ |\nabla \psi |_A^2-(\partial _t\psi )^2\right] ^2\; (>0) \end{aligned}$$

and $\phi =e^{\lambda \psi }$.

We use for notational convenience ${\mathfrak {d}}=(\Omega , t_1, t_2,\varkappa , {\mathfrak {m}}, \kappa ,\delta ,{\mathfrak {b}})$ with ${\mathfrak {b}}\ge \Vert \psi \Vert _{C^4({\overline{Q}})}$, and $\mathbf{D }_A=(\nabla _A\, \cdot , \partial _t\, \cdot )$.

Theorem 3.1

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 u^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left( {\mathcal {L}}_{A,0}^wu\right) ^2{ d}x{ d}t +\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3u^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}\mu , \end{aligned}$$

(3.1)

for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(Q,{\mathbb {R}})$.

Proof

In this proof, $\aleph _j$, $\lambda _j$ and $\tau _j$, $j=0,1,\ldots$, are positive generic constants only depending on ${\mathfrak {d}}$.

Set $\Phi =e^{-\tau \phi }$ with $\tau >0$. Elementary computations then give

$$\begin{aligned} \partial _k\Phi&=-\tau \partial _k\phi \Phi , \\ \partial _{k\ell }\Phi&=\left( -\tau \partial ^2_{k\ell }\phi +\tau ^2\partial _k\phi \partial _\ell \phi \right) \Phi , \\ \partial _t\Phi&= -\tau \partial _t\phi \Phi , \\ \partial _t^2\Phi&=\left( -\tau \partial _t^2 \phi +\tau ^2(\partial _t\phi )^2\right) \Phi . \end{aligned}$$

The preceding two first formulas can be rewritten in the following form

$$\begin{aligned} \nabla \Phi&=-\tau \Phi \nabla \phi , \\ \nabla ^2\Phi&=\Phi \left( -\tau \nabla ^2\phi +\tau ^2\nabla \phi \otimes \nabla \phi \right) . \end{aligned}$$

For $w\in H^2 (Q;{\mathbb {R}})$, we obtain

$$\begin{aligned} \Phi ^{-1}\Delta _A(\Phi w)=\Delta _Aw -2\tau (\nabla w|\nabla \phi )_A+\left[ \tau ^2 |\nabla \phi |_A^2-\tau \Delta _A\phi \right] w. \end{aligned}$$

Also,

$$\begin{aligned} \Phi ^{-1}\partial _t^2(\Phi w)=\partial _t^2w-2\tau \partial _t\phi \partial _tw +\left( -\tau \partial _t^2 \phi +\tau ^2(\partial _t\phi )^2\right) w. \end{aligned}$$

We decompose $L=\Phi ^{-1}{\mathcal {L}}_{A,0}^w\Phi$ into its self-adjoint part and skew-adjoint part:

$$\begin{aligned} L=L_++L_-. \end{aligned}$$

Simple calculations show that

$$\begin{aligned} L_+w&=\Delta _A w-\partial _t^2w+aw, \\ L_-w&= (B|\nabla w)+d \partial _tw +bw, \end{aligned}$$

with

$$\begin{aligned} a(x,t)&=\tau ^2\left( |\nabla \phi |_A^2-(\partial _t\phi )^2\right) , \\ b(x,t)&=-\tau \left( \Delta _A \phi -\partial _t^2\phi \right) , \\ B&=-2\tau A\nabla \phi , \\ d&=2\tau \partial _t\phi . \end{aligned}$$

We have

$$\begin{aligned} \langle L_+w|L_-w\rangle _{L^2(Q)}=\sum _{j=1}^9 I_j, \end{aligned}$$

(3.2)

where

$$\begin{aligned} I_1&=\int _Q\Delta _A w(\nabla w|B) { d}x{ d}t, \\ I_2&=\int _Q\Delta _A wd\partial _tw{ d}x{ d}t, \\ I_3&=\int _Q\Delta _A wbw{ d}x{ d}t, \\ I_4&=-\int _Q\partial _t^2w(\nabla w|B) { d}x{ d}t, \\ I_5&=-\int _Q\partial _t^2wd\partial _tw { d}x{ d}t, \\ I_6&=-\int _Q\partial _t^2wbw { d}x{ d}t, \\ I_7&=\int _Qaw(\nabla w|B) { d}x{ d}t, \\ I_8&=\int _Qadw\partial _tw{ d}x{ d}t, \\ I_9&=\int _Qabw^2{ d}x{ d}t. \end{aligned}$$

The most integrations by parts, with respect to the space variable, we use in this proof are often based on Green’s formula (2.1). A first integration by parts then yields

$$\begin{aligned} I_1=\int _Q\Delta _Aw (\nabla w|B) { d}x{ d}t&=-\int _Q (\nabla w|\nabla (\nabla w|B))_A{ d}x{ d}t \\&\qquad +\int _\Sigma (\nabla w|\nu )_A(\nabla w|B) { d}\sigma { d}t. \end{aligned}$$

Whence

$$\begin{aligned} I_1=-\int _Q ([\nabla ^2wB+(B')^t\nabla w]|\nabla w)_A{ d}x{ d}t +\int _\Sigma (\nabla w|\nu )_A(\nabla w|B) { d}\sigma { d}t. \end{aligned}$$

(3.3)

Now as

$$\begin{aligned} \int _Q \partial _{\ell j}^2w B_ja_{\ell k}\partial _k w{ d}x{ d}t&=-\int _Q\partial _\ell w\partial _j(B_ja_{\ell k})\partial _kw{ d}x{ d}t -\int _Q\partial _\ell wB_ja_{\ell k}\partial _{kj}^2w{ d}x{ d}t \\&\quad +\int _\Sigma \partial _\ell wB_j\nu _ja_{\ell k}\partial _k w{ d}\sigma { d}t, \end{aligned}$$

we find

$$\begin{aligned} 2\int _Q (\nabla ^2wB|\nabla w)_A{ d}x{ d}t=-\int _Q(C\nabla w|\nabla w)+\int _\Sigma (B|\nu )|\nabla w|_A^2 { d}\sigma { d}t, \end{aligned}$$

(3.4)

where $C=(\text{ div }(a_{k\ell }B))$.

Let

$$\begin{aligned} D=C/2-A(B')^t. \end{aligned}$$

We get by putting (3.4) into (3.3)

$$\begin{aligned} I_1=\int _Q(D\nabla w|\nabla w){ d}x{ d}t+\int _\Sigma \left[ (\nabla w|\nu )_A(\nabla w|B)-2^{-1}(B|\nu )|\nabla w|_A^2\right] { d}\sigma { d}t. \end{aligned}$$

(3.5)

For $I_2$, we obtain by making integrations by parts

$$\begin{aligned} I_2&=-\int _Q(\nabla w| \nabla (d\partial _tw))_A{ d}x{ d}t+\int _\Sigma (\nabla w|\nu )_A d\partial _tw{ d}\sigma { d}t \\&=- \int _Q(\nabla w|\nabla d)_A\partial _tw{ d}x{ d}t-\int _Q d (\nabla w|\nabla \partial _tw){ d}x{ d}t \\&\quad +\int _\Sigma (\nabla w|\nu )_A d\partial _tw{ d}\sigma { d}t. \end{aligned}$$

As A is symmetric, we have

$$\begin{aligned} (\nabla w|\nabla \partial _tw)_A=\partial _t|\nabla w|_A^2/2. \end{aligned}$$

Hence

$$\begin{aligned} I_2&=-\int _Q(\nabla w|\nabla d)_A \partial _tw{ d}x{ d}t+\int _Q\partial _t(d/2) |\nabla w|_A^2{ d}x{ d}t \nonumber \\&\quad +\int _\Sigma (\nabla w|\nu )_A d\partial _tw{ d}x{ d}t-\int _\Omega \left[ (d/2) |\nabla w|_A^2\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(3.6)

We have also

$$\begin{aligned} I_3&=\int _Q\Delta _A wbw=-\int _Q b|\nabla w|_A^2{ d}x{ d}t -\int _Q w(\nabla b|\nabla w)_A { d}x{ d}t \nonumber \\&\quad +\int _\Sigma (\nabla w|\nu )_Abw { d}\sigma { d}t \nonumber \\&=-\int _Q b|\nabla w|_A^2{ d}x{ d}t +\int _Q \Delta _A(b/2) w^2 { d}x{ d}t \nonumber \\&\quad -\int _\Sigma (\nabla (b/2)|\nu )_Aw^2{ d}\sigma { d}t +\int _\Sigma (\nabla w|\nu )_Abw { d}\sigma { d}t. \end{aligned}$$

(3.7)

Let $J_1=I_1+I_2+I_3$ and

$$\begin{aligned} {\mathcal {A}}_1&=D+\left[ \partial _t(d/2)-b\right] A, \\ a_1&=\Delta _A (b/2), \\ {\mathcal {B}}_1(w)&=-(\nabla w|\nabla d)_A \partial _tw, \\ g_1(w)&=(\nabla w|\nu )_A(\nabla w|B)-(B/2|\nu )|\nabla w|_A^2+(\nabla w|\nu )_A d\partial _tw \\&\quad -(\nabla (b/2)|\nu )_Aw^2{ d}\sigma + (\nabla w|\nu )_Abw, \\ h_1(w)&=-\left[ (d/2) |\nabla w|_A^2\right] _{t=t_1}^{t_2}. \end{aligned}$$

Putting together (3.5)–(3.7), we find

$$\begin{aligned} J_1&=\int _Q ({\mathcal {A}}_1\nabla w|\nabla w){ d}x{ d}t +\int _Q{\mathcal {B}}_1(w){ d}x{ d}t+\int _Qa_1w^2{ d}x{ d}t. \\&\quad +\int _\Sigma g_1(w){ d}\sigma { d}t +\int _\Omega h_1(w) { d}x. \end{aligned}$$

Straightforward computations show that

$$\begin{aligned} {\mathcal {A}}_1=2\tau A\nabla ^2\phi A+\tau \Upsilon _A(\phi ). \end{aligned}$$

Whence

$$\begin{aligned} J_1&=\tau \int _Q \left( \left[ 2A\nabla ^2\phi A+\Upsilon _A(\phi )\right] \nabla w|\nabla w\right) { d}x{ d}t +\int _Q{\mathcal {B}}_1(w){ d}x{ d}t \nonumber \\&\quad +\int _Qa_1w^2{ d}x{ d}t +\int _\Sigma g_1(w){ d}\sigma { d}t +\int _\Omega h_1(w) { d}x. \end{aligned}$$

(3.8)

We obtain, by using again an integration by parts,

$$\begin{aligned} I_4=\int _Q\partial _tw\partial _t(\nabla w|B){ d}x{ d}t -\int _\Omega \left[ \partial _tw(\nabla w|B)\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

Hence

$$\begin{aligned} I_4=\int _Q\partial _tw (\nabla w|\partial _tB) { d}x{ d}t+\int _Q\partial _tw(\nabla \partial _t w|B) { d}x{ d}t -\int _\Omega [\partial _tw(\nabla w|B)]_{t=t_1}^{t_2}{ d}x. \end{aligned}$$

But

$$\begin{aligned} \int _Q\partial _tw(\nabla \partial _t w|B) { d}x{ d}t&=-\int _Q\text{ div }(\partial _twB)\partial _tw{ d}x{ d}t+\int _\Sigma (\partial _tw)^2(B|\nu ){ d}\sigma { d}t \\&=-\int _Q\partial _tw(\nabla \partial _t w|B) { d}x{ d}t-\int _Q (\partial _tw)^2\text{ div }(B) { d}x{ d}t \\&\quad +\int _\Sigma (\partial _tw)^2(B|\nu ){ d}\sigma { d}t. \end{aligned}$$

Therefore

$$\begin{aligned} 2\int _Q\partial _tw(\nabla \partial _t w|B) { d}x{ d}t =-\int _Q (\partial _tw)^2\text{ div }(B) { d}x{ d}t +\int _\Sigma (\partial _tw)^2(B|\nu ){ d}\sigma { d}t. \end{aligned}$$

In consequence

$$\begin{aligned} I_4&=\int _Q\partial _tw (\nabla w|\partial _tB) { d}x{ d}t- \int _Q (\partial _tw)^2\text{ div }(B/2) { d}x{ d}t \nonumber \\&\quad +\int _\Sigma (\partial _tw)^2(B/2|\nu ){ d}\sigma { d}t -\int _\Omega \left[ \partial _tw(\nabla w|B)\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(3.9)

For $I_5$, we have

$$\begin{aligned} I_5&=-\int _Q (d/2)\partial _t(\partial _t w)^2 { d}x{ d}t= \int _Q\partial _t(d/2)(\partial _tw)^2 { d}x{ d}t \nonumber \\&\quad -\int _\Omega \left[ (d/2)(\partial _t w)^2)\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(3.10)

Also,

$$\begin{aligned} I_6&=-\int _Q\partial _t^2wbw{ d}x{ d}t \\&=\int _Q\partial _tb w\partial _tw { d}x{ d}t +\int _Qb (\partial _tw)^2 { d}x{ d}t -\int _\Omega \left[ b w\partial _tw\right] _{t=t_1}^{t_2}{ d}x \\&=\int _Q\partial _t(b/2) \partial _tw^2 { d}x{ d}t +\int _Qb (\partial _tw)^2 { d}x{ d}t -\int _\Omega \left[ b w\partial _tw\right] _{t=t_1}^{t_2}{ d}x \end{aligned}$$

and hence

$$\begin{aligned} I_6&=-\int _Q\partial _t^2(b/2) w^2 { d}x{ d}t +\int _Qb (\partial _tw)^2 { d}x{ d}t \nonumber \\&\quad -\int _\Omega \left[ b w\partial _tw\right] _{t=t_1}^{t_2}{ d}x+\int _\Omega \left[ (\partial _tb/2)w^2\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(3.11)

Let $J_2=I_4+I_5+I_6$ and define

$$\begin{aligned} {\mathfrak {a}}_2&=-\text{ div }(B/2)+\partial _t(d/2)+b, \\ a_2&=-\partial _t^2(b/2), \\ {\mathcal {B}}_2(w)&=\partial _tw (\nabla w| \partial _tB), \\ g_2(w)&=(\partial _tw)^2(B/2|\nu ), \\ h_2(w)&=- \left[ \partial _tw(\nabla w|B)\right] _{t=t_1}^{t_2}- \left[ d(\partial _t w)^2)\right] _{t=t_1}^{t_2}- \left[ b w\partial _tw\right] _{t=t_1}^{t_2}+\left[ (\partial _tb/2) w^2\right] _{t=t_1}^{t_2}. \end{aligned}$$

A combination of (3.9)–(3.11) gives

$$\begin{aligned} J_2&=\int _Q {\mathfrak {a}}_2(\partial _t w)^2{ d}x{ d}t +\int _Q{\mathcal {B}}_2(w){ d}x{ d}t+\int _Qa_2w^2{ d}x{ d}t \\&\quad +\int _\Sigma g_2(w){ d}\sigma { d}t +\int _\Omega h_2(w) { d}x. \end{aligned}$$

Let us observe that we have by straightforward computations

$$\begin{aligned} {\mathfrak {a}}_2=2\tau \partial _t^2\phi ,\quad {\mathcal {B}}_2={\mathcal {B}}_1. \end{aligned}$$

Hence

$$\begin{aligned} J_2&=2\tau \int _Q \partial _t^2\phi (\partial _t w)^2{ d}x{ d}t +\int _Q{\mathcal {B}}_1(w){ d}x{ d}t+\int _Qa_2w^2{ d}x{ d}t \nonumber \\&\quad +\int _\Sigma g_2(w){ d}\sigma { d}t +\int _\Omega h_2(w) { d}x. \end{aligned}$$

(3.12)

Let ${\tilde{J}}=J_1+J_2$. In light of (3.5) and (3.12) we deduce that

$$\begin{aligned} {\tilde{J}}&=\tau \int _Q \left( \left[ 2A\nabla ^2\phi A+\Upsilon _A(\phi )\right] \nabla w|\nabla w\right) { d}x{ d}t+2\tau \int _Q \partial _t^2\phi (\partial _t w)^2{ d}x{ d}t\\&\quad -4\tau \int _Q\partial _tw(\nabla w|\nabla \partial _t \phi )_A{ d}x{ d}t+\int _Q{\tilde{a}}w^2{ d}x{ d}t \\&\quad +\int _\Sigma {\tilde{g}}(w){ d}\sigma { d}t +\int _\Omega {\tilde{h}}(w) { d}x, \end{aligned}$$

where

$$\begin{aligned} {\tilde{a}}=a_1+a_2,\quad {\tilde{g}}=g_1+g_2,\quad {\tilde{h}}=h_1+h_2. \end{aligned}$$

As $\phi =e^{\lambda \psi }$, we have

$$\begin{aligned} \nabla ^2\phi&=\lambda ^2\phi (\nabla \psi \otimes \nabla \psi )+\lambda \phi \nabla ^2\psi , \\ \partial _t^2\phi&=\lambda ^2\phi (\partial _t\psi )^2+\lambda \phi \partial _t^2\psi . \end{aligned}$$

This and the fact that $\nabla \partial _t\psi =0$ imply

$$\begin{aligned}&\left( \nabla ^2\phi A\nabla w|\nabla w\right) _A+\partial _t^2\phi (\partial _t w)^2 -2\partial _tw(\nabla w|\nabla \partial _t \phi )_A \\&\quad =\lambda \phi \left[ (\nabla ^2\psi A\nabla w|\nabla w)_A+\partial _t^2\psi (\partial _tw)^2\right] \\&\qquad +\lambda ^2\phi \left[ (\nabla \psi |\nabla w)_A^2+(\partial _t\psi )^2(\partial _tw)^2-2\partial _t\psi \partial _tw(\nabla \psi |\nabla w)_A\right] . \end{aligned}$$

