Strong unique continuation for second-order hyperbolic equations with time-independent coefficients

Abstract

In this paper we prove that if u is a solution to second-order hyperbolic equation $\partial ^2_{t}u+a(x)\partial _{t}u-(\text { div}_x\left( A(x)\nabla _x u\right) +b(x)\cdot \nabla _x u+c(x)u)=0$ and u is flat on a segment $\{x_0\}\times (-T,T)$ (T finite), then u vanishes in a neighborhood of $\{x_0\}\times (-T,T)$. The novelty with respect to earlier papers on the subject is the nonvanishing damping coefficient a(x) in the hyperbolic equation.

1 Introduction

In this paper we study strong unique continuation property (SUCP) for the equation

$$\begin{aligned} \partial ^2_{t}u+a(x)\partial _{t}u-L(u)=0, \quad \hbox {in } B_{\rho _0}\times (-T,T), \end{aligned}$$

(1.1)

where ${\rho_0}, T$ are given positive numbers, $B_{\rho _0}$ is the ball of ${\mathbb {R}}^{n}$, $n\ge 2$, of radius $\rho _0$ and center at 0, $a\in L^{\infty }(\mathbb {R}^n)$, L is the second-order elliptic operator

$$\begin{aligned} L(u)=\text { div}_x\left( A(x)\nabla _x u\right) +b(x)\cdot \nabla _x u+c(x)u, \end{aligned}$$

(1.2)

$b\in L^{\infty }(\mathbb {R}^n;\mathbb {R}^n)$, $c\in L^{\infty }(\mathbb {R}^n)$ and A(x) is a real-valued symmetric $n\times n$ matrix that satisfies a uniform ellipticity condition and entries of A(x) are functions of Lipschitz class.

We say that Eq. (1.1) has the SUCP if there exists a neighborhood $\mathcal {U}$ of $\{0\}\times (-T,T)$ such that for every solution, u, to Eq. (1.1) we have

$$\begin{aligned} \left\| u\right\| _{L^2\left( B_{r}\times (-T,T)\right) }=\mathcal {O}(r^N) \text{, } \forall N\in \mathbb {N} \text{, } \text{ as } r\rightarrow 0, \quad \Longrightarrow \quad u=0 , \text{ in } \mathcal {U}. \end{aligned}$$

(1.3)

Property (1.3) was proved (if the matrix A belongs to $C^2$), under the additional condition $T=+\infty$ and u is bounded, by Masuda in 1968, [25]. Later on, in 1978, Baouendi and Zachmanoglou, [5], proved the SUCP whenever the coefficients of equation (1.1) are analytic functions. In 1999, Lebeau, [23], proved the SUCP for solution to (1.1) when $a=b=c=0$. The proof of [23] requires the symmetry of the differential operator, and there seems no obvious extension of the proof to the nonsymmetric case, in particular, to the case of damped wave equation $\partial ^2_{t}u+a(x)\partial _{t}u-\Delta u=0$. We also refer to [29, 32] where the SUCP at the boundary and the quantitative estimate of unique continuation related to property was proved when $a=0$.

The novelty of the present paper with respect to earlier papers, with finite T, is the nonvanishing damping coefficient a(x) in the hyperbolic Eq. (1.1).

It is worth noting that SUCP and the related quantitative estimates, have been extensively studied and today well understood in the context of second-order elliptic and parabolic equation. Among the extensive literature on the subject here we mention, for the elliptic equations, [3, 4, 15, 19], and, for the parabolic equations, [2, 8, 20]. In the context of elliptic and parabolic equations, the quantitative estimates of unique continuation appear in the form of three sphere inequalities [21], doubling inequalities [13], or two-sphere one-cylinder inequality [9]. We refer to [1] and [31] for a more extensive literature concerning the elliptic context and the parabolic context respectively.

In the present paper we prove (Theorem 2.1) a quantitative estimate of unique continuation from which we derive (Corollary 2.2) property (1.3) for equation (1.1). The crucial step of the proof is Theorem 3.1, in such a Theorem 3.1 we exploit in a suitable way the simple and classical idea of converting a hyperbolic equation into an elliptic equation, see, for instance, [12, Ch.6]. Formally, such a classical idea consists in substitute, in (1.1), the variable t by iy. More precisely, the idea can be told as follows. Let us define the integral transform

$$v(x,y) = \int_{{ - T}}^{T} u (x,t)\Phi (t + iy){\text{d}}t,$$

(1.4)

where the kernel $\Phi$ is a holomorphic function in variable $z=t+iy$. It is simple to check that v satisfies the elliptic equation

$$\begin{aligned} \partial ^2_{y}v+L(v)-ia(x)\partial _{y}v=F(x,y), \end{aligned}$$

(1.5)

where F is an “error term” which depends on $u(\cdot ,\pm T)$, $\partial _t u(\cdot ,\pm T)$ and $\Phi (\pm T+iy)$.

The use of converting a hyperbolic equation into an elliptic equation, in the issue of weak unique continuation property (WUCP) for finite time T, can be tracked back to Robbiano in 1991, [26], see also [27]. By WUCP for (1.1) we mean: let R be a given positive number, there exists a neighborhood $\mathcal {V}$ of $\{0\}\times (-T,T)$ such that for every solution, u to equation (1.1) we have

$$\begin{aligned} u=0, \text{ in } B_R\times (-T,T) \Longrightarrow u=0 , \text{ in } \mathcal {V}. \end{aligned}$$

(1.6)

Subsequently, in [16, 28] and [30], the WUCP was studied in the general context of equation with partially analytic coefficients (not only of hyperbolic type) and the exact dependence domain $\mathcal {V}$ was determined, see also [6, 17, 22] for the related quantitative estimates. The above mentioned papers rely on the so-called Fourier–Bros–Iagolnitzer (FBI) transform that is an integral transform like (1.4) whose kernel is the Gaussian function $\Phi (z)=\sqrt{\mu /2\pi }e^{-\mu z^2}$, where $\mu$ is a large parameter. Although the FBI transform works very well to prove the WUCP for equation (1.1), it seems no obvious whether the FBI works well to tackle the SUCP.

In the present paper to prove SUCP for (1.1) we use integral transform (1.4) with well-chosen family of polynomial kernels. More precisely, we define

$$\begin{aligned} v_k(x,y)=\int ^T_{-T}u(x,t)\varphi _k(t+iy)dt, \quad \forall k\in \mathbb {N}, \end{aligned}$$

(1.7)

where $\varphi _k(t+iy)$ is a polynomial with the following property:

• (a) $\varphi _k(t+i0)$ is an approximation of Dirac’s $\delta$-function,

• (b) $\left| \varphi _k(\pm T+iy)\right| \le C^k|y|^k$ for $k\in \mathbb {N}$ and $|y|\le 1$, where C is a constant.

