Leading and Second Order Homogenization of an Elastic Scattering Problem for Highly Oscillating Anisotropic Medium

Abstract

We consider the scattering of elastic waves by highly oscillating anisotropic periodic media with bounded support. Applying the two-scale homogenization, we first obtain a constant coefficient second-order partial differential elliptic equation that describes the wave propagation of the effective or overall wave field. We further pursue a higher-order homogenization with the help of complimentary boundary correctors and provide a detailed analysis on the rate of higher-order convergence. Finally we provide preliminary numerical examples to demonstrate the higher-order homogenization.

Appendix

In the end of this paper, we offer basic materials in analysing the elastic scattering in periodic media.

1.1 5.1 The Dirichlet to Neumann Map

Let $\mathbf{u}$ satisfy the Navier’s equation in the exterior domain

$$ \Delta ^{*}\mathbf{u}+\omega ^{2} \mathbf{u}=0\quad \mbox{in }\mathbb{R}^{d} \backslash \overline{\varOmega }, $$

and $\mathbf{u}$ has a decomposition that satisfies the Kupradze radiation condition. Let $B_{R}$ be a sufficiently large ball such that $\varOmega \subset B_{R}$. In the case that $\varOmega \subset \mathbb{R} ^{3}$, we introduce the polar coordinates $r$, $\theta$, $\phi$ and the unit vectors $\widehat{r}$, $\widehat{\theta }$, $\widehat{\phi }$. The $\theta$ coordinate corresponds to the angle from the $z$-axis, $\theta \in [0, \pi ]$, and the $\phi$ coordinate corresponds to the angle in the $(x, y)$-plane, $\phi \in [0, 2\pi ]$. Let $Y_{nm}$ be the spherical harmonic

$$ Y_{nm}(\theta , \phi ) = \sqrt{ \frac{(2n+1)(n-|m|)!}{4\pi (n+|m|!)} } P_{n}^{|m|}(\cos \theta ) e ^{im\phi }, \quad n \ge 0, \ |m|< n. $$

Now we let $U_{nm}$ and $V_{nm}$ be the vector spherical harmonics defined by

$$\begin{aligned} U_{nm} (\theta , \phi ) =& \frac{1}{\sqrt{\lambda _{n}}} \biggl( \frac{ \partial Y_{nm}}{\partial \theta } \widehat{\theta } + \frac{1}{ \sin \theta } \frac{\partial Y_{nm}}{\partial \phi } \widehat{\phi } \biggr), \quad n\ge 1, \\ V_{nm} (\theta , \phi ) =& \widehat{r} \times U_{nm} = \frac{1}{\sqrt{ \lambda _{n}}} \biggl( -\frac{1}{\sin \theta } \frac{\partial Y_{nm}}{ \partial \phi } \widehat{ \theta } + \frac{\partial Y_{nm}}{\partial \theta } \widehat{\phi } \biggr), \quad n\ge 1, \end{aligned}$$

where $\lambda _{n} = n(n+1)$. The vectors $Y_{nm} \widehat{r}$, $U_{nm}$, $V_{nm}$ form an orthonormal basis for $L^{2}(S)$ where $S$ denotes the unit sphere. Then $\mathbf{u}$ on $\partial B_{R}$ has the following series expansion

$$\begin{aligned} \mathbf{u} =& \sum_{n=0}^{\infty } \sum_{|m|< n} \bigl( (u|_{\partial B _{R}}, V_{nm}) V_{nm} + (u|_{\partial B_{R}}, U_{nm}) U_{nm} \\ & {} + (u|_{\partial B_{R}}, Y_{nm} \hat{r}) Y_{nm} \hat{r} \bigr), \end{aligned}$$

(91)

where $(\cdot ,\cdot )$ denotes the $L^{2}(S)$ inner product. One can correspondingly express $T_{\boldsymbol{\nu }} \mathbf{u}$ on $\partial B_{R}$ as (see [13])

$$\begin{aligned} T_{\boldsymbol{\nu }} \mathbf{u} = & \sum _{n=0}^{\infty }\sum_{|m|< n} \bigl( a_{n} (u|_{\partial B_{R}}, V_{nm}) V_{nm} + \bigl[ b_{n} (u|_{ \partial B_{R}}, U_{nm}) + c_{n} (u|_{\partial B_{R}}, Y_{nm} \widehat{r}) \bigr] U_{nm} \\ & {} + \bigl[ c_{n} (u|_{\partial B_{R}}, u_{nm}) + d_{n} (u|_{\partial B _{R}}, Y_{nm} \widehat{r}) \bigr] Y_{nm} \widehat{r} \bigr) \quad \mbox{on} \ B_{R}. \end{aligned}$$

(92)

The coefficients $a_{n}$, $b_{n}$, $c_{n}$, $d_{n}$ are given by

$$\begin{aligned} a_{n} =& \mu _{0} \biggl(\gamma _{s} - \frac{1}{R}\biggr), \\ b_{n} =& \biggl( 2 \mu _{0} \sqrt{\lambda _{n}} \biggl(\gamma _{p} - \frac{1}{R}\biggr) \biggr) \frac{B_{n}^{(1,1)}}{R} + \mu _{0} \biggl( 2 \gamma _{s} + R \omega _{s}^{2} + 2 (1-\lambda _{n}) \frac{1}{R} \biggr) \frac{B _{n}^{(2,1)}}{R}, \\ c_{n} =& \biggl( 2 \mu _{0} \sqrt{\lambda _{n}} \biggl(-\gamma _{s} + \frac{1}{R}\biggr) \biggr) \frac{B_{n}^{(2,1)}}{R} + \biggl( 2 \mu _{0} \biggl(- 2 \gamma _{p} + \frac{\lambda _{n}}{R}\biggr) \biggr) \frac{B_{n}^{(1,1)}}{R-\mu _{0} \omega _{s}^{2}}, \\ d_{n} =& - \biggl( 2 \mu _{0} \sqrt{\lambda _{n}} \biggl(-\gamma _{s} + \frac{1}{R}\biggr) \biggr) \frac{B_{n}^{(1,1)}}{R} + \biggl( 2 \mu _{0} \biggl(- 2 \gamma _{p} + \frac{\lambda _{n}}{R}\biggr) \biggr) \frac{B_{n}^{(1,2)}}{R-\mu _{0} \omega _{s}^{2}}, \end{aligned}$$