That is we have

$$\begin{aligned}&\left( \nabla ^2\phi A\nabla w|\nabla w\right) _A+\partial _t^2\phi (\partial _t w)^2 -2\partial _tw(\nabla w|\nabla \partial _t \phi )_A \\&\quad = \lambda \phi \left[ (\nabla ^2\psi A\nabla w|\nabla w)_A+\partial _t^2\psi (\partial _tw)^2\right] +\lambda ^2\phi \left[ (\nabla \psi |\nabla w)_A-\partial _t\psi \partial _tw\right] ^2, \end{aligned}$$

from which we deduce, by noting that $\Upsilon _A(\phi )=\lambda \phi \Upsilon _A(\psi )$,

$$\begin{aligned} {\tilde{J}}&\ge \tau \lambda \int _Q \phi \left[ \left( {\mathcal {A}}A^{1/2}\nabla w|A^{1/2}\nabla w\right) +\partial _t^2\psi (\partial _tw)^2\right] { d}x{ d}t \nonumber \\&\quad +\int _Q{\tilde{a}}w^2{ d}x{ d}t+\int _\Sigma {\tilde{g}}(w){ d}\sigma { d}t +\int _\Omega {\tilde{h}}(w) { d}x. \end{aligned}$$

(3.13)

Here

$$\begin{aligned} {\mathcal {A}}=2A^{1/2}\nabla ^2\psi A^{1/2} +A^{-1/2}\Upsilon _A(\psi )) A^{-1/2}=A^{-1/2}\Theta _A(\psi )A^{-1/2}. \end{aligned}$$

For $I_7$, we have

$$\begin{aligned} I_7&=\int _Qaw(B|\nabla w){ d}x{ d}t=\int _Qa(B/2|\nabla w^2){ d}x{ d}t \nonumber \\&=-\int _Q \text{ div }(aB/2)w^2{ d}x{ d}t + \int _\Sigma a(B/2|\nu )w^2{ d}\sigma { d}t. \end{aligned}$$

(3.14)

Finally,

$$\begin{aligned} I_8=\int _Q(ad/2)\partial _tw^2{ d}x{ d}t=-\int _Q\partial _t(ad/2)w^2{ d}x{ d}t +\int _\Omega \left[ (ad/2)w^2\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(3.15)

Define ${\hat{J}}=I_7+I_8+I_9$ and

$$\begin{aligned} {\hat{a}}&=-\text{ div }(aB/2)-\partial _t(ad/2)+ab+\Delta _A (b/2)-\partial _t^2(b/2), \\ {\hat{g}}(w)&={\tilde{g}}(w)+a(B/2|\nu )w^2, \\ {\hat{h}}(w)&={\tilde{h}}(w)+ \left[ (ad/2)w^2\right] _{t=t_1}^{t_2}. \end{aligned}$$

Then we have from (3.13) to (3.15)

$$\begin{aligned} \langle L_+w|L_-w\rangle _{L^2(Q)}&={\tilde{J}}+{\hat{J}} \nonumber \\&\ge \tau \lambda \int _Q \phi [\left( {\mathcal {A}}A^{1/2}\nabla w|A^{1/2}\nabla w\right) +\partial _t^2\psi (\partial _tw)^2]{ d}x{ d}t +\int _Q{\hat{a}}w^2{ d}x{ d}t \nonumber \\&\quad +\int _\Sigma {\hat{g}}(w){ d}\sigma { d}t +\int _\Omega {\hat{h}}(w) { d}x. \end{aligned}$$

(3.16)

We prove (see details in the end of this proof) that

$$\begin{aligned} {\hat{a}}\ge \tau ^3\lambda ^4\phi ^3\delta ,\quad \lambda \ge \lambda _1,\; \tau \ge \tau _1. \end{aligned}$$

(3.17)

Whence

$$\begin{aligned}&2\langle L_+w|L_-w\rangle _{L^2(Q)}\ge 2\varkappa ^{-1}\kappa \tau \lambda \int _Q \phi |\nabla w|_A^2{ d}x{ d}t \\&\quad +2\tau \lambda \int _Q\phi \partial _t^2\psi (\partial _tw)^2{ d}x{ d}t +2\tau ^3\lambda ^4\delta \int _Q\phi ^3w^2{ d}x{ d}t+{\mathscr {B}}_0(w), \end{aligned}$$

for $\lambda \ge \lambda _1$ and $\tau \ge \tau _1$, with

$$\begin{aligned} {\mathscr {B}}_0(w)=\int _\Sigma {\hat{g}}(w){ d}\sigma { d}t +\int _\Omega {\hat{h}}(w) { d}x. \end{aligned}$$

On the other hand, we find by making twice integration by parts

$$\begin{aligned} \int _Q(L_+w)\phi w{ d}x{ d}t&=-\int _Q\phi |\nabla w|_A^2+\int _Q\phi (\partial _tw)^2{ d}x{ d}t \\&\quad +\int _Q(a-b/2)w^2{ d}x{ d}t+{\mathscr {B}}_1(w), \end{aligned}$$

with

$$\begin{aligned} {\mathscr {B}}_1(w)&=\int _\Sigma \phi (\nabla w|\nu )_Aw{ d}\sigma { d}t-\frac{1}{2}\int _\Sigma (\nabla \phi |\nu )_Aw^2{ d}\sigma { d}t \\&\quad + \int _\Omega -[\phi w\partial _tw]_{t=t_1}^{t_2}{ d}x+\frac{1}{2}\int _\Omega \left[ w^2\partial _t\phi \right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

Let $\epsilon >0$ to be determined later. Then Cauchy–Schwarz’s inequality yields

$$\begin{aligned}&\int _Q(L_+w)^2{ d}x{ d}t+\epsilon ^2\tau ^2\lambda ^2\int _Q\phi ^2w^2{ d}x{ d}t\ge \\&\quad -\frac{\epsilon \tau \lambda }{2}\int _Q\phi |\nabla w|_A^2{ d}x{ d}t +\frac{\epsilon \tau \lambda }{2}\int _Q\phi (\partial _tw)^2{ d}x{ d}t \\&\quad +\frac{\epsilon \tau \lambda }{2}\int _Q(a-b/2)w^2{ d}x{ d}t+\frac{\epsilon \tau \lambda }{2}{\mathscr {B}}_1(w). \end{aligned}$$

Using that $a=\tau ^2\lambda ^2\phi ^2\left( |\nabla \psi |_A^2-(\partial _t\psi )^2\right)$ we get

$$\begin{aligned} \int _Q(L_+w)^2{ d}x{ d}t\ge&-\frac{\epsilon \tau \lambda }{2}\int _Q\phi |\nabla w|_A^2{ d}x{ d}t \\&+\frac{\epsilon \tau \lambda }{2}\int _Q\phi (\partial _tw)^2{ d}x{ d}t+\frac{\epsilon \tau ^3 \lambda ^3}{2}\int _Q\phi ^3\left( |\nabla \psi |_A^2-(\partial _t\psi )^2\right) w^2{ d}x{ d}t \\&-\epsilon ^2\tau ^2\lambda ^3\int _Q\phi ^2w^2{ d}x{ d}t -\frac{\epsilon \tau \lambda }{2}\int _Q(b/2)w^2{ d}x{ d}t+\frac{\epsilon \tau \lambda }{2}{\mathscr {B}}_1(w). \end{aligned}$$

Hence

$$\begin{aligned}&2\langle L_+w|L_-w\rangle _{L^2(Q)}+\int _Q(L_+w)^2{ d}x{ d}t \ge (2\varkappa ^{-1}\kappa -\epsilon /2) \tau \lambda \int _Q \phi |\nabla w|_A^2{ d}x{ d}t \\&\quad +\tau \lambda \int _Q\phi (\epsilon /2+ 2\partial _t^2\psi ) (\partial _tw)^2{ d}x{ d}t+2\tau ^3\lambda ^4\delta \int _Q\phi ^3w^2{ d}x{ d}t \\&\quad +\frac{\epsilon \tau ^3 \lambda ^3}{2}\int _Q\phi ^3(|\nabla \psi |_A^2-(\partial _t\psi )^2)w^2{ d}x{ d}t \\&\quad -\epsilon ^2\tau ^2\lambda ^2\int _Q\phi ^2w^2{ d}x{ d}t-\frac{\epsilon \tau \lambda ^2}{2}\int _Q(b/2)w^2{ d}x{ d}t+{\mathscr {B}}(w),\quad \lambda \ge \lambda _1,\; \tau \ge \tau _1. \end{aligned}$$

Here ${\mathscr {B}}(w)={\mathscr {B}}_0(w)+\frac{\epsilon \tau \lambda }{2}{\mathscr {B}}_1(w)$.

We take $\epsilon =2\varkappa ^{-1}\kappa$ in the preceding inequality and we use inequality (2.3). We obtain, by noting that in the right hand side of the last inequality, the fourth, fifth and sixth terms can be absorbed by the third term,

$$\begin{aligned}&2\langle L_+w|L_-w\rangle _{L^2(Q)}+\int _Q(L_+w)^2{ d}x{ d}t \ge \varkappa ^{-1}\kappa \tau \lambda \int _Q \phi [|\nabla w|_A^2+(\partial _tw)^2]{ d}x{ d}t\\&\quad +\tau ^3\lambda ^4\delta \int _Q\phi ^3w^2{ d}x{ d}t+{\mathscr {B}}(w) ,\quad \lambda \ge \lambda _2,\; \tau \ge \tau _2. \end{aligned}$$

We find by making elementary calculations

$$\begin{aligned} \aleph _3|{\mathscr {B}}(w)| \le \int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3u^2+\tau \lambda \phi \left( |\nabla _A u|^2+(\partial _tu)^2\right) \right] { d}\mu , \end{aligned}$$

for $\lambda \ge \lambda _3$ and $\mu \ge \mu _3$. This and

$$\begin{aligned} \Vert Lw\Vert _{L^2(Q)}^2\ge 2\langle L_+w|L_-w\rangle _{L^2(Q)}+\Vert L_+w\Vert _{L^2(Q)}^2 \end{aligned}$$

imply

$$\begin{aligned}&\aleph \int _Q\left[ \tau ^3\lambda ^4\phi ^3 w^2+\tau \lambda \phi |\mathbf{D }_A w|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Q\left( {\mathcal {L}}_{A,0}^ww\right) ^2{ d}x{ d}t +\int _{\partial Q} \left[ \tau ^3\lambda ^3\phi ^3w^2+\tau \lambda \phi |\mathbf{D }_A w|^2\right] { d}\mu . \end{aligned}$$

(3.18)

We take in this inequality $w=\Phi ^{-1}u$ with $u\in H^2 (Q;{\mathbb {R}})$. In light of the identities

$$\begin{aligned} \Phi ^{-1}\nabla u&= -\tau \lambda w\nabla \psi +\nabla w, \\ \Phi ^{-1}\partial _t u&= -\tau \lambda w\partial _t \psi +\partial _t w, \end{aligned}$$

we obtain an inequality similar to (3.1) which leads to (3.1) by observing that the additional terms in the right hand side appearing in this intermediate inequality can be absorbed by the terms in left hand side.

Proof of (3.17). Set

$$\begin{aligned} \chi =|\nabla \psi |_A^2-(\partial _t\psi )^2. \end{aligned}$$

We have

$$\begin{aligned} \Delta _A\phi&=\text{ div }(A\nabla e^{\lambda \psi })=\text{ div }(\lambda \phi A\nabla \psi )\\&=\lambda ^2\phi |\nabla \psi |_A^2+\lambda \phi \Delta _A\psi , \\ \partial _t^2\phi&=\partial _t(\lambda \phi \partial _t\psi )=\lambda ^2\phi (\partial _t\psi )^2+\lambda \phi \partial _t^2\psi . \end{aligned}$$

That is

$$\begin{aligned} {\mathcal {L}}_{A,0}^w\phi = \lambda ^2\phi \chi +\lambda \phi {\mathcal {L}}_{A,0}^w\psi . \end{aligned}$$

(3.19)

In light of (3.19) we get

$$\begin{aligned} a(x,t)&=\tau ^2\left( |\nabla \phi |_A^2-(\partial _t\phi )^2\right) =\tau ^2\lambda ^2\phi ^2\chi , \\ b(x,t)&=-\tau \left( \Delta _A \phi -\partial _t^2\phi \right) \\&=-\tau \lambda ^2\phi \chi -\tau \lambda \phi {\mathcal {L}}_{A,0}^w\psi , \\ B&=-2\tau A\nabla \phi =-2\tau \lambda \phi A\nabla \psi , \\ d&=2\tau \partial _t\phi = 2\tau \lambda \phi \partial _t\psi . \end{aligned}$$

Since

$$\begin{aligned} -aB/2= \tau ^3\lambda ^3\phi ^3\chi A\nabla \psi , \end{aligned}$$

we find

$$\begin{aligned} -\text{ div }(aB/2)=3\tau ^3\lambda ^4\phi ^3\chi |\nabla \psi |^2+\tau ^3\lambda ^3\phi ^3\text{ div }(\chi A\nabla \psi ). \end{aligned}$$

(3.20)

Also, as

$$\begin{aligned} -ad/2=-\tau ^3\lambda ^3\phi ^3\chi \partial _t\psi , \end{aligned}$$

we obtain

$$\begin{aligned} -\partial _t(ad/2)=-3\tau ^3\lambda ^4\phi ^3\chi (\partial _t\psi )^2- \tau ^3\lambda ^3\phi ^3\partial _t(\chi \partial _t\psi ). \end{aligned}$$

(3.21)

We get by putting together (3.20) and (3.21)

$$\begin{aligned} -\text{ div }(aB/2)-\partial _t(ad/2)= 3\tau ^3\lambda ^4\phi ^3\chi ^2+\tau ^3\lambda ^3\phi ^3\left[ \text{ div }(\chi A\nabla \psi )-\partial _t(\chi \partial _t\psi )\right] . \end{aligned}$$

(3.22)

On the other hand,

$$\begin{aligned} ab=-\tau ^3\lambda ^4\phi ^3\chi ^2-\tau ^3\lambda ^3\phi ^3\chi {\mathcal {L}}_{A,0}^w\psi . \end{aligned}$$

(3.23)

Set

$$\begin{aligned} \chi _1= \text{ div }(\chi A\nabla \psi )-\partial _t(\chi \partial _t\psi )-\chi {\mathcal {L}}_{A,0}^w\psi =(\nabla \chi |\nabla \psi )_A-\partial _t\chi \partial _t\psi . \end{aligned}$$

A combination of (3.22) and (3.23) yields

$$\begin{aligned} -\text{ div }(aB/2)-\partial _t(ad/2)+ab=2\tau ^3\lambda ^4\phi ^3\chi ^2+\tau ^3\lambda ^3\phi ^3\chi _1 . \end{aligned}$$

(3.24)

We have again from (3.19)

$$\begin{aligned} {\mathcal {L}}_{A,0}^wb&={\mathcal {L}}_{A,0}^w\left( -\tau \lambda ^2\phi \chi -\tau \lambda \phi {\mathcal {L}}_{A,0}^w\psi \right) \\&=- \left( \lambda ^2\phi \chi +\lambda \phi {\mathcal {L}}_{A,0}^w\psi \right) \left( \tau \lambda ^2\chi +\tau \lambda {\mathcal {L}}_{A,0}^w\psi \right) \\&\quad -\phi \left( \tau \lambda ^2{\mathcal {L}}_{A,0}^w\chi +\tau \lambda \left( {\mathcal {L}}_{A,0}^w\right) ^2\psi \right) . \end{aligned}$$

Therefore

$$\begin{aligned} {\mathcal {L}}_{A,0}^w(b/2)\ge -\tau \lambda ^4\phi \delta ,\quad \lambda \ge \lambda _4,\; \tau >0, \end{aligned}$$

(3.25)

where we used that $\chi ^2\ge \delta$.

Inequality (3.17) then follows by combining (3.24) and (3.25) and using that

$$\begin{aligned} {\hat{a}}= -\text{ div }(aB/2)-\partial _t(ad/2)+ab+{\mathcal {L}}_{A,0}^w(b/2). \end{aligned}$$

$\square$

Define $\partial _{\nu _A}\psi _0$ by

$$\begin{aligned} \partial _{\nu _A}\psi _0=(\nabla \psi _0|\nu )_A \end{aligned}$$

and set

$$\begin{aligned} \Gamma _+=\Gamma _+^{\psi _0}=\left\{ x\in \Gamma ;\; \partial _{\nu _A}\psi _0(x)>0\right\} ,\quad \Sigma _+=\Sigma _+^{\psi _0}=\Gamma _+\times (t_1,t_2). \end{aligned}$$

Let $w\in H^2(Q,{\mathbb {R}})$ satisfying $w=0$ on $\Sigma$. In that case it is straightforward to check that ${\hat{g}}(w)$ defined in the preceding proof takes the form

$$\begin{aligned} {\hat{g}}(w)=-\tau \lambda \phi (\partial _\nu w)^2|\nu |_A^2\partial _{\nu _A}\psi _0. \end{aligned}$$

(3.26)

Furthermore, if $u=\Phi w$ then

$$\begin{aligned} \partial _\nu u=\Phi \partial _\nu w. \end{aligned}$$

(3.27)

In light of identities (3.26) and (3.27), slight modifications of the last part of the preceding proof enable us to establish the following result.