In this way functions $v_k$ turn out solutions to the elliptic equation

$$\begin{aligned} \partial ^2_{y}v_k-ia(x)\partial _{y}v_k+L(v_k)=F_k(x,y), \quad \hbox {in } B_{1}\times \mathbb {R}, \end{aligned}$$

(1.8)

where $|F_k|\le C^k|y|^k$, for $k\in \mathbb {N}$ and $|y|\le 1$. This behavior of $F_k$ allows us to handle in a suitable way a Carleman estimate with singular weight for second-order elliptic operators, see Sect. 2.3, in such a way to get $u(x,0)=0$ for $x\in B_{\rho }$, where $\rho \le \rho _0/C$. Similarly, we prove for every $t\in (-T,T)$, $u(\cdot ,t)=0$ in $B_{\rho (t)}$, where $\rho (t)=(1-tT^{-1})\rho$. So that we obtain (1.3) with $\mathcal {U}=\bigcup _{t\in (-T,T)}(B_{\rho (t)}\times \{t\})$. As a consequence of this result and using the WUCP, we have that $u=0$ in the domain of dependence of $\mathcal {U}$.

The quantitative estimate of unique continuation that we prove in Theorem 2.1 can be read, roughly speaking, as a continuous dependence estimate of $u_{_{|\mathcal {U}}}$ from $u_{_{|{B_{r_0}\times (-T,T)}}}$, where $r_0$ is arbitrarily small. The sharp character of such a continuous dependence result is related to the logarithmic character of this estimate, that, at the light of counterexample of John [18], cannot be improved and to the fact that this quantitative estimate implies the SUCP property. The quantitative estimate of strong unique continuation (at the interior and at the boundary) was a crucial tool, see [33], to prove sharp stability estimate for inverse problems with unknown boundaries for wave equation $\partial ^2_tu-\text { div}_x\left( A(x)\nabla _x u\right) =0$.

Before concluding Introduction, we mention an open question (to the author knowledge). Such an open question concerns the SUCP, (1.3), for the second-order hyperbolic equation with coefficients that are analytic in variable t and smooth enough (but not analytic) in variables x. This is, for instance, the case of the equation

$$\begin{aligned} \partial ^2_{t}u+a(x,t)\partial _{t}u-\Delta _xu=0, \end{aligned}$$

where a(x, t) is smooth enough w.r.t x and analytic w.r.t. t. Concerning this topic we mention [24] in which it is proved that if u satisfies the conditions: (a) $(\{0\}\times (-T,T)) \cap \mathrm{supp}u$ is compact and (b) $D^{j}u(x,t)=\mathcal {O}\left( e^{-k/|x|}\right)$, $j=1,2$, for every k as $x\rightarrow 0$, $t\in (-T,T)$, then u vanishes in a neighborhood of $\{0\}\times (-T,T)$.

The plan of the paper is as follows: In Sect. 2 we state the main result of this paper, and in Sect. 3 we prove the main theorem.

2 The main results

2.1 Notation and definition

Let $n\in \mathbb {N}$, $n\ge 2$. For any $x\in \mathbb {R}^n$ we will denote $x=(x',x_n)$, where $x'=(x_1,\ldots ,x_{n-1})\in \mathbb {R}^{n-1}$, $x_n\in \mathbb {R}$ and $|x|=\left( \sum _{j=1}^nx_j^2\right) ^{1/2}$. Given $r>0$, we will denote by $B_r$, $B'_r$ and ${\widetilde{B}}_r$ the ball of $\mathbb {R}^{n}$, $\mathbb {R}^{n-1}$ and $\mathbb {R}^{n+1}$ of radius r centered at 0, respectively. For any open set $\Omega \subset \mathbb {R}^n$ and any function (smooth enough) u we denote by $\nabla _x u=(\partial _{x_1}u,\cdots , \partial _{x_n}u)$ the gradient of u. Also, for the gradient of u we use the notation $D_xu$. If $j=0,1,2$ we denote by $D^j_x u$ the set of the derivatives of u of order j, so $D^0_x u=u$, $D^1_x u=\nabla _x u$ and $D^2_xu$ is the hessian matrix $\{\partial _{x_ix_j}u\}_{i,j=1}^n$. Similar notation is used whenever other variables occur and $\Omega$ is an open subset of $\mathbb {R}^{n-1}$ or a subset $\mathbb {R}^{n+1}$. By $H^{\ell }(\Omega )$, $\ell =0,1,2$, we denote the usual Sobolev spaces of order $\ell$ (in particular, $H^0(\Omega )=L^2(\Omega )$), with the standard norm

$$\begin{aligned} \left\| v(x)\right\| _{H^{\ell }(\Omega )}=\left( \sum _{0\le j\le \ell }\int _{\Omega }\left| D^jv(x)\right| ^2dx\right) ^{1/2}. \end{aligned}$$

For any interval $J\subset \mathbb {R}$ and $\Omega$ as above we denote

$$\begin{aligned} \mathcal {W}\left( J;\Omega \right) =\left\{ u\in C^0\left( J;H^2\left( \Omega \right) \right) : \partial _t^\ell u\in C^0\left( J;H^{2-\ell }\left( \Omega \right) \right) , \ell =1,2\right\} . \end{aligned}$$

We shall use the letters $c, C,C_0,C_1,\cdots$ to denote constants. The value of the constants may change from line to line, but we shall specified their dependence everywhere they appear. Generally we will omit the dependence of various constants by n.

2.2 Statements of the main results

Let $\rho _0>0$, T, $\lambda \in (0,1]$, $\Lambda >0$ and $\Lambda _1>0$ be given numbers. Let $A(x)=\left\{ a^{ij}(x)\right\} ^n_{i,j=1}$ be a real-valued symmetric $n\times n$ matrix whose entries are measurable functions, and they satisfy the following conditions

$$\begin{aligned}&\lambda \left| \xi \right| ^2\le A(x)\xi \cdot \xi \le \lambda ^{-1}\left| \xi \right| ^2, \quad \hbox {for every } x, \xi \in \mathbb {R}^n, \end{aligned}$$

(2.1a)

$$\begin{aligned}&\left| A(x_{*})-A(x)\right| \le \frac{\Lambda }{\rho _0} \left| x_{*}-x \right| , \quad \hbox {for every } x_{*}, x\in \mathbb {R}^n. \end{aligned}$$

(2.1b)

Let $b\in L^{\infty }(\mathbb {R}^n; \mathbb {R}^n)$ and $a, c\in L^{\infty }(\mathbb {R}^n)$ satisfy

$$\begin{aligned} T\left| a(x)\right| +T^2\rho _0^{-1} \left| b(x)\right| +T^{2}\left| c(x)\right| \le \Lambda _1, \quad \hbox {for almost every } x\in \mathbb {R}^n, \end{aligned}$$

(2.2)