Theorem 3.2

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(Q,{\mathbb {R}})$ satisfying $u=0$ on $\Sigma$ and $u=\partial _tu=0$ in $\Omega \times \{t_1,t_2\}$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 u^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left[ {\mathcal {L}}_{A,0}^wu\right] ^2{ d}x{ d}t +\tau \lambda \int _{\Sigma _+} e^{2\tau \phi }\phi (\partial _\nu u)^2 { d}\sigma { d}t. \end{aligned}$$

(3.28)

Let us see that Theorem 3.1 remains valid whenever we add to ${\mathcal {L}}_{A,0}^w$ a first order operator. Consider then the operator

$$\begin{aligned} {\mathcal {L}}_A^w={\mathcal {L}}_{A,0}^w+q_0\partial _t+\sum _{j=1}^n q_j\partial _j+p, \end{aligned}$$

where $q_0,\ldots ,q_n$ and p belong to $L^\infty (Q,{\mathbb {C}})$ and satisfy

$$\begin{aligned} \max _{0\le i\le n}\Vert q_i\Vert _{L^\infty (Q)}\le {\mathfrak {m}},\quad \Vert p\Vert _{L^\infty (Q)} \le {\mathfrak {m}}. \end{aligned}$$

Let $u=v+iw\in H^2(Q,{\mathbb {C}})$ and apply Theorem 3.1 to both v and w. We obtain by adding side by side the inequalities we obtain by taking in (3.1) $u=v$ and then $u=w$

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }|{\mathcal {L}}_{A,0}^wu|^2{ d}x{ d}t +\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}\mu . \end{aligned}$$

(3.29)

Since

$$\begin{aligned} |{\mathcal {L}}_{A,0}^wu|^2\le 2|{\mathcal {L}}_A^wu|^2+2(n+2){\mathfrak {m}}^2(\varkappa |\nabla u|_A^2+|\partial _tu|^2+|u|^2) \end{aligned}$$

and the term

$$\begin{aligned} 2(n+2){\mathfrak {m}}^2\int _Qe^{2\tau \phi }(\varkappa |\nabla u|_A^2+|\partial _tu|^2+|u|^2){ d}x{ d}t \end{aligned}$$

can be absorbed by the left hand side of (3.29), we get the following result:

Corollary 3.1

We find three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(Q,{\mathbb {C}})$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }|{\mathcal {L}}_A^wu|^2{ d}x{ d}t+\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}\mu , \end{aligned}$$

(3.30)

Finally, we note that the preceding arguments allow us also to prove the following corollary.

Corollary 3.2

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(Q,{\mathbb {C}})$ satisfying $u=0$ on $\Sigma$ and $u=\partial _tu=0$ in $\Omega \times \{t_1,t_2\}$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\mathbf{D }_A u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left| {\mathcal {L}}_A^wu\right| ^2{ d}x{ d}t +\tau \lambda \int _{\Sigma _+} e^{2\tau \phi }\phi |\partial _\nu u|^2 { d}\sigma { d}t, \end{aligned}$$

(3.31)

We close this subsection by a one-parameter Carleman inequality that we obtain as a special case of Corollary 3.1 in which we fixed $\lambda \ge \lambda ^*$.

Theorem 3.3

There exist two constants $\aleph =\aleph ({\mathfrak {d}})>0$ and $\tau ^*=\tau ^*({\mathfrak {d}})>0$ so that, for any $\tau \ge \tau ^*$ and $u\in C_0^\infty (Q,{\mathbb {C}})$, we have

$$\begin{aligned} \sum _{|\alpha |\le 1}\tau ^{2(2-|\alpha |)}\int _Qe^{2\tau \phi }|\partial ^\alpha u|^2{ d}x{ d}t \le \aleph \tau \int _Qe^{2\tau \phi }|{\mathcal {L}}_A^wu|^2{ d}x{ d}t, \end{aligned}$$

3.2 Geometric form of the Carleman inequality

Pick $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$ and let $(a^{k\ell }(x))=(a_{k\ell }(x))^{-1}$, $x\in {\overline{\Omega }}$. Consider then on ${\overline{\Omega }}$ the Riemannian metric g defined as follows

$$\begin{aligned} g_{k\ell }(x)=|\text{ det }(A)|^{1/(n-2)}a_{k\ell }(x),\quad x\in {\overline{\Omega }}. \end{aligned}$$

Set then

$$\begin{aligned} (g^{k\ell }(x))=(g_{k\ell }(x))^{-1},\quad |g(x)|=|\text{ det }((g_{k\ell }(x))|,\quad x\in {\overline{\Omega }}. \end{aligned}$$

As usual define on $T_x{\overline{\Omega }}={\mathbb {R}}^n$ the inner product

$$\begin{aligned} (X|Y)_{g(x)}=\sum _{k,\ell =1}g_{k\ell }(x)X_kY_k,\quad X=\sum _{k=1}^nX_i\partial _i\in {\mathbb {R}}^n,\; Y=\sum _{k=1}^nY_i\partial _i\in {\mathbb {R}}^n, \end{aligned}$$

where $(\partial _1,\ldots ,\partial _n)$ is the dual basis of the Euclidean basis of ${\mathbb {R}}^n$. Set

$$\begin{aligned} |X|_{g(x)}=(X|X)_{g(x)}^{1/2},\quad X=\sum _{k=1}^nX_i\partial _i\in {\mathbb {R}}^n. \end{aligned}$$

We use for notational convenience $(X|Y)_g$ and $|X|_g$ instead of $(X|Y)_{g(x)}$ and $|X|_{g(x)}$.

Recall that the gradient of $u\in H^1(\Omega )$ is the vector field given by

$$\begin{aligned} \nabla _gu(x)=\sum _{k,\ell =1}^n g^{k\ell }(x)\partial _ku(x)\partial _\ell ,\quad x\in \Omega , \end{aligned}$$

and the divergence of a vector field $X=\sum _{\ell =1}^nX_\ell \partial _k$ with $X_\ell \in H^1(\Omega )$, $1\le \ell \le n$, is defined as follows

$$\begin{aligned} \text{ div}_g(X)(x)=\frac{1}{\sqrt{|g(x)|}}\sum _{\ell =1}\partial _\ell (\sqrt{|g(x)}|X_\ell (x)),\quad x\in \Omega . \end{aligned}$$

The usual Laplace–Betrami operator associated to the metric g is given, for $u\in H^2(\Omega )$, by

$$\begin{aligned} \Delta _gu(x)=\text{ div}_g\nabla _gu(x)=\frac{1}{\sqrt{|g(x)|}}\sum _{k,\ell =1}\partial _\ell \left( \sqrt{|g(x)|}g^{k\ell }(x)\partial _ku(x)\right) ,\quad x\in \Omega . \end{aligned}$$

Straightforward computations show that $\Delta _Au=\sqrt{|g|}\Delta _gu$, from which we deduce the following identity

$$\begin{aligned} \Delta _gu =\Delta _{\sqrt{|g|}^{-1}A}u-2\left( \nabla \sqrt{|g|}^{-1}\Big |\nabla u\right) _A-u\Delta _A\sqrt{|g|}^{-1}. \end{aligned}$$

(3.32)

We assume in this subsection that $\psi$ is a weight function for the wave operator $\Delta _{\sqrt{|g|}^{-1}A}-\partial _t^2$ with $\sqrt{|g|}^{-1}A$-pseudo convexity constant $\kappa >0$. Hereafter

$$\begin{aligned} {\mathcal {L}}_g^wu=\Delta _gu-\partial _t^2u+(P|\nabla _g u)_g+q_1\partial _tu+q_0u,\quad u\in H^2(\Omega ), \end{aligned}$$

where $P=\sum _{\ell =1}^nP_i\partial _i$, $q_0$ and q are so that $P_1,\ldots ,P_n, q_0,q_1$ belong to $L^\infty (Q;{\mathbb {C}})$ and satisfy

$$\begin{aligned} \Vert P_\ell \Vert _{L^\infty (Q)}\le {\mathfrak {m}},\; 1\le \ell \le n,\quad \Vert q_0\Vert _{L^\infty (Q)}\le {\mathfrak {m}},\quad \Vert q_1\Vert _{L^\infty (Q)}\le {\mathfrak {m}}. \end{aligned}$$

As in the preceding section ${\mathfrak {d}}=(\Omega , t_1,t_2,{\mathfrak {m}}, \varkappa ,\kappa ,\delta ,{\mathfrak {b}})$ with ${\mathfrak {b}}\ge \Vert \psi \Vert _{C^4({\overline{Q}})}$. In light of (3.32) we deduce the following Carleman inequality from Corollary 3.1 in which

$$\begin{aligned} |\mathbf{D }_gu|_g^2=|\nabla _gu|_g^2+|\partial _tu|^2, \end{aligned}$$

${ d}V_{{\overline{\Omega }}} =\sqrt{|g|}{ d}x_1\ldots { d}x_n$ is the volume form and

$$\begin{aligned} { d}\mu =\mathbbm {1}_{\Gamma \times (t_1,t_2)}{ d}V_\Gamma { d}t +\mathbbm {1}_{\Omega \times \{t_1,t_2\}}{ d}V_{{\overline{\Omega }}}\delta _t, \end{aligned}$$

where ${ d}V_\Gamma$ is the volume form on $\Gamma$.

Theorem 3.4

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(Q,{\mathbb {C}})$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\mathbf{D }_g u|_g^2\right] { d}V_{{\overline{\Omega }}}{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left| {\mathcal {L}}_g^wu\right| ^2{ d}x{ d}t+\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\mathbf{D }_g u|_g^2\right] { d}\mu , \end{aligned}$$

(3.33)

3.3 Unique continuation

We assume, for simplicity, in the present subsection that $t_1=-{\mathfrak {t}}$ and $t_2={\mathfrak {t}}$ with ${\mathfrak {t}}>0$.

We start with unique continuation across a particular convex hypersurface. To this end, we set

$$\begin{aligned} E_+({\tilde{x}},c)=\{x=(x',x_n)\in {\mathbb {R}}^n ;\; 0&\le x_n-{\tilde{x}}_n<c\quad \text{ and }\,\; x_n-{\tilde{x}}_n\quad \ge |x'-{\tilde{x}}'|^2/c\}, \end{aligned}$$

where ${\tilde{x}}=({\tilde{x}}',{\tilde{x}}_n)\in {\mathbb {R}}^n$ and $c>0$.

Theorem 3.5

Suppose, for some $r>0$, that $B({\tilde{x}},r)\Subset \Omega$. We find $0<c^*=c^*(\varkappa ,{\mathfrak {m}})$ with the property that, for any $0<c<c^*$, there exist $\tilde{{\mathfrak {t}}}=\tilde{{\mathfrak {t}}}(c,\varkappa )$ and $0<\rho =\rho (c,\varkappa )<r$ so that, if ${\mathfrak {t}}\ge \tilde{{\mathfrak {t}}}$ then there exists $0<{\mathfrak {t}}_0={\mathfrak {t}}_0(\varkappa ,{\mathfrak {m}},{\mathfrak {t}})\le {\mathfrak {t}}$ satisfying: for any $u\in H^2({\mathcal {Q}};{\mathbb {C}})$, with ${\mathcal {Q}}=B({\tilde{x}},r)\times (-{\mathfrak {t}},{\mathfrak {t}})$, such that ${\mathcal {L}}_A^wu=$ in ${\mathcal {Q}}$ and $\text{ supp }(u(\cdot ,t))\cap B({\tilde{x}},r)\subset E_+({\tilde{x}},c)$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$, we have $u=0$ in $B({\tilde{x}},\rho )\times (-{\mathfrak {t}}_0,{\mathfrak {t}}_0)$.

Proof

Let ${\tilde{x}}\in \Omega$ and ${\mathfrak {e}}_n=(0,1)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}}$. Set $x_0={\tilde{x}}+c{\mathfrak {e}}_n$ and let $0<r_0\le \min (r,c/2)$. Define

$$\begin{aligned} \psi _0(x)=\psi _0(x',x_n)=|x-x_0|^2/2. \end{aligned}$$

As

$$\begin{aligned} |x-x_0|\le |x-{\tilde{x}}|+|{\tilde{x}}-x_0|\le r_0+c\le 3c/2,\quad x\in B({\tilde{x}},r_0), \end{aligned}$$

we find in a straightforward manner that, for some constant $\aleph =\aleph ({\mathfrak {m}})$, we have

$$\begin{aligned} (\Theta (\psi _0)(x)\xi |\xi )\ge 2\varkappa ^2-\aleph c,\quad x\in B({\tilde{x}},r_0),\; \xi \in {\mathbb {R}}^n, \end{aligned}$$

We fix $0<c<c^*=\varkappa ^2/\aleph$. With this choice of c we get

$$\begin{aligned} (\Theta (\psi _0)(x)\xi |\xi )\ge \varkappa ^2|\xi |^2,\quad x\in B({\tilde{x}},r_0),\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

It is then not difficult to check that, where $E_+=E_+({\tilde{x}},c)$,

$$\begin{aligned} (E_+\cap B({\tilde{x}},r_0)) {\setminus }\{{\tilde{x}}\}\subset \left\{ x\in B({\tilde{x}},r_0){\setminus }\{{\tilde{x}}\};\; \psi _0 (x)<\psi _0({\tilde{x}})=c^2/2\right\} . \end{aligned}$$

Pick $\chi \in C_0^\infty (B({\tilde{x}},r_0))$ satisfying $\chi =1$ in $B({\tilde{x}},\rho _1)$, for some fixed $0<\rho _1<r_0$. Fix then $\epsilon >0$ in such a way that

$$\begin{aligned} E_+\cap [B({\tilde{x}},r){\setminus } {\overline{B}}({\tilde{x}},\rho _1 )]\subset \left\{ x\in B({\tilde{x}},r);\; \psi _0 (x)<\psi _0 ({\tilde{x}})-\epsilon \right\} . \end{aligned}$$

Also, choose $0<\rho _0<\rho _1$ such that

$$\begin{aligned} E_+\cap B({\tilde{x}},\rho _0)\subset \left\{ x\in B({\tilde{x}},r);\; \psi _0 (x)>\psi _0 ({\tilde{x}})-\epsilon /2 \right\} . \end{aligned}$$

We are going to apply Corollary 3.1 with Q substituted by ${\mathcal {Q}}=B({\tilde{x}},r)\times (-{\mathfrak {t}},{\mathfrak {t}})$. Let

$$\begin{aligned} \psi (x,t)=\psi _0(x)-\gamma t^2/2 +C, \end{aligned}$$

where the constant $C >0$ is chosen sufficiently large in order to guarantee that $\psi \ge 0$. From Example 2.1 we can easily see that $\phi =e^{\lambda \psi }$ is a weight function for ${\mathcal {L}}_{A,0}^w$ in ${\mathcal {Q}}$ provided that $0<\gamma <\min (c/(2{\mathfrak {t}}),\varkappa /4)$.

Suppose that ${\mathfrak {t}}$ is chosen sufficiently large in such a way $c/(2{\mathfrak {t}})<\varkappa /4$. In that case we can take $\gamma =c/(4{\mathfrak {t}})$.

Fix $\gamma$ as above and let $0<\varrho \le 1$ to be determined later. Let $\lambda ^*$ and $\tau ^*$ be as in Corollary 3.1. In the sequel we fix $\lambda \ge \lambda ^*$. Set

$$\begin{aligned} \mathbf{Q }_0&=[B({\tilde{x}},\rho _0)\cap E_+]\times (-\varrho {\mathfrak {t}},\varrho {\mathfrak {t}}), \\ \mathbf{Q }_1&= \left\{ E_+\cap \left[ B({\tilde{x}},r_0){\setminus } {\overline{B}}({\tilde{x}},\rho _1)\right] \right\} \times (-{\mathfrak {t}},{\mathfrak {t}}), \\ \mathbf{Q }_2&=[B({\tilde{x}},r_0)\cap E_+]\times [(-{\mathfrak {t}},- {\mathfrak {t}}/2)\cup ({\mathfrak {t}}/2,{\mathfrak {t}})]. \end{aligned}$$

Then straightforward computations show

$$\begin{aligned} e^{\lambda \psi }\ge c_0&=e^{\lambda (c^2/2-\epsilon /2-\gamma \varrho ^2 {\mathfrak {t}}^2/2+\delta )}\quad \text{ in }\; \mathbf{Q }_0, \\ e^{\lambda \psi }\le c_1&=e^{\lambda (c^2/2-\epsilon +\delta )}\quad \text{ in }\; \mathbf{Q }_1, \\ e^{\lambda \psi }\le c_2&=e^{\lambda (c^2/2-\gamma {\mathfrak {t}}^2/8+\delta )}\quad \text{ in }\; \mathbf{Q }_2. \end{aligned}$$

In these inequalities we substitute $\gamma {\mathfrak {t}}$ by c/4 in order to get

$$\begin{aligned} e^{\lambda \psi }\ge c_0&=e^{\lambda (c^2/2-\epsilon /2- c \varrho ^2 {\mathfrak {t}}/8+\delta )}\quad \text{ in }\; \mathbf{Q }_0, \\ e^{\lambda \psi }\le c_1&=e^{\lambda (c^2/2-\epsilon +\delta )}\quad \text{ in }\; \mathbf{Q }_1, \\ e^{\lambda \psi }\le c_2&=e^{\lambda (c^2/2-c {\mathfrak {t}}/32+\delta )}\quad \text{ in }\; \mathbf{Q }_2. \end{aligned}$$

If ${\mathfrak {t}}\ge 2\epsilon /c$ we choose $\varrho$ so that $c\varrho ^2{\mathfrak {t}}=2\epsilon$. In that case we have

$$\begin{aligned} e^{\lambda \psi }\ge c_0&=e^{\lambda (c^2/2-3\epsilon /4+\delta )}\quad \text{ in }\; \mathbf{Q }_0, \\ e^{\lambda \psi }\le c_1&=e^{\lambda (c^2/2-\epsilon +\delta )}\quad \text{ in }\; \mathbf{Q }_1, \\ e^{\lambda \psi }\le c_2&=e^{\lambda (c^2/2-c {\mathfrak {t}}/32+\delta )}\quad \text{ in }\; \mathbf{Q }_2. \end{aligned}$$

In consequence $c_1<c_0$. Furthermore, if ${\mathfrak {t}}> 24\epsilon /c$ then we have also $c_2<c_0$.

Let $u\in H^2({\mathcal {Q}};{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^wu=0$ in ${\mathcal {Q}}$ and $\text{ supp }(u(\cdot ,t))\cap B({\tilde{x}},r)\subset E_+$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$. Define $v=\chi (x)\vartheta (t)u$ with $\vartheta \in C_0^\infty ((-{\mathfrak {t}},{\mathfrak {t}}))$ satisfying $\vartheta =1$ in $[- {\mathfrak {t}}/2,{\mathfrak {t}}/2]$.

As ${\mathcal {L}}_A^wu=0$ in ${\mathcal {Q}}$, elementary computations show that ${\mathcal {L}}_A^wv=f_1+f_2$ in ${\mathcal {Q}}$, where

$$\begin{aligned} f_1&=2\vartheta (\nabla \chi |\nabla u)_A+\vartheta u\text{ div }(A\nabla \chi ) +\vartheta u\sum _{j=1}^n q_j\partial _j\chi , \\ f_2&= 2\chi \partial _tu\partial _t\vartheta +\chi u\partial _t^2\vartheta +q_0\chi u\partial _t\vartheta . \end{aligned}$$

Taking into account that $\text{ supp }(f_1)\subset \mathbf{Q }_1$ and $\text{ supp }(f_2)\subset \mathbf{Q }_2$, we find by applying Corollary 3.1, where $\tau \ge \tau ^*$,

$$\begin{aligned}&\int _{B({\tilde{x}},\rho _0)\times (-\varrho {\mathfrak {t}},\varrho {\mathfrak {t}})}|u|^2{ d}x{ d}t =\int _{\mathbf{Q }_0}|v|^2{ d}x{ d}t \nonumber \\&\quad \le \aleph \tau ^{-3}\left[ e^{-\tau (c_0 -c_1)}\int _{\mathbf{Q }_1}|f_1|^2{ d}x{ d}t+e^{-\tau (c_0 -c_2)}\int _{\mathbf{Q }_2}|f_2|^2{ d}x{ d}t\right] . \end{aligned}$$

(3.34)

Passing then to the limit, as $\tau$ tends to $\infty$, in (3.34) in order to obtain that $u=0$ in $B({\tilde{x}},\rho _0)\times (-\varrho {\mathfrak {t}},\varrho {\mathfrak {t}})$. $\square$

Definition 3.1

We will say that ${\mathcal {L}}_A^w$ has the weak unique continuation property in Q if we can find $0<\tau \le {\mathfrak {t}}$ so that, for any open subset ${\mathcal {O}}$ with $\overline{{\mathcal {O}}}\subsetneqq \Omega$, there exists an open subset ${\mathcal {O}}_0$ with ${\mathcal {O}}_0\supsetneqq \overline{{\mathcal {O}}}$ so that, for any $u\in H^2(Q;{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^wu=0$ in Q and $u=0$ in ${\mathcal {O}}\times (-{\mathfrak {t}},{\mathfrak {t}})$, we have $u=0$ in ${\mathcal {O}}_0\times (-\tau ,\tau )$.

Theorem 3.6

There exist a universal constant ${\mathfrak {t}}^*>0$ and a neighborhood ${\mathcal {N}}$ of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }};{\mathbb {R}}^n\times {\mathbb {R}}^n)$ so that, for each ${\mathfrak {t}}\ge {\mathfrak {t}}^*$ and $A\in {\mathcal {N}}$, ${\mathcal {L}}_A^w$ has the weak unique continuation property in Q.

Proof

Set

$$\begin{aligned} H=\left\{ (x',x_n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}};\; x_n=0\right\} . \end{aligned}$$

Let ${\mathcal {N}}_0$ be the neighborhood of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }};{\mathbb {R}}^n\times {\mathbb {R}}^n)$ given by Lemma 2.2. Pick $A\in {\mathcal {N}}_0$ and let $u\in H^2(Q;{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^w=0$ in Q and $u=0$ in ${\mathcal {O}}\times (-{\mathfrak {t}},{\mathfrak {t}})$, where ${\mathcal {O}}$ is an open subset satisfying $\overline{{\mathcal {O}}}\subsetneqq \Omega$.

Fix $y\in \partial {\mathcal {O}}\cap \Omega$ and $r>0$ so that $B(y,r)\subset \Omega$. Pick then $y_0\in {\mathcal {O}}\cap B(y,r)$ sufficiently close to y in such a way that $\partial B(y_0, d)\cap \partial {\mathcal {O}}\ne \emptyset$, with $d=\text{ dist }(y_0,\partial {\mathcal {O}})$. Let then $z\in \partial B(y_0, d)\cap \partial {\mathcal {O}}$. Making a translation and a change of coordinates if necessary, we may assume that $z=0$ and $B(y_0,d)\subset \{(y',y_n)\in {\mathbb {R}}^n;\; x_n<0\}$. We still denote, for notational convenience, the new matrix obtained after this translation and this change of coordinates by A. In that case, according to Lemma 2.2, A belongs to the neighborhood ${\mathcal {N}}$ appearing in this lemma. Whence, $\text{ supp }(u(\cdot , t))\cap B(z,\rho )\subset H_+$, for some $\rho >0$, with

$$\begin{aligned} H_+=\left\{ (x',x_n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}};\; x_n\ge 0\right\} . \end{aligned}$$

Let $\varphi$ given by (2.4) with $\vartheta =0$ and ${\tilde{A}}=A_H$. If v is defined in a neighborhood ${\tilde{\omega }}$ by $v(y,\cdot )=u(\varphi ^{-1}(y),\cdot )$ then straightforward computations give ${\mathcal {L}}^w_{{\tilde{A}}}v=0$ in ${\tilde{\omega }}$ and $\text{ supp }(v(\cdot ,t))\subset E_+$ with

$$\begin{aligned} E_+=\left\{ (y',y_n)\in {\tilde{\omega }};\; 0<y_n<1, y_n\ge |y'|^2\right\} . \end{aligned}$$

We complete the proof similarly to that of Theorem 3.5, with ${\tilde{x}}=0$, $B(0,r)\Subset {\tilde{\omega }}$, $\varkappa =1/4$ and $c=1$. Note that in the present case

$$\begin{aligned} \psi (y,t)=(y_n-1)^2/2+|y'|^2/2-\gamma t^2/2+C, \end{aligned}$$

We get that there exists a universal constant ${\mathfrak {t}}^*$ so that, for any ${\mathfrak {t}}\ge {\mathfrak {t}}^*$, we find $0<\tau \le {\mathfrak {t}}$ for which $v=0$ in $\tilde{{\mathcal {U}}}\times (-\tau ,\tau )$, for some neighborhood $\mathcal {{\tilde{U}}}$ of 0. In consequence $u=0$ in ${\mathcal {U}}\times (-\tau ,\tau )$, where ${\mathcal {U}}$ is a neighborhood of z in $\Omega$. In other words we proved that $u=0$ in ${\mathcal {O}}_0\times (-\tau ,\tau )$, where ${\mathcal {O}}_0={\mathcal {O}}\cup {\mathcal {U}}\supsetneqq {\mathcal {O}}$. The proof is then complete. $\square$

The result of Theorem 3.6 is false without the condition that ${\mathfrak {t}}$ is sufficiently large as shows the non uniqueness result in [2]. The authors show that, in the two dimensional case, there exists ${\mathcal {U}}$, a neighborhood of the origin in ${\mathbb {R}}^2\times {\mathbb {R}}$, $p\in C^\infty ({\mathcal {U}})$, $u\in C^\infty ({\mathcal {U}})$ so that $(\Delta -\partial _t^2+p(x,t))u=0$ in ${\mathcal {U}}$ and $\text{ supp }(u)\subset {\mathcal {U}}\cap \{(x_1,x_2,t)\in {\mathbb {R}}^2\times {\mathbb {R}};\; x_2\ge 0\}$.

A better result than that in Theorem 3.6 can be obtained in the case of operators with time-independent coefficients. Let $\dot{{\mathcal {L}}}_A^w$ be the operator ${\mathcal {L}}_A^w$ when $q_0=0$, $q_j=q_j(x)$, $j=1,\ldots ,n$, and $p=p(x)$.

Theorem 3.7

[31] There exits ${\mathfrak {t}}^*={\mathfrak {t}}(\Omega ,\varkappa , {\mathfrak {m}})$ so that, for any ${\mathfrak {t}}>{\mathfrak {t}}^*$ and $\omega \Subset \Omega$, if $u\in H^2(Q)$ satisfies $\dot{{\mathcal {L}}}_A^w=0$ in Q and $u=0$ in $\omega \times (-{\mathfrak {t}},{\mathfrak {t}})$ then $u=0$ in $\Omega \times (-\tau ,\tau )$, where $\tau ={\mathfrak {t}}-{\mathfrak {t}}^*$.

The main idea in the proof of Theorem 3.7 consists in transforming, via the Fourier–Bros–Iagolnitzer transform, the wave operator $\dot{{\mathcal {L}}}_A^w$ into an elliptic operator for which uniqueness of continuation property result is known.

Theorem 3.6 can be used to establish a result on local uniqueness of continuation from Cauchy data on a part of the boundary.

Corollary 3.3

Let ${\mathfrak {t}}^*$ and ${\mathcal {N}}$ be as in Theorem 3.6 with $\Omega$ substituted by larger domain ${\hat{\Omega }}\Supset \Omega$. Let $\Gamma _0$ be a nonempty open subset of $\Gamma$ and $\Sigma _0=\Gamma _0\times (-{\mathfrak {t}},{\mathfrak {t}})$ with ${\mathfrak {t}}\ge {\mathfrak {t}}^*$. There exist ${\mathcal {U}}$ a neighborhood of a point of $\Gamma _0$ in $\Omega$ and $0<\tau \le {\mathfrak {t}}$ so that if $A\in {\mathcal {N}}$, and if $u\in H^2(Q;{\mathbb {C}})$ satisfies ${\mathcal {L}}_A^wu=0$ in Q and $u=\partial _\nu u=0$ on $\Sigma _0$ then $u=0$ in ${\mathcal {U}}\times (-\tau ,\tau )$.

Proof

Pick $A\in {\mathcal {N}}$ and $u\in H^2(Q;{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^wu=0$ in Q and $u=\partial _\nu u=0$ on $\Sigma _0$. Then there exists ${\mathcal {V}}\subset {\hat{\Omega }}$, a neighborhood of a point in $\Gamma _0$, so that ${\hat{u}}$, the extension of u by zero in ${\mathbb {R}}^n{\setminus } {\overline{\Omega }}$, belongs to $H^2(\Omega '\times (-{\mathfrak {t}},{\mathfrak {t}}))$, with $\Omega '=\Omega \cup {\mathcal {V}}$, satisfies ${\mathcal {L}}_A^w{\hat{u}}=0$ in $\Omega '\times (-{\mathfrak {t}},{\mathfrak {t}})$ and ${\hat{u}}=0$ in $[(\Omega '{\setminus } {\overline{\Omega }}) \cap {\mathcal {V}}]\times (-{\mathfrak {t}},{\mathfrak {t}})$. Theorem 3.6 allows us to conclude that there exist ${\mathcal {U}}$, a neighborhood of a point of $\Gamma _0$ in $\Omega$, and $0<\tau \le {\mathfrak {t}}$ so that $u=0$ in ${\mathcal {U}}\times (-\tau ,\tau )$. $\square$

It is worth mentioning that the following global unique continuation result from boundary data can be deduced from [6, Theorem 1.1].

Theorem 3.8

Let $\Gamma _0$ an arbitrary non empty open subset of $\Gamma$. There exits ${\mathfrak {t}}^*={\mathfrak {t}}(\Omega ,\varkappa , {\mathfrak {m}})$ so that, for any ${\mathfrak {t}}>{\mathfrak {t}}^*$ we find $0<{\mathfrak {t}}_0<{\mathfrak {t}}$ with the property that if $u\in C^\infty ({\overline{\Omega }}\times [-{\mathfrak {t}},{\mathfrak {t}}])$ satisfies ${\mathcal {L}}_{A,0}^wu=0$ and

$$\begin{aligned} u=\partial _\nu u=0\quad \text{ on }\quad \{\Gamma _0\times (-{\mathfrak {t}},{\mathfrak {t}})\}\cup \{\Gamma \times [(-{\mathfrak {t}},-{\mathfrak {t}}_0)\cup ({\mathfrak {t}}_0,{\mathfrak {t}})]\} \end{aligned}$$

then u is identically equal to zero.

We end this subsection by remarking that we can proceed similarly to Theorem 3.6 to prove the unique continuation property across a pseudo-convex hypersurface.

Theorem 3.9

Let $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$ and $H=\{x\in \omega ;\;\theta (x)=\theta ({\tilde{x}})\}$ be a A-pseudo-convex hypersurface defined in a neighborhood $\omega$ of ${\tilde{x}}\in \Omega$ with $\theta \in C^{3,1}({\overline{\omega }})$. Then there exist ${\mathcal {B}}$, a neighborhood of ${\tilde{x}}$, and ${\mathfrak {t}}^*>0$ so that, for each ${\mathfrak {t}}\ge {\mathfrak {t}}^*$, we find $0<\tau \le {\mathfrak {t}}$ with the property that if $u\in H^2(\omega \times (-{\mathfrak {t}},{\mathfrak {t}}))$ satisfies ${\mathcal {L}}_A^wu=0$ in $\omega \times (-{\mathfrak {t}},{\mathfrak {t}})$ and $\text{ supp }(u(\cdot ,t))\subset H_+=\{ x\in \omega ;\; \theta (x)\ge \theta ({\tilde{x}})\}$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$, then $u=0$ in ${\mathcal {B}}\times (-\tau ,\tau )$.

3.4 Observability inequality

We suppose in this subsection that $t_1=0$ and $t_2={\mathfrak {t}}>0$.

We shall need in the sequel the following technical lemma.

Lemma 3.1

Fix $0<\alpha <1$ and let $0\le \psi _0\in C^1({\overline{\Omega }})$ satisfying

$$\begin{aligned} \min _{x\in {\overline{\Omega }}}|\nabla \psi _0 |_A^2:=\delta _0>0. \end{aligned}$$

Let $\mathbf{m }=\Vert \psi _0\Vert _{L^\infty (\Omega )}$ and define, for an arbitrary constant $C>0$,

$$\begin{aligned} \psi (x,t)=\psi _0(x)-{\mathfrak {t}}^{-2+\alpha }(t-{\mathfrak {t}}/2)^2+C,\quad x\in {\overline{\Omega }},\; t\in [0,{\mathfrak {t}}]. \end{aligned}$$

(3.35)

If ${\mathfrak {t}}> {\mathfrak {t}}_\alpha =\max \left( \delta _0^{-1/[2(1-\alpha )]}, (64{\mathfrak {m}}/2)^{1/\alpha }\right)$ then

$$\begin{aligned}&\min _{{\overline{Q}}}\left( |\nabla \psi |_A^2-(\partial _t\psi )^2\right) ^2:=\delta >0, \end{aligned}$$

(3.36)

$$\begin{aligned}&\psi (x,t) \ge - {\mathfrak {t}}^\alpha /64+C,\quad (x,t)\in {\overline{\Omega }}\times [3{\mathfrak {t}}/8,5{\mathfrak {t}}/8], \end{aligned}$$

(3.37)

$$\begin{aligned}&\psi (x,t)\le -2{\mathfrak {t}}^\alpha /64+C,\quad (x,t)\in {\overline{\Omega }}\times \left( [0,{\mathfrak {t}}/4]\cup [3{\mathfrak {t}}/4,{\mathfrak {t}}]\right) . \end{aligned}$$

(3.38)

Proof

If ${\mathfrak {t}}\ge {\mathfrak {t}}_\alpha$ then

$$\begin{aligned} |\nabla \psi _0|^2-(\partial _t\psi )^2\ge \delta _0 -{\mathfrak {t}}^{-4+2\alpha } {\mathfrak {t}}^2=\delta _0-{\mathfrak {t}}^{-2(1-\alpha )}:=\delta >0. \end{aligned}$$

That is we proved (3.36).

Inequality (3.37) is straightforward. On the other hand, we have

$$\begin{aligned} \psi (x,t) \le \mathbf{m }-{\mathfrak {t}}^\alpha /16+C< 2{\mathfrak {t}}^\alpha /64 -{\mathfrak {t}}^\alpha /16+C=-2{\mathfrak {t}}^\alpha /64+C \end{aligned}$$

if $(x,t)\in {\overline{\Omega }}\times \left( [0,{\mathfrak {t}}/4]\cup [3{\mathfrak {t}}/4,{\mathfrak {t}}]\right)$. That is we proved (3.38). $\square$

We consider in this subsection the following wave operator

$$\begin{aligned} {\mathcal {L}}_A^w=\Delta _A-\partial _t^2 +p\partial _t +q, \end{aligned}$$

with $p,q\in L^\infty (\Omega ;{\mathbb {C}})$. We associate to ${\mathcal {L}}_A^w$ the IBVP

$$\begin{aligned} \left\{ \begin{array}{ll} {\mathcal {L}}_A^w=0\quad \text{ in }\; Q, \\ (u(\cdot , 0),\partial _tu(\cdot ,0))=(u_0,u_1), \\ u_{|\Sigma }=0. \end{array} \right. \end{aligned}$$

(3.39)

According to the semigroup theory, for all $(u_0,u_1)\in H_0^1(\Omega )\times L^2(\Omega )$, the IBVP (3.39) admits unique solution

$$\begin{aligned} u\in C([0,{\mathfrak {t}}];H_0^1(\Omega ))\cap C^1([0,{\mathfrak {t}}];L^2(\Omega )). \end{aligned}$$

We also know that $\partial _\nu u\in L^2(\Sigma )$ (hidden regularity).

From usual energy estimate for wave equations, if

$$\begin{aligned} {\mathcal {E}}_u(t)=\Vert \mathbf{D }_Au(\cdot ,t)\Vert _{L^2(\Omega ;{\mathbb {C}}^{n+1})},\quad 0\le t\le {\mathfrak {t}}, \end{aligned}$$

(3.40)

then

$$\begin{aligned} {\mathcal {E}}_u(t)\le \aleph _0 {\mathcal {E}}_u(0),\quad 0\le t\le {\mathfrak {t}}, \end{aligned}$$

(3.41)

where the constant $\aleph _0>0$ only depends of $\Omega$, A, ${\mathfrak {t}}$, p and q.

We apply (3.41) to $v(\cdot ,t)=u(\cdot ,s-t)$, with fixed $0<t\le s$. We find

$$\begin{aligned} {\mathcal {E}}_u(s-t)={\mathcal {E}}_v(t)\le \aleph _1 {\mathcal {E}}_v(0)=\aleph _1{\mathcal {E}}_u(s),\quad 0\le t\le s, \end{aligned}$$

where the constant $\aleph _1>0$ only depends of $\Omega$, A, ${\mathfrak {t}}$, p and q. We have in particular

$$\begin{aligned} {\mathcal {E}}_u(0)\le \aleph _1 {\mathcal {E}}_u(s),\quad 0\le t\le s. \end{aligned}$$

(3.42)

In light of (3.41) and (3.42) we get

$$\begin{aligned} \aleph ^{-1}{\mathcal {E}}_u(0)\le {\mathcal {E}}_u(t)\le \aleph {\mathcal {E}}_u(0),\quad 0\le t\le {\mathfrak {t}}, \end{aligned}$$

(3.43)

for some constant $\aleph >1$ only depending of $\Omega$, A, ${\mathfrak {t}}$, p and q.