Let

$$\begin{aligned} L(u)=\text { div}_x\left( A(x)\nabla _x u\right) +b(x)\cdot \nabla _x u+c(x)u. \end{aligned}$$

(2.3)

Let $u\in \mathcal {W}\left( [-T,T];B_{\rho _0}\right)$ be a solution to

$$\begin{aligned} \partial ^2_{t}u+a(x)\partial _{t}u-L(u)=0, \quad \hbox {a.e. } \hbox {in } B_{\rho _0}\times (-T,T). \end{aligned}$$

(2.4)

Let $\varepsilon$ and H be given positive numbers and let $r_0\in (0,\rho _0]$. We assume

$$\begin{aligned} \rho _0^{-n}T^{-1}\int _{-T}^{T}\int _{B_{r_0}}\left| u(x,t)\right| ^2dxdt\le \varepsilon ^2 \end{aligned}$$

(2.5)

and

$$\begin{aligned} \max _{t\in [-T,T]}\left( \rho _0^{-n}\int _{B_{\rho _0}}\left| u(x,t)\right| ^2 dx+\rho _0^{-n+1}\int _{B_{\rho _0}}\left| \partial _tu(x,t)\right| ^2dx\right) \le H^2. \end{aligned}$$

(2.6)

Theorem 2.1

Let $u\in \mathcal {W}\left( [-T,T];B_{\rho _0}\right)$ be a weak solution to (2.4) and let (2.1), (2.2), (2.5) and (2.6) be satisfied. For every $\alpha \in (0,1/2)$ there exist constants $s_0\in (0,1)$ and $C\ge 1$ depending on $\lambda$, $\Lambda$, $\Lambda _1$, $\alpha$ and $T\rho _0^{-1}$ only such that for every $t_0\in (-T,T)$ and every $0<r_0\le \rho \le s_0\rho _0$ the following inequality holds true

$$\begin{aligned} \rho _0^{-n}\int _{B_{\rho (t_0)}}\left| u(x,t_0) \right| ^2dx\le C\frac{\left( \log \rho _0/r_0)\right) ^{\alpha }(H+\varepsilon )^2}{\left( \log \left( e+H\varepsilon ^{-1}\right) \right) ^{\alpha }}, \end{aligned}$$

(2.7)

where

$$\begin{aligned} \rho (t_0)=(1-|t_0|T^{-1})\rho . \end{aligned}$$

The proof of Theorem 2.1 is given in Sect. 3.

The proof of the following corollary is standard (see, for instance, [32, Remark 2.2]), but we give it for the reader convenience.

Corollary 2.2

(Strong Unique Continuation Property) Let $u\in \mathcal {W}\left( [-T,T];B_{\rho _0}\right)$ be a weak solution to (2.4). Assume that (2.1) and (2.2) be satisfied. We have that, if

$$\begin{aligned} \left( \rho _0^{-n}T^{-1}\int ^T_{-T}\int _{B_{r_0}}\left| u(x,t)\right| ^2dxdt\right) ^{1/2}=O(r_0^N)\text{, }\quad \forall N\in \mathbb {N} \text{, } \text{ as } r_0\rightarrow 0, \end{aligned}$$

then

$$\begin{aligned} u(\cdot ,t)=0 \text{, } \text{ for } |x|+\frac{\rho _0|t|}{T}\le s_0\rho _0, \end{aligned}$$

(2.8)

where $s_0$ is defined in Theorem 2.1.

Proof

We consider the case $t=0$; similarly, we could proceed for $t\ne 0$. If $\left\| u(\cdot ,0) \right\| _{L^2\left( B_{s_0\rho _0}\right) }=0$ there is nothing to proof, otherwise, if

$$\begin{aligned} \left\| u(\cdot ,0) \right\| _{L^2\left( B_{s_0\rho _0}\right) }>0 \end{aligned}$$

(2.9)

we argue by contradiction. By (2.9) it is not restrictive to assume that

$$\begin{aligned} \max _{t\in [-T,T]}\left( \rho _0^{-n}\int _{B_{\rho _0}}\left| u(x,t)\right| ^2 dx+\rho _0^{-n+1}\int _{B_{\rho _0}}\left| \partial _tu(x,t)\right| ^2dx\right) =1. \end{aligned}$$

(2.10)

Now we apply inequality (2.7) with $\varepsilon =C_Nr_0^N$, $H=1$ and passing to the limit as $r_0\rightarrow 0$ we derive

$$\begin{aligned} \left\| u(\cdot ,0) \right\| _{L^2\left( B_{s_0\rho _0}\right) }\le CN^{-\alpha /2}, \quad \forall N\in \mathbb {N}, \end{aligned}$$

(2.11)

by passing again to the limit as $N\rightarrow 0$ in (2.11) we get $\left\| u(\cdot ,0) \right\| _{L^2\left( B_{s_0\rho _0}\right) }=0$ that contradicts (2.9). $\square$

2.3 Auxiliary result: Carleman estimate with singular weight

In order to prove Theorem 2.1, we need a Carleman estimate proved by several authors; here we recall [3, 15]. In order to control the dependence of the various constants, we use here a version of such a Carleman estimate proved, in the context of parabolic operator, in [11], see also [7, Sect. 8].

First we introduce some notation. Let $\mathcal {P}$ be the elliptic operator

$$\begin{aligned} \mathcal {P}(w):=\partial ^2_{y}w+L(w)-ia(x)\partial _{y}w. \end{aligned}$$

(2.12)

Denote

$$\begin{aligned}&\varrho (x,y)=\left( A^{-1}(0)x\cdot x+y^2\right) ^{1/2}, \end{aligned}$$

(2.13)

$$\begin{aligned}&{\widetilde{B}}^{\varrho }_{r}= \left\{ (x,y)\in \mathbb {R}^{n+1}: \varrho (x,y)< r\right\} , \quad \hbox { } r>0. \end{aligned}$$

(2.14)

Notice that

$$\begin{aligned} {\widetilde{B}}^{\varrho }_{\sqrt{\lambda } r}\subset {\widetilde{B}}_r \subset {\widetilde{B}}^{\varrho }_{r/\sqrt{\lambda }}, \quad \hbox { }\forall r>0. \end{aligned}$$

(2.15)

Theorem 2.3

Let $\mathcal {P}$ be the operator (2.12) and assume that (2.1) is satisfied. There exists constants $C_{*}>1$ depending on $\lambda$, and $\Lambda$ only and $\tau _0>1$ depending on $\lambda$, $\Lambda$ and $\Lambda _1$ only such that, denoting

$$\begin{aligned}&\Psi (r)=r\exp \left( \int ^r_0\frac{e^{-C_{*}\eta }-1}{\eta }d\eta \right) , \end{aligned}$$

(2.16a)

$$\begin{aligned}&\psi (x,y)=\Psi \left( \varrho (x,y)/2\sqrt{\lambda }\right) , \end{aligned}$$