We recall that $\Sigma _+=\Gamma _+\times (0,{\mathfrak {t}})$.

Theorem 3.10

Fix $0<\alpha <1$ and assume that $0\le \psi _0\in C^4({\overline{\Omega }})$ is A-pseudo-convex with constant $\kappa >0$ and let $\Gamma _+=\{x\in \Gamma ;\; \partial _{\nu _A}\psi _0(x)>0\}$. If $\tilde{{\mathfrak {t}}}_\alpha =\min \left( {\mathfrak {t}}_\alpha , (8\varkappa /\kappa )^{1/(2-\alpha )}\right)$, where ${\mathfrak {t}}_\alpha$ be as in Lemma 3.1, then, for any ${\mathfrak {t}}\ge \tilde{{\mathfrak {t}}}_\alpha$ and $(u_0,u_1)\in H_0^1(\Omega )\times L^2(\Omega )$, we have

$$\begin{aligned} \Vert (u_0,u_1)\Vert _{H_0^1(\Omega )\times L^2(\Omega )}\le \aleph \Vert \partial _\nu u\Vert _{L^2(\Sigma _+)}, \end{aligned}$$

where the constant $\aleph >0$ only depends of $\Omega$, ${\mathfrak {t}}$, $\varkappa$, $\kappa$, $\Gamma _+$ and u is the solution of the IBVP (3.39) corresponding to $(u_0,u_1)$.

Proof

Fix ${\mathfrak {t}}>\tilde{{\mathfrak {t}}}_\alpha$ and let $\psi$ defined as in (3.35) in which the constant $C>0$ is chosen sufficiently large to guarantee that $\psi \ge 0$. In that case we easily check that $\phi =e^{\lambda \psi }$ is a weight function for the operator ${\mathcal {L}}_{A,0}^w$ in Q.

Pick $\varrho \in C_0^\infty ({\mathfrak {t}}/8,7{\mathfrak {t}}/8)$ so that $\varrho =1$ in $[{\mathfrak {t}}/4,3{\mathfrak {t}}/4]$. Clearly, a simple density argument shows that Corollary 3.2 remains valid for $\varrho u$ for any solution u of (3.39) with $(u_0,u_1)\in H_0^1(\Omega )\times L^2(\Omega )$. According to this Corollary we have, for fixed $\lambda \ge \lambda ^*$ and any $\tau \ge \tau ^*$,

$$\begin{aligned} \aleph \int _Qe^{2\tau \phi } |\mathbf{D }_A (\varrho u)|^2{ d}x{ d}t\le&\int _Qe^{2\tau \phi }\left| {\mathcal {L}}_A^w(\varrho u)\right| ^2{ d}x{ d}t \nonumber \\&+\int _{\Sigma _+} e^{2\tau \phi }|\partial _\nu (\varrho u)|^2 { d}\sigma { d}t, \end{aligned}$$

(3.44)

But

$$\begin{aligned} {\mathcal {L}}_A^w(\varrho u)=2\varrho 'u+\varrho ''u. \end{aligned}$$

Hence (3.44) together with Poincaré’s inequality ($u(\cdot , t)\in H_0^1(\Omega )$) give

$$\begin{aligned} \aleph \int _Qe^{2\tau \phi } |\mathbf{D }_A (\varrho u)|^2{ d}x{ d}t\le&\int _{Q\cap { supp}(\varrho ')}e^{2\tau \phi }|\mathbf{D }_A u|^2{ d}x{ d}t \nonumber \\&+\int _{\Sigma _+} e^{2\tau \phi }|\partial _\nu u|^2 { d}\sigma { d}t. \end{aligned}$$

(3.45)

Define

$$\begin{aligned} c_0 =e^{\lambda (-\gamma {\mathfrak {t}}^2/64+C)}\quad \text{ and }\quad c_1 =e^{\lambda (-2\gamma {\mathfrak {t}}^2/64+C)}. \end{aligned}$$

If ${\mathcal {E}}_u$ is given by (3.40) then we get from (3.37), (3.38) and (3.45), where $\tau \ge \tau ^*$,

$$\begin{aligned} \aleph e^{\tau c_0}\int _{3{\mathfrak {t}}/8}^{5{\mathfrak {t}}/8} {\mathcal {E}}_u(t){ d}t \le e^{\tau c_1} \int _0^{\mathfrak {t}}{\mathcal {E}}_u(t){ d}t +\int _{\Sigma _+} e^{2\tau \phi }|\partial _\nu u|^2 { d}\sigma { d}t, \end{aligned}$$

This inequality together with (3.43) imply

$$\begin{aligned} \left( \aleph e^{\tau c_0} -e^{\tau c_1}\right) {\mathcal {E}}_u(0) \le \int _{\Sigma _+} e^{2\tau \phi }|\partial _\nu u|^2 { d}\sigma { d}t,\quad \tau \ge \tau ^*. \end{aligned}$$

As $c_0 >c_1$, we fix $\tau$ sufficiently large in such a way that ${\tilde{\aleph }}=\aleph e^{\tau c_0} -e^{\tau c_1}>0$. That is we have

$$\begin{aligned} {\tilde{\aleph }}{\mathcal {E}}_u(0) \le \int _{\Sigma _+} e^{2\tau \phi }|\partial _\nu u|^2 { d}\sigma { d}t,\quad \tau \ge \tau ^*. \end{aligned}$$

(3.46)

The expected inequality follows readily from (3.46). $\square$

From the calculations in Example 2.1 when $\psi _0(x)=|x-x_0|^2/2$, with $x_0\in {\mathbb {R}}^n{\setminus }{\overline{\Omega }}$, there exists a neighborhood ${\mathcal {N}}$ of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }};{\mathbb {R}}^{n\times n})$ so that that for any $A\in {\mathcal {N}}$, $\varkappa =1/2$ and $\psi _0$ is A-pseudo-convex with constant $\kappa =1/4$. In this case

$$\begin{aligned} \tilde{{\mathfrak {t}}}_\alpha =\max \left( d_0^{-1/(1-\alpha )},[16(d+d_0)]^{1/\alpha },16^{1/(2-\alpha )}\right) , \end{aligned}$$

with $d_0=\text{ dist }(x_0,{\overline{\Omega }})$ and $d=\text{ diam }(\Omega )$.

A result in the variable coefficients case was already established in [37, Theorem 1.1]. This result is based on a generalization of the multiplier method in which a vector field is used as an alternative to the multiplier. This vector field satisfies a certain convexity condition. Note however that the lower bound in ${\mathfrak {t}}$ appearing in [37, Theorem 1.1] is not easily comparable to that we used in Theorem 3.10. The minimal time guaranteeing observability was estimated in precise way in [14] for wave equations with $C^1$ variable coefficients. Recently an observability result was established in [36] for constant coefficients wave equation in the case of time-dependent domains. The minimal time in [36] is explicit.

4 Elliptic equations

We show briefly how we can modify the calculations we carried out for wave equations in order to retrieve Carleman inequalities for elliptic equations and the corresponding property of unique continuation. In this section

$$\begin{aligned} {\mathcal {L}}_A^e=\Delta _A+\sum _{\ell =1}^np_\ell \partial _\ell +q, \end{aligned}$$

where $p_1,\ldots p_n$ and q belong to $L^\infty (\Omega ;{\mathbb {C}})$ and satisfy

$$\begin{aligned} \Vert p_\ell \Vert _{L^\infty (\Omega )}\le {\mathfrak {m}},\; 1\le \ell \le n\quad \text{ and }\quad \Vert q\Vert _{L^\infty (\Omega )}\le {\mathfrak {m}}. \end{aligned}$$

Also, $0\le \psi \in C^4({\overline{\Omega }})$ is fixed so that

$$\begin{aligned} |\nabla \psi |\ge \delta \quad \text{ in }\; {\overline{\Omega }}, \end{aligned}$$

for some constant $\delta >0$.

4.1 Carleman inequality

Let $\phi =e^{\lambda \psi }$ and set ${\mathfrak {d}}=(\Omega ,\varkappa , \delta ,{\mathfrak {m}} )$.

Theorem 4.1

We find three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^2(\Omega ,{\mathbb {C}})$, we have

$$\begin{aligned}&\aleph \int _\Omega e^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda ^2 \phi |\nabla u|^2\right] { d}x \nonumber \\&\quad \le \int _\Omega e^{2\tau \phi }|{\mathcal {L}}_A^eu|^2{ d}x +\int _\Gamma e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\nabla u|^2\right] { d}\sigma . \end{aligned}$$

(4.1)

Proof

In this proof, $\lambda _k$ and $\tau _k$, $k=1,2,\ldots$, denote generic constants only depending on ${\mathfrak {d}}$.

Let $\Phi =e^{-\tau \phi }$, $\tau >0$. We have from the calculations of the preceding section

$$\begin{aligned} L=\Phi ^{-1}\Delta _A(\Phi w)=\Delta _Aw -2\tau (\nabla w|\nabla \phi )_A+\left[ \tau ^2 |\nabla \phi |_A^2-\tau \Delta _Aw\right] w. \end{aligned}$$

We decompose L in the following special form

$$\begin{aligned} L=L_0+L_1+c, \end{aligned}$$

with, for $w\in H^2(\Omega ,{\mathbb {R}})$,

$$\begin{aligned} L_0w&=\Delta _Aw+aw, \\ L_1w&=(B|\nabla w)+bw. \end{aligned}$$

The coefficients of $L_0$ and $L_1$ and c are given as follows

$$\begin{aligned} a&=\tau ^2|\nabla \phi |_A^2, \\ b&=-2\tau \Delta _A\phi , \\ c&=\tau \Delta _A\phi , \\ B&=-2\tau A\nabla \phi . \end{aligned}$$

We have

$$\begin{aligned} \langle L_0|L_1\rangle _{L^2(\Omega )}=\sum _{k=1}^4I_k, \end{aligned}$$

(4.2)

where

$$\begin{aligned} I_1&=\int _\Omega \Delta _A w(\nabla w|B) { d}x, \\ I_2&=\int _\Omega \Delta _A wbw{ d}x, \\ I_3&=\int _\Omega aw(\nabla w|B) { d}x, \\ I_4&=\int _\Omega abw^2{ d}x. \end{aligned}$$

Let

$$\begin{aligned} D=C/2-A(B')^t. \end{aligned}$$

Then straightforward modifications of the computations of the preceding section yield

$$\begin{aligned} I_1&=\int _\Omega (D\nabla w|\nabla w){ d}x+\int _\Gamma \left[ (\nabla w|\nu )_A(\nabla w|B)-(B/2|\nu )|\nabla w|_A^2\right] { d}\sigma , \end{aligned}$$

(4.3)

$$\begin{aligned} I_2&=-\int _\Omega b|\nabla w|_A^2{ d}x +\int _\Omega \Delta _A(b/2) w^2 { d}x \nonumber \\&\quad -\int _\Gamma (\nabla (b/2)|\nu )_Aw^2{ d}\sigma +\int _\Gamma (\nabla w|\nu )_Abw { d}\sigma , \end{aligned}$$

(4.4)

$$\begin{aligned} I_3&=-\int _\Omega \text{ div }(aB/2)w^2{ d}x+ \int _\Gamma a(B/2|\nu )w^2{ d}\sigma . \end{aligned}$$

(4.5)

Identities (4.3)–(4.5) in (4.2) give

$$\begin{aligned} \langle L_0|L_1\rangle _{L^2(\Omega )}=\int _\Omega ({\mathfrak {A}}\nabla w|\nabla w){ d}x+\int _\Omega {\mathfrak {a}}w^2{ d}x+\int _\Gamma g(w) { d}\sigma , \end{aligned}$$

(4.6)

with

$$\begin{aligned} {\mathfrak {A}}&=D-bA, \\ {\mathfrak {a}}&=ab+\Delta _A(b/2)-\text{ div }(aB/2), \\ g(w)&=(\nabla w|\nu )_A(\nabla w|B)-(B/2|\nu )|\nabla w|_A^2 \\&\quad -(\nabla (b/2)|\nu )_Aw^2+(\nabla w|\nu )_Abw+a(B/2|\nu )w^2. \end{aligned}$$

We have

$$\begin{aligned} ({\mathfrak {A}}\xi |\xi )=\tau \lambda ^2\left[ |\nabla \psi |^2(A\xi |\xi )+(\nabla \psi |A\xi )^2\right] +\tau \lambda (\tilde{{\mathfrak {A}}}\xi |\xi ), \end{aligned}$$

where $\tilde{{\mathfrak {A}}}$ is a matrix depending only on A and $\psi$. Therefore

$$\begin{aligned} ({\mathfrak {A}}\xi |\xi )\ge \tau \lambda ^2\delta ^2\varkappa |\xi |^2/2,\quad \lambda \ge \lambda _1. \end{aligned}$$

(4.7)

We have also

$$\begin{aligned} {\mathfrak {a}}=\tau ^3\lambda ^4\phi ^3|\nabla \psi |_A^4+\tilde{{\mathfrak {a}}}, \end{aligned}$$

where the reminder term $\tilde{{\mathfrak {a}}}$ contains, as for the wave equation, only terms with factors $\tau ^k\lambda ^\ell \phi ^m$, $1\le k,\ell ,m \le 3$ and terms with factor $\tau \lambda ^4\phi$. Hence

$$\begin{aligned} {\mathfrak {a}}\ge \tau ^3\lambda ^4\delta ^4\phi ^3/2,\quad \lambda \ge \lambda _2,\; \tau \ge \tau _2. \end{aligned}$$

(4.8)

The rest of the proof is almost similar to that of the wave equation. $\square$

Remark 4.1

The symbol of the principal part of the operator ${\mathcal {L}}_A^e$ is given by

$$\begin{aligned} p(x,\xi )=|\xi |_A^2(x)=(A(x)\xi |\xi ),\quad x\in {\overline{\Omega }},\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

Therefore if $\phi \in C^4({\overline{\Omega }})$ then we have $p(x,\xi +i\tau \nabla \phi )=p_0+ip_1$ with

$$\begin{aligned} p_0=|\xi |_A^2-\tau ^2|\nabla \phi |_A,\quad p_1=2\tau (\xi |\nabla \phi )_A. \end{aligned}$$

When $\phi =e^{\lambda \psi }$ we find, for $\tau \ge 1$,

$$\begin{aligned} \{p_0,p_1\}:=\sum _{j=1}^n\left[ \partial _{\xi _j}p_0\partial _{x_j}p_1-\partial _{x_j}p_0\partial _{x_j}p_1\right] =\tau ^2\left[ 2\lambda ^3\phi ^2|\nabla \psi |_A^4+O(\lambda ^2)\right] . \end{aligned}$$

In consequence $\phi$ satisfies the sub-ellipticity condition in [18, Theorem 8.3.1, page 190] if $\lambda$ is sufficiently large and hence the following Carleman inequality holds: there exist $\aleph >0$ and $\tau ^*>0$ only depending on $\Omega$ and bounds on the coefficients of ${\mathcal {L}}_A^e$ so that

$$\begin{aligned} \sum _{|\alpha |\le 1}\tau ^{2(2-|\alpha |)}\int _\Omega e^{2\tau \phi }|\partial ^\alpha u|^2{ d}x\le \aleph \tau \int _\Omega e^{2\tau \phi }|{\mathcal {L}}_A^eu|^2{ d}x,\quad u\in C_0^\infty (\Omega ),\; \tau \ge \tau ^*. \end{aligned}$$

In other words, if $\phi =e^{\lambda \psi }$ is a weight function for the elliptic operator ${\mathcal {L}}_{A,0}^e$ then $\phi$ possesses the sub-ellipticity condition for large $\lambda$.

4.2 Unique continuation

We use a similar method as for the wave equation. For sake of completeness, we provide some details.

We start with a unique continuation result across a convex hypersurface. To this end, we set

$$\begin{aligned} \psi (x',x_n)=(x_n-1)^2+|x'|^2,\quad (x',x_n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}}. \end{aligned}$$

As $|\nabla \psi (0,0)|=2$, there exists $r>0$ so that $|\nabla \psi |\ge 1$ in B(0, r). Consider then the set

$$\begin{aligned} E_+=\left\{ (x',x_n)\in B(0,r) ;\; 0\le x_n<1\; \text{ and }\; x_n\ge |x'|^2\right\} . \end{aligned}$$

We have clearly

$$\begin{aligned} E_+{\setminus }\{(0,0)\}=\left\{ (x',x_n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}};\; \psi (x',x_n)<\psi (0,0)=1\right\} . \end{aligned}$$

Pick $\chi \in C_0^\infty (B(0,r))$ satisfying $\chi =1$ in $B(0,\rho _1)$, where $0<\rho _1 <r$ is fixed arbitrary. Let then $\epsilon >0$ so that

$$\begin{aligned} E_+\cap \left[ B(0,r){\setminus }{\overline{B}}(0,\rho _1)\right] \subset \left\{ (x',x_n)\in B(0,r);\; \psi (x',x_n)<\psi (0,0)-\epsilon \right\} . \end{aligned}$$

Lemma 4.1

There exists $0<\rho _0<\rho _1$ so that if $u\in H^2(B(0,r);{\mathbb {C}})$ satisfies ${\mathcal {L}}_A^eu=0$ in B(0, r) and $\text{ supp }(u)\subset E_+$ then $u=0$ in $B(0,\rho _0)$.