(2.16b)

for every $\tau \ge \tau _0$ and $w\in C^{\infty }_0\left( {\widetilde{B}}^{\varrho }_{2\sqrt{\lambda }/C_{*}}\setminus \{0\}\right)$ we have

$$\begin{aligned} \int _{\mathbb {R}^{n+1}}\left( \tau \psi ^{1-2\tau }\left| \nabla _{x,y}w\right| ^2+ \tau ^3\psi ^{-1-2\tau } \left| w\right| ^2\right) dxdy \le C_{*}\int _{{\mathbb {R}^{n+1}}}\psi ^{2-2\tau } \left| \mathcal {P}(w)\right| ^2dxdy. \end{aligned}$$

(2.17)

Remark 2.4

We emphasize that

$$\begin{aligned} \Psi (r)\simeq r, \quad \hbox { as } r\rightarrow 0. \end{aligned}$$

Moreover, $\Psi$ is an increasing and concave function and there exists $C>1$ depending on $\lambda$, and $\Lambda$ such that

$$\begin{aligned} C^{-1}r\le \Psi \left( r\right) \le r, \quad \hbox { } \forall r\in (0,1]. \end{aligned}$$

(2.18)

3 Proof of Theorem 2.1

The primary step to achieve Theorem 2.1 consists in proving the following

Theorem 3.1

Let us assume $\rho _0=1$ and $T=1$. Let $u\in \mathcal {W}\left( [-1,1];B_1\right)$ be a weak solution to (2.4) and let (2.1), (2.2), (2.5) and (2.6) be satisfied. For every $\alpha \in (0,1/2)$ there exist constants $s_0\in (0,1)$ and $C\ge 1$ depending on $\lambda$, $\Lambda$, $\Lambda _1$ and $\alpha$ only such that for every $0<r_0\le s\le s_0$ the following inequality holds true

$$\begin{aligned} \int _{B_{s}}\left| u(x,0) \right| ^2dx\le C\frac{\left( \log 1/r_0\right) ^{\alpha }(H+\varepsilon )^2}{\left( \log \left( e+H\varepsilon ^{-1}\right) \right) ^{\alpha }}. \end{aligned}$$

(3.1)

In order to prove Theorem 3.1, we define

$$\begin{aligned}&v_k(x,y)=\int ^1_{-1}u(x,t)\varphi _k(t+iy)dt, \quad \forall k\in \mathbb {N}, \end{aligned}$$

(3.2)

where

$$\begin{aligned}&\varphi _k(z)=\mu _k\left( 1-z^2\right) ^k, \quad z=t+iy\in \mathbb {C}, \end{aligned}$$

(3.3)

and

$$\begin{aligned} \mu _k=\left( \int ^1_{-1}\left( 1-t^2\right) ^kdt\right) ^{-1}, \end{aligned}$$

(3.4)

so that we have

$$\begin{aligned} \int ^1_{-1}\varphi _k(t)dt=1, \quad \forall k\in \mathbb {N}. \end{aligned}$$

(3.5)

It is easy to check that

$$\begin{aligned} \mu _k\simeq \sqrt{\frac{k}{\pi }}, \quad \text{ as } k\rightarrow \infty . \end{aligned}$$

(3.6)

We need some simple lemmas to state the properties of functions $v_k$.

Lemma 3.2

We have

$$\begin{aligned} \left\| v_k(\cdot ,0)-u(\cdot ,0)\right\| _{L^2(B_1)}\le c\frac{\log k}{ \sqrt{k}}, \quad \forall k\in \mathbb {N}, \end{aligned}$$

(3.7)

where c depends on n only.

Proof

By (3.2) and (3.5) we have

$$\begin{aligned} v_k(x,0)-u(x,0)=\int ^1_{-1}\left( u(x,t)-u(x,0)\right) \varphi _k(t)dt, \quad \forall x\in B_1; \end{aligned}$$

hence, by Schwarz inequality and integrating over $B_1$, we have

$$\begin{aligned}&\int _{B_1}\left| v_k(x,0)-u(x,0)\right| ^2dx\le \int _{B_1}dx\int ^1_{-1}\left| u(x,t)-u(x,0)\right| ^2\varphi _k(t)dt=\nonumber \\&\quad =\int _{[-\gamma ,\gamma ]}\varphi _k(t)dt\int _{B_1}\left| u(x,t)-u(x,0)\right| ^2dx+\int _{[-1,1]\setminus [-\gamma ,\gamma ]}\varphi _k(t)dt\int _{B_1}\left| u(x,t)-u(x,0)\right| ^2dx, \end{aligned}$$

(3.8)

where $\gamma \in (0,1)$ is a number that we will choose. Now we have

$$\begin{aligned} \varphi _k(t)\le \mu _k(1-\gamma ^2)^{k/2}, \quad \forall t\in [-1,1]\setminus [-\gamma ,\gamma ] \end{aligned}$$

and, by (2.6),

$$\begin{aligned} \int _{B_1}\left| u(x,t)-u(x,0)\right| ^2dx\le t^2H^2 . \end{aligned}$$

Hence, by (2.6), (3.6) and (3.8), we have

$$\begin{aligned} \left\| v_k(\cdot ,0)-u(\cdot ,0)\right\| _{L^2(B_1)}\le c H \left( \gamma +k^{1/4}(1-\gamma ^2)^{k/2}\right) , \quad \text{ for } \text{ every } \gamma \in (0,1), \end{aligned}$$

(3.9)

where c depends on n only. Now, we choose $\gamma =k^{-1/2}\log k$ and we get (3.7). $\square$

Lemma 3.3

Let u be a solution to (2.4), and let (2.1) and (2.2) be satisfied, then $v_k\in H^2\left( B_1\times (-1,1)\right)$ is a solution to the equation

$$\begin{aligned} \partial ^2_{y}v_k-ia(x)\partial _{y}v_k+L(v_k)=F_k(x,y), \quad \hbox {in } B_{1}\times \mathbb {R}, \end{aligned}$$

(3.10)

where $F_k\in L^{\infty }(-1,1;L^2(B_1))$ and it satisfies

$$\begin{aligned} \left\| F_k(\cdot ,y)\right\| _{L^2(B_1)}\le CH k\mu _k |\sqrt{5}y|^{k-1}, \quad \forall y\in [-1,1], \end{aligned}$$

(3.11)

C depending on $\Lambda _1$ only.

In addition, $v_k$ satisfies the following properties

$$\begin{aligned}&\sup _{y\in [-1,1]}\left\| v_k(\cdot ,y)\right\| _{L^2(B_1)}\le 2^k\mu _k H, \end{aligned}$$

(3.12)

$$\begin{aligned}&\int _{{\widetilde{B}}_{r_0/2}} \left( \left| v_k\right| ^2+r_0^2\left| \nabla _{x,y} v_k\right| ^2\right) dxdy\le C\left( r_04^kk\varepsilon ^2+H^2k^3\left( \sqrt{5}r_0\right) ^{2(k+2)}\right) , \end{aligned}$$

(3.13)

where C depends on $\lambda$ and $\Lambda _1$ only.