Proof

Let us choose $0<\rho _0 <\rho _1$ in such a way that

$$\begin{aligned} E_+\cap B(0,\rho _0)\subset \left\{ (x',x_n)\in B(0,r);\; \psi (x',x_n)>\psi (0,0)-\epsilon /2\right\} . \end{aligned}$$

Pick $u\in H^2(B(0,r);{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^eu=0$ in B(0, r) and $\text{ supp }(u)\subset E_+$. Let $v=\chi u$, and $\lambda ^*$ and $\tau ^*$ be as in Theorem 4.1. Fix then $\lambda \ge \lambda ^*$ and set

$$\begin{aligned} c_0 =e^{\lambda (1-\epsilon /2)}, \quad c_1 = e^{\lambda (1-\epsilon )} . \end{aligned}$$

Theorem 4.1 yields

$$\begin{aligned} \int _{B(0,\rho _0)}u^2{ d}x&=\int _{B(0,\rho _0)\cap E_+}v^2{ d}x \\&\le \aleph \tau ^{-3} e^{-\tau (c_0 -c_1)}\int _{E_+\cap (B(0,r){\setminus } B(0,\rho _1))}({\mathcal {L}}_A^ev)^2{ d}x,\quad \tau \ge \tau ^*. \end{aligned}$$

Noting that $c_0 >c_1$, we obtain that $u=0$ in $B(0,\rho _0)$ by taking in the right hand side of the last inequality the limit, as $\tau$ tends to $\infty$. $\square$

Let $\vartheta =\vartheta (x')$ be in $C^{3,1}({\overline{B}}(0,r))$ satisfying $\vartheta (0)=0$ and $\nabla '\vartheta (0)=0$, and consider

$$\begin{aligned} \varphi :(x',x_n)\in \omega \mapsto (y',y_n)=(x',x_n-\vartheta (x')+|x'|^2). \end{aligned}$$

As $\varphi '(0,0)=\mathbf{I }$, we deduce that $\varphi$ is a diffeomorphism from $\omega \subset B(0,r)\times {\mathbb {R}}$, a neighborhood of 0 in ${\mathbb {R}}^n$, onto ${\tilde{\omega }}=\varphi (\omega )$.

Pick $u\in H^2(\omega ,{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^eu=0$ in $\omega$ and $\text{ supp }(u)\subset \omega _+=\{(x',x_n)\in \omega ;\; x_n\ge \vartheta (x')\}$. Define v by $v(y',y_n)=u(\varphi ^{-1}(y',y_n))$, $(y',y_n)\in {\tilde{\omega }}$. Then it is straightforward to check that ${\mathcal {L}}_{{\tilde{A}}}^e v=0$ in ${\tilde{\omega }}$. Here ${\mathcal {L}}_{{\tilde{A}}}^e$ is of the same form as ${\mathcal {L}}_A^e$. Its principal part is given by

$$\begin{aligned} {\mathcal {L}}_{{\tilde{A}},0}^e= \Delta _{{\tilde{A}}} \end{aligned}$$

with

$$\begin{aligned} {\tilde{A}} (y)=\varphi '\left( \varphi ^{-1}(y)\right) A\left( \varphi ^{-1}(y)\right) (\varphi ')^t\left( \varphi ^{-1}(y)\right) . \end{aligned}$$

Furthermore, $\text{ supp }(v)\subset {\tilde{\omega }}_+=\{ (y',y_n)\in {\tilde{\omega }};\; y_n\ge |y'|^2\}$.

Similar calculations as in Sect. 2.4 show, by reducing $\omega$ if necessary, that

$$\begin{aligned} \left( {\tilde{A}} (y)\xi |\xi \right) \ge \varkappa |\xi |^2/4,\quad y\in {\tilde{\omega }},\; \xi \in {\mathbb {R}}^n. \end{aligned}$$

We apply Lemma 4.1 in order to get $v=0$ in $\tilde{{\mathcal {V}}}$, where $\tilde{{\mathcal {V}}}$ is a neighborhood of the origin, and hence $u=0$ in ${\mathcal {V}}$ with ${\mathcal {V}}=\varphi ^{-1}(\tilde{{\mathcal {V}}})$. In other words, we proved the following result.

Lemma 4.2

There exists a neighborhood ${\mathcal {V}}$ of the origin in $\omega$ so that if $u\in H^2(\omega ;{\mathbb {C}} )$ satisfies ${\mathcal {L}}_A^eu=0$ in $\omega$ and $\text{ supp }(u)\subset \omega _+$ then $u=0$ in ${\mathcal {V}}$.

The global uniqueness of continuation result is based on the following lemma.

Lemma 4.3

Let $\Omega _0\subset \Omega$ so that $\partial \Omega _0\cap \Omega \ne \emptyset$. There exists $z\in \partial \Omega _0\cap \Omega$ and ${\mathcal {W}}$ a neighborhood of z in $\Omega$ so that if $u\in H^2(\Omega ;{\mathbb {C}})$ satisfies ${\mathcal {L}}_A^eu=0$ in $\Omega$ together with $u=0$ in $\Omega _0$ then $u=0$ in ${\mathcal {W}}$.

Proof

Fix $y\in \partial \Omega _0\cap \Omega$ and $r>0$ so that $B(y,r)\subset \Omega$. Pick then $y_0\in \Omega _0\cap B(y,r)$ sufficiently close to y in such a way that $\partial B(y_0, d)\cap \partial \Omega _0\ne \emptyset$, with $d=\text{ dist }(y_0,\partial \Omega _0)$. Pick then $z\in \partial B(y_0, d)\cap \partial \Omega _0$. Making a translation we may assume that $z=0$. As the $\partial B(y_0, d)$ can be represented locally by a graph $x_n=\vartheta (x')$. Making a change of coordinates we may assume that $\text{ supp }(u)\subset \{(x',x_n);\; x_n\ge \vartheta (x')\}$. This orthogonal transformation modify A, but the new matrix has the same properties as A. We then complete the proof by using Lemma 4.2 with A substituted by this new matrix. $\square$

Theorem 4.2

Let $u\in H^2(\Omega ;{\mathbb {C}})$ satisfying ${\mathcal {L}}_A^eu=0$ in $\Omega$ and $u=0$ in $\omega$, for some nonempty open subset $\omega$ of $\Omega$. Then $u=0$ in $\Omega$.

Proof

Let $\Omega _0$ be the maximal domain in which $u=0$. If $\Omega {\setminus } \overline{\Omega _0}\ne \emptyset$ then we would have $\partial \Omega _0\cap \Omega \ne \emptyset$. Therefore we would find, by Lemma 4.3, $z\in \partial \Omega _0\cap \Omega$ and ${\mathcal {W}}$ a neighborhood of z in $\Omega$ so that $u=0$ in ${\mathcal {W}}$. That is $u=0$ in $\Omega _0\cup {\mathcal {W}}$ which contains strictly $\Omega _0$ and hence contradicts the maximality of $\Omega _0$. $\square$

It is worth mentioning that Theorem 4.2 can also be obtained as a consequence [11, Proposition 2.28, page 28] that quantifies the uniqueness of continuation from a subdomain of $\Omega$ to another subdomain of $\Omega$. The proof of [11, Proposition 2.28, page 28] relies on three-ball inequality which is itself a consequence of the Carleman inequality of Theorem 4.1.

As an immediate consequence of Theorem 4.2 we get uniqueness of continuation from the Cauchy data on a subboundary.

Corollary 4.1

Assume that ${\mathcal {L}}_A^e$ is defined in ${\hat{\Omega }}\Supset \Omega$. Let $\Gamma _0$ be an arbitrary nonempty open subset of $\Gamma$. If $u\in H^2(\Omega ;{\mathbb {C}})$ satisfies ${\mathcal {L}}_A^eu=0$ in $\Omega$ together with $u=\partial _\nu u= 0$ in $\Gamma _0$ then $u=0$ in $\Omega$.

Proof

If $u\in H^2(\Omega ;{\mathbb {C}})$ satisfies $u=\partial _\nu u= 0$ in $\Gamma _0$ then we find ${\mathcal {V}}\subset {\hat{\Omega }}$, a neighborhood of a point in $\Gamma _0$, so that ${\tilde{u}}$, the extension of u by zero in ${\mathbb {R}}^n{\setminus } {\overline{\Omega }}$, belongs to $H^2(\Omega \cup {\mathcal {V}})$, satisfies ${\mathcal {L}}_A^e{\tilde{u}}=0$ in $\Omega \cup {\mathcal {V}}$ and ${\tilde{u}}=0$ in ${\mathcal {V}}{\setminus } {\overline{\Omega }}$. Theorem 4.2 allows us to conclude that ${\tilde{u}}=0$ and hence $u=0$. $\square$

We observe once again that Corollary 4.1 can be deduced from [11, Proposition 2.28 in page 28 and Proposition 2.30 in page 29]. We point out that [11, Proposition 2.30 in page 29] quantifies the uniqueness of continuation from the Cauchy data on a subboundary to an interior subdomain.

5 Parabolic equations

We fix in this section $0\le \psi \in C^4({\overline{Q}})$ of the form $\psi (x,t)=\psi _0(x)+\psi _1(t)$, where

$$\begin{aligned} |\nabla \psi _0|\ge \delta \quad \text{ in }\; {\overline{\Omega }}, \end{aligned}$$

for some constant $\delta >0$. Let $\phi =e^{\lambda \psi }$, $\lambda >0$, and consider the parabolic operator

$$\begin{aligned} {\mathcal {L}}_A^p=\Delta _A-\partial _t+\sum _{\ell =1}^np_\ell \partial _\ell +q, \end{aligned}$$

where $A\in {\mathscr {M}}(\Omega ,\varkappa ,{\mathfrak {m}})$, $p_1,\ldots p_n$ and q belong to $L^\infty (Q ;{\mathbb {C}})$ and satisfy

$$\begin{aligned} \Vert p_\ell \Vert _{L^\infty (Q)}\le {\mathfrak {m}},\; 1\le \ell \le n\quad \text{ and }\quad \Vert q\Vert _{L^\infty (Q)}\le {\mathfrak {m}}. \end{aligned}$$

5.1 Carleman inequality

Recall that ${\mathcal {L}}_{A,0}^p$ represents the principal part of ${\mathcal {L}}_A^p$:

$$\begin{aligned} {\mathcal {L}}_{A,0}^p=\Delta _A-\partial _t. \end{aligned}$$

We will use in the sequel the notation ${\mathfrak {d}}=(\Omega ,t_1,t_2,\varkappa , \delta ,{\mathfrak {m}} )$.

Theorem 5.1

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^{2,1}(Q,{\mathbb {C}})$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda ^2 \phi |\nabla u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }|{\mathcal {L}}_A^pu|^2{ d}x{ d}t+\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\nabla u|^2\right] { d}\mu \nonumber \\&\qquad +\int _\Sigma e^{2\tau \phi }(\tau \lambda \phi )^{-1}|\partial _tu|^2{ d}\sigma { d}t, \end{aligned}$$

(5.1)

Proof

As for the wave equation, we set $\Phi =e^{-\tau \phi }$, $\tau >0$. We also recall that

$$\begin{aligned} \partial _k\Phi&=-\tau \partial _k\phi \Phi , \\ \partial _{k\ell }\Phi&=\left( -\tau \partial ^2_{k\ell }\phi +\tau ^2\partial _k\phi \partial _\ell \phi \right) \Phi , \\ \partial _t\Phi&= -\tau \partial _t\phi \Phi . \end{aligned}$$

We have, for $w\in H^{2,1}(Q,{\mathbb {R}})$,

$$\begin{aligned} \Phi ^{-1}\Delta _A(\Phi w)=\Delta _Aw -2\tau (\nabla w|\nabla \phi )_A+\left[ \tau ^2 |\nabla \phi |_A^2-\tau \Delta _Aw\right] w. \end{aligned}$$

Also,

$$\begin{aligned} \Phi ^{-1}\partial _t(\Phi w)=\partial _tw-\tau \partial _t\phi w. \end{aligned}$$

We decompose $L=\Phi ^{-1}{\mathcal {L}}_{A,0}^p\Phi$ as in the elliptic case. That is in the form

$$\begin{aligned} L=L_0+L_1+c, \end{aligned}$$

with

$$\begin{aligned} L_0w&=\Delta _A w+aw, \\ L_1w&= (B|\nabla w)- \partial _tw +bw, \end{aligned}$$

where

$$\begin{aligned} a(x,t)&=\tau ^2|\nabla \phi |_A^2, \\ b(x,t)&=-2\tau \Delta _A \phi , \\ c(x,t)&=\tau \Delta _A \phi +\tau \partial _t\phi , \\ B&=-2\tau A\nabla \phi . \end{aligned}$$

We have

$$\begin{aligned} \langle L_0w|L_1w\rangle _{L^2(Q)}=\sum _{j=1}^6 I_j. \end{aligned}$$

(5.2)

Quantities $I_j$, $1\le j\le 6$, are given as follows

$$\begin{aligned} I_1&=\int _Q\Delta _A w(\nabla w|B) { d}x{ d}t, \\ I_2&=-\int _Q\Delta _A w\partial _tw{ d}x{ d}t, \\ I_3&=\int _Q\Delta _A wbw{ d}x{ d}t, \\ I_4&=\int _Qaw(\nabla w|B) { d}x{ d}t, \\ I_5&=-\int _Qaw\partial _tw{ d}x{ d}t, \\ I_6&=\int _Qabw^2{ d}x{ d}t. \end{aligned}$$

Let $C=(\text{ div }(a_{k\ell }B))$ and

$$\begin{aligned} D=C/2-A(B')^t. \end{aligned}$$

We already proved that

$$\begin{aligned} I_1=\int _Q(D\nabla w|\nabla w){ d}x{ d}t+\int _\Sigma \left[ (\nabla w|\nu )_A(\nabla w\cdot B)-(B/2|\nu )|\nabla w|_A^2\right] { d}\sigma { d}t. \end{aligned}$$

(5.3)

On the other hand inequality (3.6) with $d=-1$ gives

$$\begin{aligned} I_2=\int _\Sigma (\nabla w|\nu )_A \partial _tw{ d}x{ d}t+\int _\Omega \left[ |\nabla w|_A^2/2\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(5.4)

$I_3$ is the same as in (3.7):

$$\begin{aligned} I_3&=-\int _Q b|\nabla w|^2{ d}x{ d}t +\int _Q \Delta _A(b/2) w^2 { d}x{ d}t \nonumber \\&\quad -\int _\Sigma (\nabla (b/2)|\nu )_Aw^2{ d}\sigma { d}t +\int _\Sigma (\nabla w|\nu )_Abw { d}\sigma { d}t. \end{aligned}$$

(5.5)

Let $J_1=I_1+I_2+I_3$ and

$$\begin{aligned} {\mathfrak {A}}&=D-bA, \\ a_1&=\Delta _A (b/2), \\ g_1(w)&=(\nabla w|\nu )_A(\nabla w|B)-(B/2|\nu )|\nabla w|_A^2-(\nabla w|\nu )_A \partial _tw \\&\quad -(\nabla (b/2)|\nu )_Aw^2{ d}\sigma + (\nabla w|\nu )_Abw, \\ h_1(w)&=\left[ |\nabla w|_A^2/2\right] _{t=t_1}^{t_2}. \end{aligned}$$

We find by putting together (5.3)–(5.5)

$$\begin{aligned} J_1&=\int _Q ({\mathfrak {A}}\nabla w|\nabla w){ d}x{ d}t +\int _Qa_1w^2{ d}x{ d}t. \nonumber \\&\quad +\int _\Sigma g_1(w){ d}\sigma { d}t +\int _\Omega h_1(w) { d}x. \end{aligned}$$

(5.6)

$I_4$ was calculated in (3.14). Precisely, we have

$$\begin{aligned} I_4=-\int _Q \text{ div }(aB/2)w^2{ d}x{ d}t + \int _\Sigma a(B/2|\nu )w^2{ d}\sigma { d}t. \end{aligned}$$

(5.7)

On the other hand an integration by parts, with respect to t, gives

$$\begin{aligned} I_5=-\int _Q (a/2)\partial _t(w^2){ d}x{ d}t= \int _Q\partial _t (a/2) w^2{ d}x{ d}t-\int _\Omega \left[ (a/2)w^2\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

(5.8)

Set

$$\begin{aligned} a_2&=-\text{ div }(aB/2)+\partial _t(a/2)+ab, \\ g_2(w)&=a(B/2|\nu )w^2, \\ h_2(w)&=-\left[ (a/2)w^2\right] _{t=t_1}^{t_2}, \end{aligned}$$

and let $J_2=I_4+I_5+I_6$. In light of (5.7) and (5.8) we get

$$\begin{aligned} J_2=\int _Q a_2w^2{ d}x{ d}t+\int _\Sigma g_2(w){ d}\sigma { d}t +\int _\Omega h_2(w) { d}x. \end{aligned}$$

(5.9)

Putting together (5.2), (5.6) and (5.9) in order to obtain

$$\begin{aligned} \langle L_0w|L_1w\rangle _{L^2(Q)}&=2\tau \int _Q ({\mathfrak {A}}\nabla w|\nabla w){ d}x{ d}t \nonumber \\&\quad +\int _Q{\mathfrak {a}}w^2{ d}x{ d}t +\int _\Sigma g(w){ d}\sigma { d}t +\int _\Omega h(w) { d}x, \end{aligned}$$

(5.10)

where ${\mathfrak {a}}=a_1+a_2$, $g=g_1+g_2$ and $h=h_1+h_2$.

As $\partial _t a=0$ (which is a consequence of $\partial _t\nabla \psi =0$), we see that ${\mathfrak {A}}$ and ${\mathfrak {a}}$ has exactly the same form as in the elliptic case. Therefore we can mimic the proof of the elliptic case to complete the proof. $\square$

We already defined $\Gamma _+=\Gamma _+^{\psi _0}=\{x\in \Gamma ;\; \partial _{\nu _A} \psi _0>0\}$ and $\Sigma _+=\Sigma _+^{\psi _0}=\Gamma _+\times (t_1,t_2)$. Similarly to the wave equation we have the following result.

Theorem 5.2

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^{2,1}(Q,{\mathbb {C}})$ satisfying $u=0$ on $\Sigma$ and $u(\cdot ,t)=0$, $t\in \{t_1,t_2\}$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda ^2 \phi |\nabla u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left| {\mathcal {L}}_A^pu\right| ^2{ d}x{ d}t +\tau \lambda \int _{\Sigma _+} e^{2\tau \phi }\phi |\partial _\nu u|^2 { d}\sigma { d}t, \end{aligned}$$

(5.11)

5.2 Unique continuation

An adaptation of the proof in the case of wave equations enables us to establish the following result.