Proof

The fact that $v_k$ belongs to $H^2\left( B_1\times (-1,1)\right)$ is an immediate consequence of differentiation under the integral sign. Actually we have

$$\begin{aligned} \partial _y^m D^j_xv_k(x,y)=\int ^1_{-1}D^j_x u(x,t)\partial _y^m\left( \varphi _k(t+iy)\right) dt, \quad \text{ for } j,m=0,1,2; \end{aligned}$$

(3.14)

hence, by Schwarz inequality and taking into account that $u\in \mathcal {W}\left( [-1,1];B_1\right)$, we have $v_k\in H^2\left( B_1\times (-1,1)\right)$.

Now we prove (3.10).

By integration by parts and taking into account that

$$\begin{aligned} \partial _t \varphi _k(t+iy)=\frac{1}{i}\partial _y \varphi _k(t+iy), \end{aligned}$$

,we have

$$\begin{aligned} \begin{aligned} \int ^1_{-1}\partial _t u(x,t)\varphi _k(t+iy)dt&=\int ^1_{-1}\partial _t u(x,t)\varphi _k(t+iy)dt\\&=\left. (u(x,t)\varphi _k(t+iy))\right| _{t=-1}^{t=1}-\frac{1}{i}\int ^1_{-1}u(x,t)\partial _y \varphi _k(t+iy)dt\\&=\left. (u(x,t)\varphi _k(t+iy))\right| _{t=-1}^{t=1}+i\partial _yv_k(x,y). \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} -i\partial _yv_k(x,y)=-\int ^1_{-1}\partial _t u(x,t)\varphi _k(t+iy)dt+\left. (u(x,t)\varphi _k(t+iy))\right| _{t=-1}^{t=1}. \end{aligned}$$

(3.15)

Similarly, we have

$$\begin{aligned} \begin{aligned} \partial ^2_yv_k(x,y)&=\partial ^2_yv_k(x,y)=-\int ^1_{-1}\partial ^2_tu(x,t)\varphi _k(t+iy)dt+\\ {}&+\left. (\partial _tu(x,t)\varphi _k(t+iy))\right| _{t=-1}^{t=1}-\left. (u(x,t)\varphi '_k(t+iy))\right| _{t=-1}^{t=1}. \end{aligned} \end{aligned}$$

(3.16)

Now, by (2.4), (3.14), (3.15) and (3.16) we have

$$\begin{aligned}&\partial ^2_{y}v_k-ia(x)\partial _{y}v_k+L(v_k)=\nonumber \\&\quad =-\int ^1_{-1}\left\{ \partial ^2_tu(x,t)+a(x)\partial _tu(x,t)-L(u)(x,t)\right\} \varphi _k(t+iy)dt+F_k(x,y)=F_k(x,y), \end{aligned}$$

(3.17)

where

$$\begin{aligned} \begin{aligned} F_k(x,y)&=\varphi _k(1+iy)\left( a(x)u(x,1)+\partial _t u(x,1)\right) -\varphi '_k(1+iy)u(x,1)\\&\quad -\left[ \varphi _k(-1+iy)\left( a(x)u(x,-1)+\partial _t u(x,-1)\right) -\varphi '_k(-1+iy)u(x,-1)\right] . \end{aligned} \end{aligned}$$

(3.18)

and (3.10) is proved.

Now we prove (3.11).

It is easy to check that for every $k\in \mathbb {N}$ we have

$$\begin{aligned}&\left| \varphi _k(\pm 1+iy)\right| =\mu _k\left( 4y^2+y^4\right) ^{k/2}, \quad \forall y\in \mathbb {R}, \end{aligned}$$

(3.19a)

$$\begin{aligned}&\left| \varphi '_k(\pm 1+iy)\right| =2k\mu _k\left( 1+y^2\right) ^{1/2}\left( 4y^2+y^4\right) ^{(k-1)/2}, \quad \forall y\in \mathbb {R}. \end{aligned}$$

(3.19b)

In addition, since

$$\begin{aligned} \left| \varphi _k(t+iy)\right| =\mu _k\left[ t^4-2t^2(1-y^2)+(1+y^2)^2\right] ^{\frac{k}{2}}, \end{aligned}$$

we have

$$\begin{aligned} \left| \varphi _k(t+iy)\right| \le 2^k\mu _k, \quad \forall (t,y)\in [-1,1]\times [-1,1]. \end{aligned}$$

(3.20)

By (2.2), (2.6), (3.19a) and (3.19b) we have (3.11).

By Schwarz inequality and (3.20) we have, for any $R\in (0,1]$,

$$\begin{aligned} \sup _{y\in [-1,1]}\left\| v_k(\cdot ,y)\right\| _{L^2(B_{R})}\le 2^k\mu _k\left( \int _{-1}^{1}\int _{B_{R}}u^2(x,t)dxdt\right) ^{1/2}; \end{aligned}$$

(3.21)

hence, for $R=1$, taking into account (2.6), we obtain (3.12).

Finally, let us prove (3.13). For this purpose, we firstly observe that applying (3.21) for $R=r_0$ and taking into account (2.5), we have

$$\begin{aligned} \sup _{y\in [-1,1]}\left\| v_k(\cdot ,y)\right\| _{L^2(B_{r_0})}\le C 2^{k}\mu _k\varepsilon . \end{aligned}$$

(3.22)

Afterward, since $v_k$ is solution to elliptic equation (3.10), the following Caccioppoli inequality, [10, 14], holds

$$\begin{aligned} \int _{{\widetilde{B}}_{r_0/2}} \left| \nabla _{x,y} v_k\right| ^2dxdy\le Cr_0^{-2}\int _{{\widetilde{B}}_{r_0}} \left( \left| v_k\right| ^2+r_0^{4}\left| F_k\right| ^2\right) dxdy, \end{aligned}$$

(3.23)

where C depends on $\lambda$ only. Finally, by (3.6), (3.11), (3.22) and (3.23) we get (3.13). $\square$

Proof of Theorem 3.1

Set

$$\begin{aligned} r_1=\frac{\sqrt{\lambda } r_0}{8}. \end{aligned}$$

By (3.13) we have

$$\begin{aligned} \int _{{\widetilde{B}}^{\varrho }_{4r_1}} \left( \left| v_k\right| ^2+r_1^2\left| \nabla _{x,y} v_k\right| ^2\right) dxdy\le C r_1\sigma _k(\varepsilon , r_1), \end{aligned}$$

(3.24)

where C depends on $\lambda$ and $\Lambda _1$ only and

$$\begin{aligned} \sigma _k(\varepsilon , r_1)=4^kk\varepsilon ^2+H^2k^3\left( C_1r_1\right) ^{2k+2}, \end{aligned}$$

(3.25)

where $C_1=2\sqrt{5}\lambda ^{-1/2}$.