Theorem 5.3

Let $u\in H^{2,1}(Q)$ satisfying ${\mathcal {L}}_A^pu=0$ in Q and $u=0$ in $\omega \times (t_1,t_2)$, for some nonempty open subset $\omega$ of $\Omega$. Then $u=0$ in Q.

We remark that Theorem 5.3 can be also obtained from [13, Proposition 3.2].

Similarly to the case of wave equations Theorem 5.3 can serve to prove uniqueness of continuation from the Cauchy data on a subboundary.

Corollary 5.1

Assume that ${\mathcal {L}}_A^p$ is defined in ${\hat{\Omega }}\Supset \Omega$. Let $\Gamma _0$ be an arbitrary nonempty open subset of $\Gamma$. If $u\in H^1((t_1,t_2), H^2(\Omega ))$ satisfies ${\mathcal {L}}_A^pu=0$ in Q and $u=\partial _\nu u= 0$ in $\Sigma _0=\Gamma _0\times (t_1,t_2)$ then $u=0$ in Q.

We remark that Corollary 5.1 follows also from [13, Proposition 3.2 and Proposition 4.1].

5.3 Final time observability inequality

We assume in the present subsection that $t_1=0$ and $t_2={\mathfrak {t}}>0$, and we consider the IBVP

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta _Au-\partial _tu=f\quad \text{ in }\; Q, \\ u(\cdot ,0)=u_0, \\ u_{|\Sigma }=0. \end{array} \right. \end{aligned}$$

(5.12)

Define the unbounded operator ${\mathscr {A}}:L^2(\Omega )\rightarrow L^2(\Omega )$ by

$$\begin{aligned} {\mathscr {A}}u=-\Delta _Au,\quad D({\mathscr {A}})=H_0^1(\Omega )\cap H^2(\Omega ). \end{aligned}$$

It is known that $-{\mathscr {A}}$ generates an analytic semigroup $e^{-t{\mathscr {A}}}$. In particular, for any $(u_0,f)\in L^2(\Omega )\times L^1((0,T);L^2(\Omega ))$, the IBVP (5.12) has a unique (mild) solution $u={\mathscr {S}}(u_0,f)\in C([0,T];L^2(\Omega ))$ so that

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^2(\Omega )}\le \Vert u_0\Vert _{L^2(\Omega )}+\Vert f\Vert _{L^1((0,T);L^2(\Omega ))},\quad 0\le t\le {\mathfrak {t}}. \end{aligned}$$

(5.13)

This solution is given by Duhamel’s formula

$$\begin{aligned} u(t)=e^{-t{\mathscr {A}}}u_0+\int _0^te^{-(t-s){\mathscr {A}}}f(s){ d}s,\quad 0\le t\le {\mathfrak {t}}. \end{aligned}$$

(5.14)

Note that if $u_0\in D({\mathscr {A}})$ then $u={\mathcal {S}}(u_0,0)$ satisfies

$$\begin{aligned} u\in C([0,{\mathfrak {t}}];D({\mathscr {A}}))\cap C^1([0,{\mathfrak {t}}];L^2(\Omega )). \end{aligned}$$

We refer to [30, Chapter 11] for a concise introduction to semigroup theory.

Lemma 5.1

Let $f\in L^2(Q)$ and $\zeta \in C_0^\infty ([0,{\mathfrak {t}}))$. Then $u({\mathfrak {t}})={\mathscr {S}}(0,\zeta f)({\mathfrak {t}})\in H_0^1(\Omega )$ and

$$\begin{aligned} \Vert u({\mathfrak {t}})\Vert _{H_0^1(\Omega )}\le \aleph \Vert \zeta f\Vert _{L^2(Q)}, \end{aligned}$$

(5.15)

where $\aleph >0$ is a constant only depending on $\Omega$, A, ${\mathfrak {t}}$ and $\zeta$.

Proof

We have from [29, Theorem 8.1 in page 254] that $D({\mathscr {A}}^{1/2})=H_0^1(\Omega )$. Therefore, in light of (5.14), we obtain

$$\begin{aligned} {\mathscr {A}}^{1/2}u({\mathfrak {t}})=\int _0^{\mathfrak {t}}{\mathscr {A}}^{1/2}e^{-({\mathfrak {t}}-s){\mathscr {A}}}(\zeta (s)f(s)){ d}s. \end{aligned}$$

But

$$\begin{aligned} \left\| {\mathscr {A}}^{1/2}e^{-t{\mathscr {A}}}\right\| _{L^2(\Omega )}\le \aleph _0t^{-1/2},\quad t>0, \end{aligned}$$

where the constant $\aleph _0>0$ only depends on $\Omega$, A. If $\text{ supp }(\zeta )\subset [0,{\mathfrak {t}}-\epsilon ]$, $\epsilon >0$, we find

$$\begin{aligned} \left\| {\mathscr {A}}^{1/2}u({\mathfrak {t}})\right\| _{L^2(\Omega )}\le \aleph _0\int _0^{{\mathfrak {t}}-\epsilon }({\mathfrak {t}}-s)^{-1/2}\Vert \zeta (s)f(s)\Vert _{L^2(\Omega )}{ d}s. \end{aligned}$$

Whence Cauchy-Schwarz’s inequality yields

$$\begin{aligned} \left\| {\mathscr {A}}^{1/2}u({\mathfrak {t}})\right\| _{L^2(\Omega )}&\le \aleph _0\left( \int _0^{{\mathfrak {t}}-\epsilon }({\mathfrak {t}}-s)^{-1}{ d}s\right) ^{1/2}\Vert \zeta f\Vert _{L^2(Q)} \\&\le \aleph _0 \ln \left[ ({\mathfrak {t}}/\epsilon )^{1/2}\right] \Vert \zeta f\Vert _{L^2(Q)}. \end{aligned}$$

The expected inequality then follows. $\square$

Theorem 5.4

Let $0\le \psi _0 \in C^4(\Omega )$ with no critical point in ${\overline{\Omega }}$. Set $\Gamma _+=\{x\in \Gamma ;\; \partial _{\nu _A}\psi _0(x)>0\}$ and $\Sigma _+=\Gamma _+\times (0,{\mathfrak {t}})$. For any $u=e^{-t{\mathscr {A}}}u_0$ with $u_0\in D({\mathscr {A}})$ we have

$$\begin{aligned} \Vert u({\mathfrak {t}})\Vert _{H_0^1(\Omega )}\le \aleph \Vert \partial _\nu u\Vert _{L^2(\Sigma _+)}, \end{aligned}$$

where the constant $\aleph >0$ only depends of $\Omega$, A and ${\mathfrak {t}}$.

Proof

Let $\chi \in C_0^\infty ((0,{\mathfrak {t}}))$ satisfying $\chi =1$ in $[{\mathfrak {t}}/4,3{\mathfrak {t}}/4]$. As $(\Delta _A-\partial _t)(\chi u)=-\chi 'u$, we get by applying Theorem 5.2, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$,

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |\chi u|^2+\tau \lambda ^2 \phi |\nabla (\chi u)|^2\right] { d}x{ d}t \\&\quad \le \int _Qe^{2\tau \phi }\left| u\right| ^2{ d}x{ d}t +\tau \lambda \int _{\Sigma _+} e^{2\tau \phi }\phi |\partial _\nu u|^2 { d}\sigma { d}t, \end{aligned}$$

where the notations are those of Theorem 5.2.

Since the first term in the right hand side of this inequality can be absorbed by the left hand side, provided that $\lambda$ and $\tau$ are sufficiently large, we obtain in a straightforward manner

$$\begin{aligned} \Vert u\Vert _{L^2(\Omega \times ({\mathfrak {t}}/4,3{\mathfrak {t}}/4))}\le \aleph \Vert \partial _\nu u\Vert _{L^2(\Sigma _+)}. \end{aligned}$$

(5.16)

Pick $\varphi \in C^\infty ([0,{\mathfrak {t}}])$ so that that $\varphi =0$ in $[0,{\mathfrak {t}}/4]$ and $\varphi =1$ in $[3{\mathfrak {t}}/4,{\mathfrak {t}}]$. We easily check that $\varphi u={\mathscr {S}}(0,\varphi 'u)$. In light of Lemma 5.1, we then conclude that

$$\begin{aligned} \Vert u({\mathfrak {t}})\Vert _{H_0^1(\Omega )}\le \aleph \Vert \varphi 'u\Vert _{L^2(Q)}=\aleph \Vert \varphi 'u\Vert _{L^2(\Omega \times ({\mathfrak {t}}/4,3{\mathfrak {t}}/4))}. \end{aligned}$$

This and (5.16) imply the expected inequality. $\square$

6 Schrödinger equations

Let $\phi =e^{\lambda \psi }$ be a weight function for the Schrödinger operator

$$\begin{aligned} {\mathcal {L}}_{A,0}^s=\Delta _A+i\partial _t \end{aligned}$$

and set

$$\begin{aligned} {\mathcal {L}}_A^s=\Delta _A+i\partial _t+\sum _{\ell =1}^np_\ell \partial _\ell +q, \end{aligned}$$

where $p_1,\ldots p_n$ and q belong to $L^\infty (Q ;{\mathbb {C}})$ and satisfy

$$\begin{aligned} \Vert p_\ell \Vert _{L^\infty (Q)}\le {\mathfrak {m}},\; 1\le \ell \le n\quad \text{ and }\quad \Vert q\Vert _{L^\infty (Q)}\le {\mathfrak {m}}. \end{aligned}$$

6.1 Carleman inequality

Let

$$\begin{aligned} \delta =\min _{{\overline{Q}}}|\nabla \psi |_A\; (>0) \end{aligned}$$

and ${\mathfrak {d}}=(\Omega ,t_1,t_2,\varkappa , \delta ,{\mathfrak {m}} )$.

Theorem 6.1

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^{2,1}(Q,{\mathbb {C}})$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\nabla u|^2\right] { d}x{ d}t \\&\quad \le \int _Qe^{2\tau \phi }|{\mathcal {L}}_A^su|^2{ d}x{ d}t+\int _{\partial Q} e^{2\tau \phi }\left[ \tau ^3\lambda ^3\phi ^3|u|^2+\tau \lambda \phi |\nabla u|^2\right] { d}\mu \\&\qquad +\int _\Sigma e^{2\tau \phi }(\tau \lambda \phi )^{-1}|\partial _tu|^2{ d}\sigma { d}t , \end{aligned}$$

Proof

In this proof, $\aleph$, $\lambda _j$, $\tau _j$, $j=1,2,\ldots$, denote positive generic constants only depending on ${\mathfrak {d}}$.

As in the preceding section, if $\Phi =e^{-\tau \phi }$, $\tau >0$, then

$$\begin{aligned} \partial _k\Phi&=-\tau \partial _k\phi \Phi , \\ \partial _{k\ell }\Phi&=\left( -\tau \partial ^2_{k\ell }\phi +\tau ^2\partial _k\phi \partial _\ell \phi \right) \Phi , \\ \partial _t\Phi&= -\tau \partial _t\phi \Phi . \end{aligned}$$

We have, where $w\in H^2 (Q;{\mathbb {C}})$,

$$\begin{aligned} \Phi ^{-1}\Delta _A(\Phi w)=\Delta _Aw -2\tau (\nabla w|\nabla \phi )_A+\left[ \tau ^2 |\nabla \phi |_A^2-\tau \Delta _A\phi \right] w \end{aligned}$$

and

$$\begin{aligned} i\Phi ^{-1}\partial _t(\Phi w)=i\partial _tw -i\tau \partial _t\phi w. \end{aligned}$$

We decompose $L=\Phi ^{-1}{\mathcal {L}}_{A,0}^s\Phi$ as follows

$$\begin{aligned} L=L_0+L_1+c \end{aligned}$$

with

$$\begin{aligned} L_0w&=\Delta _A w+i\partial _tw+aw, \\ L_1w&= (B|\nabla w) +bw, \end{aligned}$$

where we set

$$\begin{aligned} a&=\tau ^2|\nabla \phi |_A^2, \\ b&=-\tau \Delta _A \phi , \\ c&= i\partial _t\phi , \\ B&=-2\tau A\nabla \phi . \end{aligned}$$

We have

$$\begin{aligned} \langle L_0w|L_1w\rangle _{L^2(Q)}=\int _QL_0w\overline{L_1w}{ d}x{ d}t=\sum _{j=1}^6 I_j, \end{aligned}$$

(6.1)

with

$$\begin{aligned} I_1&=\int _Q\Delta _A w(\nabla {\overline{w}}|B) { d}x{ d}t, \\ I_2&=\int _Q\Delta _A wb{\overline{w}}{ d}x{ d}t, \\ I_3&=i\int _Q\partial _tw(\nabla {\overline{w}}|B) { d}x{ d}t, \\ I_4&=i\int _Q\partial _twb{\overline{w}} { d}x{ d}t, \\ I_5&=\int _Qaw(\nabla {\overline{w}}|B) { d}x{ d}t, \\ I_6&=\int _Qab|w|^2{ d}x{ d}t. \end{aligned}$$

Some parts of the proof are quite similar to that of the wave equation and therefore we omit their details. We have

$$\begin{aligned} I_1=\int _Q(D\nabla w|\nabla {\overline{w}}){ d}x{ d}t+\int _\Sigma \left[ (\nabla w|\nu )_A(\nabla {\overline{w}}| B)-(B/2|\nu )|\nabla w|_A^2\right] { d}\sigma { d}t, \end{aligned}$$

(6.2)

where

$$\begin{aligned} D=C/2-(B')^t, \end{aligned}$$

with $C=(\text{ div }(a_{k\ell }B))$.

Also,

$$\begin{aligned} \mathfrak {R}I_2=\int _Q\Delta _A wb{\overline{w}}&=-\int _Q b|\nabla w|_A^2{ d}x{ d}t -\mathfrak {R}\int _Q {\overline{w}}(\nabla b|\nabla w)_A { d}x{ d}t \\&\quad +\mathfrak {R}\int _\Sigma (\nabla w|\nu )_Ab{\overline{w}} { d}\sigma { d}t. \end{aligned}$$

But $\mathfrak {R}({\overline{w}}\nabla w)=\nabla |w|^2/2$. Therefore

$$\begin{aligned} \mathfrak {R}I_2&=-\int _Q b|\nabla w|_A^2{ d}x{ d}t +\int _Q \Delta _A(b/2) |w|^2 { d}x{ d}t \nonumber \\&\quad -\int _\Sigma (\nabla (b/2)|\nu )_A|w|^2{ d}\sigma { d}t +\mathfrak {R}\int _\Sigma (\nabla w|\nu )_Ab{\overline{w}} { d}\sigma { d}t. \end{aligned}$$

(6.3)

Let $J=\mathfrak {R}(I_1+I_2)$ and define

$$\begin{aligned} {\mathcal {A}}&= D-bA, \\ a_1&=\Delta _A(b/2), \\ g_1(w)&=\mathfrak {R}\left[ (\nabla w|\nu )_A(\nabla {\overline{w}}| B) +(\nabla w|\nu )_Ab{\overline{w}}\right] \\&\quad -(\nabla (b/2)|\nu )_A|w|^2-(B/2|\nu )|\nabla w|_A^2. \end{aligned}$$

We combine (6.2) and (6.3) in order to obtain

$$\begin{aligned} J= \int _Q\mathfrak {R}({\mathcal {A}}\nabla w|\nabla {\overline{w}}){ d}x{ d}t +\int _Qa_1|w|^2{ d}x{ d}t +\int _\Sigma g_1(w){ d}\sigma { d}t. \end{aligned}$$

(6.4)

We have once again from the calculations we done for the wave equation

$$\begin{aligned} {\mathcal {A}}=2\tau A\nabla ^2\phi A+\tau \Upsilon _A(\phi (\cdot ,t)). \end{aligned}$$

This identity together with the following ones

$$\begin{aligned} \nabla ^2\phi&=\lambda ^2\phi (\nabla \psi _0\otimes \nabla \psi _0)+\lambda \phi \nabla ^2\psi _0, \\ \Upsilon _A(\phi (\cdot ,t))&=\lambda \phi \Upsilon _A(\psi _0) \end{aligned}$$

imply

$$\begin{aligned} {\mathcal {A}}=\Theta _A(\psi _0)+2\tau \lambda ^2\phi A(\nabla \psi _0\otimes \nabla \psi _0)A. \end{aligned}$$

As $A(\nabla \psi _0\otimes \nabla \psi _0)A$ is non negative and $\psi _0$ is A-pseudo-convex with constant $\kappa >0$, we get

$$\begin{aligned} \mathfrak {R}({\mathcal {A}}\nabla w|\nabla {\overline{w}})\ge \kappa \varkappa ^2|\nabla w|^2. \end{aligned}$$

This inequality in (6.4) yields

$$\begin{aligned} J\ge \kappa \varkappa ^2\int _Q|\nabla w|^2{ d}x{ d}t +\int _Qa_1|w|^2{ d}x{ d}t +\int _\Sigma g_1(w){ d}\sigma { d}t. \end{aligned}$$

(6.5)

We find, by making an integration by parts with respect to t and then with respect to x,

$$\begin{aligned} \int _Q\partial _tw(\nabla {\overline{w}}|B){ d}x{ d}t&=\int _Q\partial _t{\overline{w}}(\nabla w|B){ d}x{ d}t-\int _Q\partial _tw{\overline{w}}\text{ div }(B){ d}x{ d}t \\&\quad +\int _Q(\nabla w|\partial _tB){\overline{w}}{ d}x{ d}t \\&\quad -\int _\Sigma w\partial _t{\overline{w}}(B|\nu ){ d}\sigma { d}t+ \int _\Omega \left[ w(\nabla {\overline{w}}|B)\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