Now we apply Theorem 2.3.

Denote

$$\begin{aligned}\psi _0(r):=\Psi (r/2\sqrt{\lambda })\quad \hbox {, for every } r>0\end{aligned}$$

and

$$\begin{aligned} R=\frac{\sqrt{\lambda }}{2C_{*}}. \end{aligned}$$

Let us define

$$\begin{aligned} \zeta (x,y)=h\left( \psi (x,y)\right) . \end{aligned}$$

where h belongs to $C^2_0\left( 0, \psi _0\left( 2R\right) \right)$ and satisfies

$$\begin{aligned}&0\le h\le 1, \\&\quad h(r)=1 ,\quad \hbox { } \forall r\in \left[ \psi _0\left( 2r_1\right) , \psi _0\left( R\right) \right] , \\&h(r)=0, \quad \hbox { } \forall r\in \left[ 0,\psi _0\left( r_1\right) \right] \cup \left[ \psi _0\left( 3R/2\right) , \psi _0\left( 2R\right) \right] , \\&r_1\left| h'(r)\right| +r_1^2\left| h''(r)\right| \le c, \quad \hbox { } \forall r\in \left[ \psi _0\left( r_1\right) , \psi _0\left( 2r_1\right) \right] , \\&\quad \left| h'(r)\right| +\left| h''(r)\right| \le c, \quad \hbox { } \forall r\in \left[ \psi _0\left( R\right) , \psi _0\left( 3R/2\right) \right] , \end{aligned}$$

where c depends on $\lambda$ and $\Lambda$ only. Notice that if $2r_1\le \varrho (x,y)\le R$, then $\zeta (x,y)=1$ and if $\varrho (x,y)\ge 2R$ or $\varrho (x,y)\le r_1$, then $\zeta (x,y)=0$.

By density, we can apply (2.17) to the function $w=\zeta v_k$ and we have, for every $\tau \ge \tau _0$,

$$\begin{aligned} \int _{{\widetilde{B}}^{\varrho }_{2R}}\left( \tau \psi ^{1-2\tau }\left| \nabla _{x,y}\left( \zeta v_k\right) \right| ^2+ \tau ^3\psi ^{-1-2\tau }\left| \zeta v_k\right| ^2\right) dxdy \le C\left( I_1+I_2+I_2\right) , \end{aligned}$$

(3.27)

where C depends on $\lambda$, $\Lambda$ and $\Lambda _1$ only and

$$\begin{aligned}&I_1=\int _{{\widetilde{B}}^{\varrho }_{2R}}\psi ^{2-2\tau } \left| F_k\right| ^2 \zeta ^2dxdy, \end{aligned}$$

(3.28a)

$$\begin{aligned}&I_2=\int _{{\widetilde{B}}^{\varrho }_{2R}}\psi ^{2-2\tau } \left| \mathcal {P}(\zeta )\right| ^2 \left| v_k \right| ^2dxdy, \end{aligned}$$

(3.28b)

$$\begin{aligned}&I_3=\int _{{\widetilde{B}}^{\varrho }_{2R}}\psi ^{2-2\tau } \left| \nabla _{x,y} v_k\right| ^2\left| \nabla _{x,y}\zeta \right| ^2dxdy. \end{aligned}$$

(3.28c)

$\square$

Estimate of $I_1$.

By (2.18) we have

$$\begin{aligned} C_2^{-1}\le \left( |x|^2+y^2\right) ^{-1/2}\psi (x,y)\le C_2, \quad \forall (x,y)\in {\widetilde{B}}_1, \end{aligned}$$

(3.29)

where $C_2>2$ depends on $\lambda$ and $\Lambda$ only.

By (3.11), (3.28a) and (3.29) we have

$$\begin{aligned} \begin{aligned} I_1&=\int _{{\widetilde{B}}^{\varrho }_{2R}}\psi ^{2-2\tau } \left| F_k\right| ^2 \zeta ^2dxdy\le C_2^{2(\tau -1)}\int _{{\widetilde{B}}^{\varrho }_{2R}}\left( |x|^2+y^2\right) ^{1-\tau } \left| F_k\right| ^2 dxdy\le \\ {}&\le CH^2k^35^kC_2^{2(\tau -1)}\int _{{\widetilde{B}}^{\varrho }_{2R}}\left( |x|^2+y^2\right) ^{k-\tau }dxdy, \end{aligned} \end{aligned}$$

(3.30)

where C depends on $\lambda$ and $\Lambda$ only.

Now let k and $\tau$ satisfy

$$\begin{aligned} k\ge \tau \ge \tau _0. \end{aligned}$$

(3.31)

By (3.30) and (3.31) we have

$$\begin{aligned} I_1\le C H^2 k^3 5^kC_2^{2(k-1)}. \end{aligned}$$

(3.32)

Estimate of $I_2$.

By (3.12), (3.24) and (3.28b) we have

$$\begin{aligned} \begin{aligned} I_2&\le Cr_1^{-4}\int _{{\widetilde{B}}^{\varrho }_{2r_1}\setminus {\widetilde{B}}^{\varrho }_{r_1}}\psi ^{2-2\tau } \left| v_k\right| ^2 dxdy +C\int _{{\widetilde{B}}^{\varrho }_{3R/2}\setminus {\widetilde{B}}^{\varrho }_{R}}\psi ^{2-2\tau }\left| v_k\right| ^2 dxdy \le \\ {}&\le C\left( r_1^{-3}\psi ^{2-2\tau }_0(r_1)\sigma _k(\varepsilon , r_1)+H^2k4^k\psi ^{2-2\tau }_0(R)\right) ; \end{aligned} \end{aligned}$$

hence, by (3.29) we have

$$\begin{aligned} I_2 \le C\left( \psi ^{-1-2\tau }_0(r_1)\sigma _k(\varepsilon , r_1)+H^2k4^k\psi ^{1-2\tau }_0(R)\right) , \end{aligned}$$

(3.33)

where C depends on $\lambda$ and $\Lambda$ only.

Estimate of $I_3$.

By (3.28c) we have

$$\begin{aligned} \begin{aligned} I_3&\le Cr_1^{-2}\psi ^{2-2\tau }_0(r_1)\int _{{\widetilde{B}}^{\varrho }_{2r_1}\setminus {\widetilde{B}}^{\varrho }_{r_1}} \left| \nabla _{x,y} v_k\right| ^2 dxdy\\&+C\psi ^{2-2\tau }_0(R)\int _{{\widetilde{B}}^{\varrho }_{3R/2}\setminus {\widetilde{B}}^{\varrho }_{R}} \left| \nabla _{x,y} v_k\right| ^2 dxdy. \end{aligned} \end{aligned}$$

(3.34)

Now in order to estimate from above the right-hand side of (3.34), we use the Caccioppoli inequality, (3.11), (3.12) and (3.24) and we get

$$\begin{aligned} \begin{aligned} I_3&\le Cr_1^{-2}\psi ^{2-2\tau }_0(r_1)\left( r_1^{-2}\int _{{\widetilde{B}}^{\varrho }_{4r_1}\setminus {\widetilde{B}}^{\varrho }_{r_1/2}} \left| v_k\right| ^2 dxdy+r_1^2\int _{{\widetilde{B}}^{\varrho }_{4r_1}\setminus {\widetilde{B}}^{\varrho }_{r_1/2}} \left| F_k\right| ^2 dxdy\right) \\& \quad +C\psi ^{2-2\tau }_0(R)\left( R^{-2}\int _{{\widetilde{B}}^{\varrho }_{2R}\setminus {\widetilde{B}}^{\varrho }_{R/2}} \left| v_k\right| ^2 dxdy +R^2\int _{{\widetilde{B}}^{\varrho }_{2R}\setminus {\widetilde{B}}^{\varrho }_{R/2}} \left| F_k\right| ^2 dxdy \right) \le \\& \quad \le C \sigma _k(\varepsilon ,r_1) \psi ^{-1-2\tau }_0(r_1)+CH^25^{k}k^3\psi ^{1-2\tau }_0(R):={\widetilde{I}}_3, \end{aligned} \end{aligned}$$

(3.35)

where C depends on $\lambda$, $\Lambda$ and $\Lambda _1$ only.

Let $r_1\le \frac{R}{2}$ and let s be such that $\frac{2r_1}{\sqrt{\lambda }}\le s\le \frac{R}{\sqrt{\lambda }}$. Denote

$$\begin{aligned} {\widetilde{s}}=\sqrt{\lambda }s. \end{aligned}$$

By estimating from below trivially the left-hand side of (3.27) and taking into account (3.35), we get

$$\begin{aligned} \psi ^{-1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{{\widetilde{s}}}\setminus {\widetilde{B}}^{\varrho }_{2r_1}}\left| v_k\right| ^2+\psi ^{1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{{\widetilde{s}}}\setminus {\widetilde{B}}^{\varrho }_{2r_1}}\left| \nabla _{x,y} v_k\right| ^2 \le C\left( I_1+I_2+{\widetilde{I}}_3\right) , \end{aligned}$$

(3.36)

where C depends on $\lambda$, $\Lambda$ and $\Lambda _1$ only.

Now, by (2.15), (3.24) and into account that $\psi _0({\widetilde{s}})\ge \psi _0(r_1)$ we have

$$\begin{aligned} \psi ^{-1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{2r_1}} \left| v_k\right| ^2+\psi ^{1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{2r_1}}\left| \nabla _{x,y} v_k\right| ^2 dxdy \nonumber \\ \le C\psi ^{-1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{2r_1}} \left( \left| v_k\right| ^2+r_1^2\left| \nabla _{x,y} v_k\right| ^2 \right) dxdy\le C r_1\sigma _k(\varepsilon , r_1)\psi ^{-1-2\tau }_0(r_1). \end{aligned}$$

(3.37)

Now let us add at both the sides of (3.36) the quantity

$$\begin{aligned} \psi ^{-1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{2r_1}} \left| v_k\right| ^2+\psi ^{1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{2r_1}}\left| \nabla _{x,y} v_k\right| ^2 dxdy \end{aligned}$$

and by (3.37) we have

$$\begin{aligned} \psi ^{-1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{{\widetilde{s}}}}\left| v_k\right| ^2+\psi ^{1-2\tau }_0({\widetilde{s}})\int _{{\widetilde{B}}^{\varrho }_{{\widetilde{s}}}}\left| \nabla _{x,y} v_k\right| ^2 \le C\left( I_1+I_2+{\widetilde{I}}_3\right) , \end{aligned}$$

(3.38)

where C depends on $\lambda$, $\Lambda$ and $\Lambda _1$ only. Moreover, by (3.32), (3.33) and (3.35) we have

$$\begin{aligned} I_1+I_2+{\widetilde{I}}_3\le C \sigma _k(\varepsilon ,r_1) \psi ^{-1-2\tau }_0(r_1)+CH^2k^3 5^kC_2^{2k}\psi ^{1-2\tau }_0(R). \end{aligned}$$

(3.39)

Now by (3.29), (3.32), (3.33), (3.35) and (3.39) we have that if (3.31) is satisfied, then

$$\begin{aligned} \int _{{\widetilde{B}}_{\lambda s}} \left| v_k\right| ^2+s^2\int _{{\widetilde{B}}_{\lambda s}}\left| \nabla _{x,y} v_k\right| ^2 \le C\omega _{k,\tau }\quad , \end{aligned}$$

(3.40)

where C depends on $\lambda$, $\Lambda$ and $\Lambda _1$ only and

$$\begin{aligned} \omega _{k,\tau }(\varepsilon ,r_1)=\sigma _k(\varepsilon ,r_1)\left( \frac{\psi _0({\widetilde{s}})}{\psi _0(r_1)}\right) ^{1+2\tau } +H^2k^3 5^kC_2^{2k}\left( \frac{\psi _0({\widetilde{s}})}{\psi _0(R)}\right) ^{1+2\tau }. \end{aligned}$$

(3.41)

By a standard trace inequality, we have

$$\begin{aligned} s\int _{B_{\lambda s/2}} \left| v_k(\cdot ,0) \right| ^2 \le C\omega _{k,\tau }(\varepsilon ,r_1) \end{aligned}$$

(3.42)

and Lemma (3.2) implies

$$\begin{aligned} s\int _{B_{\lambda s/2}} \left| u(\cdot ,0) \right| ^2\le C\left( \frac{\log k}{ \sqrt{k}}+\omega _{k,\tau }(\varepsilon ,r_1)\right) , \end{aligned}$$

(3.43)

where C depend on $\lambda$, $\Lambda$ and $\Lambda _1$ only.