We then obtain, by noting that $\text{ div }(B)=2b$,

$$\begin{aligned} \int _Q\partial _tw(\nabla {\overline{w}}|B){ d}x{ d}t&=\int _Q\partial _t{\overline{w}}(\nabla w|B){ d}x{ d}t-2\int _Q\partial _tw{\overline{w}}b{ d}x{ d}t \\&\quad +\int _Q(\nabla w|\partial _tB){\overline{w}}{ d}x{ d}t \\&\quad -\int _\Sigma w\partial _t{\overline{w}}(B|\nu ){ d}\sigma { d}t+ \int _\Omega \left[ w(\nabla {\overline{w}}|B)\right] _{t=t_1}^{t_2}{ d}x. \end{aligned}$$

From the identity

$$\begin{aligned} \partial _tw(\nabla {\overline{w}}|B)-\partial _t{\overline{w}}(\nabla w|B)=2i\mathfrak {I}[\partial _tw(\nabla {\overline{w}}|B)] \end{aligned}$$

we deduce that

$$\begin{aligned} 2i\mathfrak {I}\int _Q\partial _tw(\nabla {\overline{w}}|B) { d}x{ d}t&= -2\int _Q\partial _tw{\overline{w}}b{ d}x{ d}t+\int _Q(\nabla w|\partial _tB){\overline{w}}{ d}x{ d}t \\&\quad -\int _\Sigma w\partial _t{\overline{w}}(B|\nu ){ d}\sigma { d}t+ \int _\Omega [w(\nabla {\overline{w}}|B)]_{t=t_1}^{t_2}{ d}x. \end{aligned}$$

Or equivalently

$$\begin{aligned} -\mathfrak {I}\int _Q\partial _tw(\nabla {\overline{w}}|B) { d}x{ d}t&= -\mathfrak {R}\left( i\int _Q\partial _tw{\overline{w}}b{ d}x{ d}t\right) -\mathfrak {I}\int _Q(\nabla w|\partial _tB/2){\overline{w}}{ d}x{ d}t \\&\quad +\mathfrak {I}\int _\Sigma w\partial _t{\overline{w}}(B/2|\nu ){ d}\sigma { d}t-\mathfrak {I}\int _\Omega [w(\nabla {\overline{w}}|B/2)]_{t=t_1}^{t_2}{ d}x. \end{aligned}$$

Observing that

$$\begin{aligned} \mathfrak {R}I_3=-\mathfrak {I}\int _Q\partial _tw(\nabla {\overline{w}}|B) { d}x{ d}t\quad \text{ and }\quad I_4= i\int _Q\partial _tw{\overline{w}}b{ d}x{ d}t, \end{aligned}$$

we obtain

$$\begin{aligned} \mathfrak {R}(I_3+I_4)= -\mathfrak {I}\int _Q(\nabla w|\partial _tB/2){\overline{w}}{ d}x{ d}t +\int _\Sigma g_2(w){ d}\sigma { d}t+\int _\Omega h(w){ d}x, \end{aligned}$$

(6.6)

with

$$\begin{aligned} g_2(w)&=\mathfrak {I}(w\partial _t{\overline{w}}(B/2|\nu )), \\ h(w)&=-\mathfrak {I}\left( \left[ w(\nabla {\overline{w}}|B/2)\right] _{t=t_1}^{t_2}\right) . \end{aligned}$$

We find, by using once again the identity $\mathfrak {R}w\nabla {\overline{w}}=\nabla |w|^2/2$,

$$\begin{aligned} \mathfrak {R}I_5=-\int _Q \text{ div }(aB/2)|w|^2{ d}x{ d}t+\int _\Sigma a(B/2|\nu )|w|^2{ d}\sigma { d}t \end{aligned}$$

and hence

$$\begin{aligned} \mathfrak {R}(I_5+I_6)=\int _Qa_2|w|^2{ d}x{ d}t+\int _\Sigma g_3(w){ d}\sigma { d}t, \end{aligned}$$

(6.7)

with

$$\begin{aligned} a_2&=-\text{ div }(aB/2)+ab, \\ g_3(w)&= a(B/2|\nu )|w|^2. \end{aligned}$$

Let

$$\begin{aligned} {\mathfrak {a}}=a_1+a_2=-\text{ div }(aB/2)+ab+\Delta _A(b/2). \end{aligned}$$

We can carry out the same calculations as for the wave equation in order to obtain

$$\begin{aligned} {\mathfrak {a}}\ge \tau ^3\lambda ^4\phi ^3\delta ^4,\quad \lambda \ge \lambda _1,\; \tau \ge \tau _1. \end{aligned}$$

We end up getting, by combining (6.1), (6.5), (6.6) and (6.7), the following inequality

$$\begin{aligned} \mathfrak {R}\langle L_0w|L_1w\rangle _{L^2(Q)}&\ge \tau \lambda \kappa \int _Q|\nabla w|^2{ d}x{ d}t +\tau ^3\lambda ^4\delta ^4\int _Q\phi ^3|w|^2{ d}x{ d}t \nonumber \\&\quad -\mathfrak {I}\int _Q(\nabla w|\partial _tB/2){\overline{w}}{ d}x{ d}t +\int _\Sigma g(w){ d}\sigma { d}t+\int _\Omega h(w){ d}x, \end{aligned}$$

(6.8)

where we set $g=g_1+g_2$.

Let $\epsilon >0$. Then an elementary convexity inequality yields

$$\begin{aligned} |(\nabla w|\partial _tB){\overline{w}}|\le \aleph \phi (\epsilon \tau \lambda [\nabla w|^2+\epsilon ^{-1}\tau \lambda ^2|w|^2). \end{aligned}$$

In consequence the third term in (6.8) can be absorbed by the first two ones, provided that $\lambda \ge \lambda _2$ and $\tau \ge \tau _2$. That is we have

$$\begin{aligned} \mathfrak {R}\langle L_0w|L_1w\rangle _{L^2(Q)}&\ge \tau \lambda \kappa \int _Q|\nabla w|^2{ d}x{ d}t +\tau ^3\lambda ^4\delta ^4\int _Q\phi ^3|w|^2{ d}x{ d}t \\&\quad +\int _\Sigma g(w){ d}\sigma { d}t+\int _\Omega h(w){ d}x. \end{aligned}$$

The rest of the proof is almost similar to that of the wave equation. $\square$

Recall that $\Gamma _+=\Gamma _+^{\psi _0}=\{x\in \Gamma ;\; \partial _{\nu _A}\psi _0>0\}$ and $\Sigma _+=\Sigma _+^\psi =\Gamma _+\times (t_1,t_2)$. As for the wave equation we have

Theorem 6.2

There exist three constants $\aleph =\aleph ({\mathfrak {d}})$, $\lambda ^*=\lambda ^*({\mathfrak {d}})$ and $\tau ^*=\tau ^*({\mathfrak {d}})$ so that, for any $\lambda \ge \lambda ^*$, $\tau \ge \tau ^*$ and $u\in H^{2,1}(Q,{\mathbb {C}})$ satisfying $u=0$ on $\Sigma$ and $u(\cdot ,t)=0$, $t\in \{t_1,t_2\}$, we have

$$\begin{aligned}&\aleph \int _Qe^{2\tau \phi }\left[ \tau ^3\lambda ^4\phi ^3 |u|^2+\tau \lambda \phi |\nabla u|^2\right] { d}x{ d}t \nonumber \\&\quad \le \int _Qe^{2\tau \phi }\left| {\mathcal {L}}_A^su\right| ^2{ d}x{ d}t +\tau \lambda \int _{\Sigma _+} e^{2\tau \phi }\phi |\partial _\nu u|^2 { d}\sigma { d}t, \end{aligned}$$

(6.9)

6.2 Unique continuation

In this subsection, $t_1=-{\mathfrak {t}}$ and $t_2={\mathfrak {t}}$, where ${\mathfrak {t}}>0$ is fixed. We recall that

$$\begin{aligned} E_+({\tilde{x}},c)&=\{x=(x',x_n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}} ;\; 0\le x_n-{\tilde{x}}_n<c\,\, \text{ and }\; x_n-{\tilde{x}}_n\,\,\ge |x'-{\tilde{x}}'|^2/c\}, \end{aligned}$$

with ${\tilde{x}}\in \Omega$, $c>0$.

Theorem 6.3

Suppose that $B({\tilde{x}},r)\Subset \Omega$, for some $r>0$. There exists $c^*=c^*(\varkappa ,{\mathfrak {m}})$ with the property that, for any $0<c<c^*$, we find $0<\rho =\rho (c,\varkappa )<r$ so that if $u\in H^{2,1}(Q)$ satisfies ${\mathcal {L}}_A^su=$ in Q and $\text{ supp }(u(\cdot ,t))\cap B({\tilde{x}},r)\subset E_+({\tilde{x}},c)$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$, then $u=0$ in $B({\tilde{x}},\rho )\times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$.

Proof

We proceed similarly to the proof of Theorem 3.5. We keep the same notations as in Theorem 3.5. Let $c^*=c^*(\varkappa ,{\mathfrak {m}})$ defined as in Theorem 3.5 and $0<c<c^*$.

Fix $0<\eta <1$ and take instead of $\mathbf{Q }_j$, $j=0,1,2$, in Theorem 3.5 the following sets

$$\begin{aligned} \mathbf{Q }_0&=[B({\tilde{x}},\rho _0)\cap E_+]\times (-\eta {\mathfrak {t}}/2,\eta {\mathfrak {t}}/2), \\ \mathbf{Q }_1&= \left\{ E_+\cap \left[ B({\tilde{x}},r_0){\setminus } {\overline{B}}({\tilde{x}},\rho _1)\right] \right\} \times (-{\mathfrak {t}},{\mathfrak {t}}), \\ \mathbf{Q }_2&=[B({\tilde{x}},r_0)\cap E_+]\times [(-{\mathfrak {t}},- \eta {\mathfrak {t}})\cup (\eta {\mathfrak {t}},{\mathfrak {t}})]. \end{aligned}$$

Also, the constants $c_j$, $j=0,1,2$ are substituted by the following ones

$$\begin{aligned} c_0&=e^{\lambda (c^2/2-\epsilon /2-\gamma \eta ^2 {\mathfrak {t}}^2/8+\delta )}\quad \text{ in }\; \mathbf{Q }_0, \\ c_1&=e^{\lambda (c^2/2-\epsilon +\delta )}\quad \text{ in }\; \mathbf{Q }_1, \\ c_2&=e^{\lambda (c^2/2-\gamma \eta ^2 {\mathfrak {t}}^2/2+\delta )}\quad \text{ in }\; \mathbf{Q }_2. \end{aligned}$$

Straightforward computations show that choosing $\gamma$ so that

$$\begin{aligned} \frac{4\epsilon }{3\eta ^2{\mathfrak {t}}^2}<\gamma <\frac{4\epsilon }{\eta ^2{\mathfrak {t}}^2} \end{aligned}$$

guarantee that $c_1<c_0$ and $c_2<c_0$. We can then mimic the last part of Theorem 3.5 to derive that if $u\in H^{2,1}(Q)$ satisfies ${\mathcal {L}}_A^su=$ in Q and $\text{ supp }(u(\cdot ,t))\cap B({\tilde{x}},r)\subset E_+({\tilde{x}},c)$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$, then $u=0$ in $B({\tilde{x}},\rho _0)\times (-\eta {\mathfrak {t}}/2,\eta {\mathfrak {t}}/2)$. Since $0<\eta <1$ is chosen arbitrarily we get, as expected, $u=0$ in $B({\tilde{x}},\rho _0)\times (-{\mathfrak {t}}/2, {\mathfrak {t}}/2)$. $\square$

We say that ${\mathcal {L}}_A^s$ has the reduced unique continuation property if, for any non empty open subset ${\mathcal {O}}\subset \Omega$ and for any $u\in H^{2,1}(Q)$ satisfying ${\mathcal {L}}_A^su=0$ in Q and $u=0$ in ${\mathcal {O}} \times (-{\mathfrak {t}},{\mathfrak {t}})$, we must have $u=0$ in $\Omega \times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$.

Theorem 6.4

There exists a neighborhood ${\mathcal {N}}$ of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }};{\mathbb {R}}^n\times {\mathbb {R}}^n)$ so that ${\mathcal {L}}_A^s$ has the reduced unique continuation property for any $A\in {\mathcal {N}}$.

Proof

Let ${\mathcal {N}}$ be the neighborhood of $\mathbf{I }$ in $C^{2,1}({\overline{\Omega }},{\mathbb {R}}^{n\times n})$ given in Lemma 2.2. Pick $u\in H^{2,1}(Q)$ satisfying ${\mathcal {L}}_A^su=0$ in Q and $u=0$ in ${\mathcal {O}} \times (-{\mathfrak {t}},{\mathfrak {t}})$ for some non empty open subset ${\mathcal {O}}\subset \Omega$. Define $\Omega _0$ as the maximal subdomain of $\Omega$ so that $u=0$ in $\Omega _0\times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$. We claim that $\Omega {\setminus } \overline{\Omega _0}$ is empty which is sufficient to give the expected result. Indeed if $\Omega {\setminus } \overline{\Omega _0}$ is nonempty then we can proceed as in the proof of Theorem 3.6 to derive that u vanishes in ${\mathcal {U}}\times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$, for some ${\mathcal {U}}$, a neighborhood of a point in $\partial \Omega _0\cap \Omega$. But this contradicts the maximality of $\Omega _0$. $\square$

The uniqueness of continuation from the Cauchy data on a subboundary is given in the following corollary.

Corollary 6.1

Let ${\mathcal {N}}$ be as in Theorem 6.4 with $\Omega$ substituted by larger domain ${\hat{\Omega }}\Supset \Omega$. Let $\Gamma _0$ a nonempty open subset of $\Gamma$ and $\Sigma _0=\Gamma _0\times (-{\mathfrak {t}},{\mathfrak {t}})$. For $A\in {\mathcal {N}}$, let $u\in H^{2,1}(Q)$ satisfying ${\mathcal {L}}_A^wu=0$ in Q and $u=\partial _\nu u=0$ on $\Sigma _0$. Then $u=0$ in $\Omega \times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$.

Also, the unique continuation across a A-pseudo-convex hypersurface is contained in the following theorem.

Theorem 6.5

Let $H=\{x\in \omega ;\;\theta (x)=\theta ({\tilde{x}})\}$ be a A-pseudo-convex hypersurface defined in a neighborhood of ${\tilde{x}}\in \Omega$ with $\theta \in C^{3,1}({\overline{\omega }})$. Then there exists ${\mathcal {B}}$, a neighborhood of ${\tilde{x}}$, so that if $u\in H^{2,1}(\omega \times (-{\mathfrak {t}},{\mathfrak {t}}))$ satisfies ${\mathcal {L}}_A^wu=0$ in $\omega \times (-{\mathfrak {t}},{\mathfrak {t}})$ and $\text{ supp }(u(\cdot ,t))\subset H_+=\{ x\in \omega ;\; \theta (x)\ge \theta ({\tilde{x}})\}$, $t\in (-{\mathfrak {t}},{\mathfrak {t}})$, then $u=0$ in ${\mathcal {B}}\times (-{\mathfrak {t}}/2,{\mathfrak {t}}/2)$.

6.3 Observability inequality

In this subsection $t_1=0$ and $t_2={\mathfrak {t}}>0$.

Let ${\mathscr {A}}:L^2(\Omega )\rightarrow L^2(\Omega )$ be the unbounded operator introduced in the preceding section. That is

$$\begin{aligned} {\mathscr {A}}u=-\Delta _Au,\quad D({\mathscr {A}})=H_0^1(\Omega )\cap H^2(\Omega ). \end{aligned}$$

It is known that $u(t)=e^{it{\mathscr {A}}}u_0$, $u_0\in L^2(\Omega )$ is the solution of the following IBVP

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta _Au+i\partial _tu=0\quad \text{ in }\; Q, \\ u(\cdot ,0)=u_0, \\ u_{|\Sigma }=0. \end{array} \right. \end{aligned}$$

(6.10)

Furthermore, u belongs to $C([0,{\mathfrak {t}}];D({\mathscr {A}}))\cap C^1([0,{\mathfrak {t}}];L^2(\Omega ))$ whenever $u_0\in D({\mathscr {A}})$ and, for $0\le t\le {\mathfrak {t}}$, we have

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^2(\Omega )}=\Vert u_0\Vert _{L^2(\Omega )},\quad \Vert \nabla _A u(\cdot ,t)\Vert _{L^2(\Omega )}=\Vert \nabla _Au_0\Vert _{L^2(\Omega )}. \end{aligned}$$

(6.11)

Theorem 6.6

Suppose that $0\le \psi _0\in C^4({\overline{\Omega }})$ is A-pseudo-convex with constant $\kappa >0$ and let $\Gamma _+=\{x\in \Gamma ;\; \partial _{\nu _A}\psi _0(x)>0\}$. Then there exists a constant $\aleph$ only depending $\Omega$, ${\mathfrak {t}}$, $\varkappa$, $\kappa$ and $\Gamma _+$, so that, for any $u_0\in D({\mathscr {A}})$, we have

$$\begin{aligned} \Vert u_0\Vert _{H_0^1(\Omega )}\le \aleph \Vert \partial _\nu u\Vert _{L^2(\Sigma _+)}, \end{aligned}$$

where $\Sigma _+=\Gamma _+\times (0,{\mathfrak {t}})$ and $u=e^{it{\mathscr {A}}}u_0$.

Proof

In light of Theorem 6.1 and identities (6.11), the expected inequality can be proved by modifying slightly that of the wave equation. $\square$

Remark 6.1

It is worth mentioning that the results for the elliptic, wave and Schrödinger equations can be extended to the case where $\Delta _A$ is substituted by the associated magnetic operator defined by

$$\begin{aligned} \Delta _{A,\mathbf{b }}u=\sum _{k, \ell =1}^n(\partial _k +ib_k)a_{k\ell }(\partial _\ell +ib_\ell )u, \end{aligned}$$

with $\mathbf{b }=(b_1,\ldots ,b_n)\in W^{1,\infty }(\Omega ;{\mathbb {R}}^n)$.

Note that $\Delta _{A,\mathbf{b }}u$ can be rewritten in the following form

$$\begin{aligned} \Delta _{A,\mathbf{b }}u=\Delta _A u +2i(\nabla u|\mathbf{b })_A+\left( -|\mathbf{b }|_A^2+\text{ div }(A\mathbf{b })\right) u. \end{aligned}$$