Now, we choose $k=\tau$ in (3.43) and using trivial inequality we have that for any $0<\alpha <\frac{1}{2}$ there exist constants $C_3>1$ and $k_0$ depending on $\lambda$, $\Lambda$, $\Lambda _1$ and $\alpha$ only such that for every $k\ge k_0$ we have

$$\begin{aligned} s\int _{B_{\lambda s/2}} \left| u(\cdot ,0) \right| ^2\le C_3 H_1^2\left[ \left( C_3s r_1^{-1}\right) ^{ 2k+1}\varepsilon _1^2+\left( C_3s\right) ^{2k+1}+k^{-\alpha }\right] , \end{aligned}$$

(3.44)

where

$$\begin{aligned} H_1:=H+e\varepsilon \quad \text{ and } \quad \varepsilon _1:=\frac{\varepsilon }{H+e\varepsilon }. \end{aligned}$$

Let us denote

$$\begin{aligned}{\overline{s}}=\frac{1}{2C_3}\end{aligned}$$

and put $s={\overline{s}}$, by (3.44) we have trivially

$$\begin{aligned}&{\overline{s}}\int _{B_{\lambda {\overline{s}}/2}} \left| u(\cdot ,0) \right| ^2\le C_3 H_1^2\left[ \left( 2r_1\right) ^{ -(2k+1)}\varepsilon _1^2+2^{-(2k+1)}+k^{-\alpha }\right] , \nonumber \\&k_{*}= \min \left\{ p\in \mathbb {Z}:p \ge \frac{\log \varepsilon _1}{2\log r_1}\right\} .\end{aligned}$$

(3.45)

If $k_{*}\ge k_0$, then we choose $k=k_{*}$ and by (3.45) we have

$$\begin{aligned} {\overline{s}}\int _{B_{\lambda {\overline{s}}/2}} \left| u(\cdot ,0) \right| ^2\le 2C_3 H_1^2\left( \varepsilon _1^{2\theta _0}+\left( \frac{2\log (1/r_1)}{\log (1/\varepsilon _1)}\right) ^{\alpha }\right) , \end{aligned}$$

(3.46)

where

$$\begin{aligned} \theta _0=\frac{\log 2}{2\log (1/r_1)}. \end{aligned}$$

(3.47)

Otherwise, if $k_{*} < k_0$, then $\frac{\log \varepsilon _1}{2\log r_1}<k_0,$ hence

$$\begin{aligned} \theta _0 \log (1/\varepsilon _1)=\frac{\log \varepsilon _1}{2\log r_1}\log 2<k_0 \log 2. \end{aligned}$$

This implies

$$\begin{aligned}2^{-2k_0}\varepsilon _1^{2\theta _0}\ge 1,\end{aligned}$$

that, in turns, taking into account (2.6), gives trivially

$$\begin{aligned} \int _{B_{\lambda {\overline{s}}/2}} \left| u(\cdot ,0) \right| ^2\le H^2 \le 4^{k_0}\varepsilon _1^{2\theta _0} H^2\le 4^{k_0} (H+e\varepsilon )^{2(1-\theta _0)}\varepsilon ^{2\theta _0}. \end{aligned}$$

(3.48)

Finally, by (3.46) and (3.48) we obtain (3.1), with $s_0=2\lambda ^{-1}{\overline{s}}$. $\Box$

Conclusion of the proof of Theorem 2.1.

Let $t_0\in (-T,T)$. It is not restrictive to assume $t_0\ge 0$. Denote

$$\begin{aligned} \rho (t_0)=\left( 1-T^{-1}t_0\right) \rho _0, \quad T(t_0)=\left( 1-T^{-1}t_0\right) T \end{aligned}$$

and

$$\begin{aligned} U(y,\eta )=u\left( \rho (t_0)y,\eta T(t_0)+t_0\right) , \quad \text{ for } (y,\eta )\in B_1\times (-1,1). \end{aligned}$$

It is easy to check that ${\widetilde{u}}$ is a solution to

$$\begin{aligned} \partial ^2_{\eta }U+{\widetilde{a}}(y)\partial ^2_{\eta }U-\mathcal {L}U=0, \quad \text{ for } (y,\eta )\in B_1\times (-1,1), \end{aligned}$$

where

$$\begin{aligned}&\mathcal {L}U=\text { div}_y\left( {\widetilde{A}}(y)\nabla _yU\right) +{\widetilde{b}}(y)\cdot \nabla _yU+{\widetilde{c}}(y)U, \\&{\widetilde{A}}(y)=\left( T\rho ^{-1}_0\right) ^2A\left( \rho (t_0)y\right) ,\quad {\widetilde{a}}(y)=(T-t_0)a\left( \rho (t_0)y\right) , \\&{\widetilde{b}}(y)=\left( T(T-t_0)\rho ^{-1}_0\right) b\left( \rho (t_0)y\right) , \quad {\widetilde{c}}(y)=\left( T-t_0\right) ^2c\left( \rho (t_0)y\right) . \end{aligned}$$

By (2.1a) and (2.1b) we have, respectively,

$$\begin{aligned}&\lambda _0\left| \xi \right| ^2\le {\widetilde{A}}(y)\xi \cdot \xi \le \lambda _0^{-1}\left| \xi \right| ^2, \quad \hbox {for every } x, \xi \in \mathbb {R}^n, \\&\left| {\widetilde{A}}(y_{*})-A(y)\right| \le \frac{\Lambda _0}{\rho _0} \left| y_{*}-y \right| , \quad \hbox {for every } y_{*}, y\in \mathbb {R}^n, \end{aligned}$$

where

$$\begin{aligned}&\lambda _0=\lambda \min \{\left( T\rho ^{-1}_0\right) ^2,\left( T\rho ^{-1}_0\right) ^{-2}\} , \text{ and } \Lambda _0=T^2\rho ^{-1}_0\Lambda . \end{aligned}$$

By (2.2) we have

$$\begin{aligned} \left| {\widetilde{a}}(y)\right| +\left| {\widetilde{b}}(y)\right| +\left| {\widetilde{c}}(y)\right| \le \Lambda _1, \quad \hbox {for almost every } y\in \mathbb {R}^n. \end{aligned}$$

In addition, by (2.5), (2.6) we have, respectively,

$$\begin{aligned} \int _{-1}^{1}\int _{B_{r_0\rho _0^{-1}}}\left| U(y,\eta )\right| ^2dyd\eta \le \varepsilon ^2\left( 1-t_0T^{-1}\right) ^{-n} \end{aligned}$$

and

$$\begin{aligned} \max _{\eta \in [-1,1]}\left( \int _{B_1}\left| U(y,\eta )\right| ^2dy+\int _{B_1}\left| \partial _{\eta }U(y,\eta )\right| ^2dy\right) \le H^2\left( 1-t_0T^{-1}\right) ^{-n}. \end{aligned}$$

Now we apply Theorem 3.1. Denoting $s=\rho \rho _0^{-1}$ we have $0<r_0\rho _0^{-1}<s\le s_0$; therefore,

$$\begin{aligned} \int _{B_s}\left| U(y,0) \right| ^2dy\le \frac{C}{\left( 1-t_0T^{-1}\right) ^{n}}\frac{\left( \log (\rho _0/r_0\right) ^{\alpha }(H+e\varepsilon )^2}{\left( \log \left( e+H\varepsilon ^{-1}\right) \right) ^{\alpha }}. \end{aligned}$$

Finally, come back to the variables x and t we get (2.7). $\square$