where

$$\begin{aligned} & \gamma _{s} = \omega _{s} \frac{h_{n}^{\prime }(\omega _{s} R)}{h_{n}(\omega _{s} R)}, \qquad \gamma _{p} = \omega _{p} \frac{h_{n}^{\prime }(\omega _{p} R)}{h _{n}(\omega _{p} R)}, \qquad B_{n}^{(1,1)} = -\frac{ \sqrt{\lambda _{n}} R}{R \gamma _{p} (R\gamma _{s} + 1)-\lambda _{n}}, \\ &B_{n}^{(1,2)} = \frac{ R(1+ R \gamma _{s}) }{R \gamma _{p} (R\gamma _{s} + 1)-\lambda _{n}}, \qquad B_{n}^{(1,1)} = -\frac{R^{2} \gamma _{p}}{R \gamma _{p} (R\gamma _{s} + 1)-\lambda _{n}}. \end{aligned}$$

Now for any functions $\mathbf{w}$ and $\mathbf{u}$ that satisfy the Kupradze radiation condition (4), one can directly obtain from (91) and (92) that

$$\begin{aligned} \int _{\partial B_{R}}T_{\boldsymbol{\nu }} \mathbf{u} \cdot \mathbf{w} \,dS - \int _{\partial B_{R}}T_{\boldsymbol{\nu }} \mathbf{w} \cdot \mathbf{u}\, dS =0. \end{aligned}$$

We remark that when $\varOmega \subset \mathbb{R}^{2}$, the above equality can be derived in a similar way [4].

Let $B_{R}\subset \mathbb{R}^{d}$ be a ball of radius $R>0$, then the Dirichlet to Neumann (DN) map was given by [4].

Definition 1

For any $\mathbf{g}\in (H^{1/2}(\partial B_{R}) )^{d}$, the DN map

$$\begin{aligned} \varLambda :\quad \bigl(H^{1/2}(\partial B_{R}) \bigr)^{d} \to \bigl(H^{-1/2}( \partial B_{R}) \bigr)^{d} \quad \text{with}\ \varLambda \mathbf{g}|_{ \partial B_{R}} = T_{\boldsymbol{\nu }}\mathbf{u}|_{\partial B_{R}}, \end{aligned}$$

(93)

where $\mathbf{u}\in (H^{1}_{\mathit{loc}}(\mathbb{R}^{d}\setminus \overline{B}_{R}) )^{d}$ is a solution of the Navier’s equation $\Delta ^{*}\mathbf{u}+\omega ^{2}\mathbf{u}=0$ in $\mathbb{R}^{d} \setminus \overline{B_{R}}$ and $\mathbf{u}$ satisfies the Kupradze radiation condition (4) at infinity.

Notice that the DN map $\varLambda$ is a bounded operator, so that it helps to reduce the scattering problem in unbounded domain to a bounded domain, and we refer readers to [4, Sect. 2] for detailed discussions.

1.2 5.2 Derivation of the Homogenized Equation

Consider the simplest linear elliptic system of the homogenization theory. The periodic homogenization theory was studied by [10, 14] and we refer readers to these references for the comprehensive study. We are concerned with the divergence form second order elliptic operators with rapidly oscillating periodic coefficients,

$$\mathcal{L}_{\epsilon }:=-\nabla \cdot \biggl(\mathbf{A} \biggl(\frac{x}{ \epsilon }\biggr)\nabla \biggr)=-\dfrac{\partial }{\partial x_{i}} \biggl( a _{\mathit{ijk}\ell } \biggl(\dfrac{x}{\epsilon }\biggr)\dfrac{\partial }{\partial x_{k}} \biggr), \quad \epsilon >0. $$

We assume the coefficients $\mathbf{A}(y)= (a_{\mathit{ijk}\ell }(y) )$ with $1\leq i,j,k, \ell \leq d$ for the dimension $d\geq 2$ is real, bounded and measurable such that $\mathbf{A}$ satisfies

$$ \mbox{ellipticity:} \quad \mu \sum_{i,j=1}^{d}| \varepsilon _{ij}|^{2}\leq a _{\mathit{ijk}\ell }(y)\varepsilon _{ij}\varepsilon _{k\ell }\leq \dfrac{1}{ \mu }\sum _{i,j=1}^{d}|\varepsilon _{ij}|^{2}, $$

(94)

for all symmetric matrix $(\varepsilon _{ij})_{1\leq i,j\leq d}$, and

$$ \mbox{Y-periodicity:} \quad \mathbf{A}(y+z)=\mathbf{A}(y)\quad \mbox{for all }y \in \mathbb{R}^{d},\ z \in Y:=[0,1]^{d}, $$

for some constant $\mu >0$.

Given $\mathbf{F}\in (H^{-1}(\varOmega ) )^{d}$, let $\mathbf{u} ^{\epsilon }\in (H_{0}^{1}(\varOmega ) )^{d}$ be a solution of

$$ \mathcal{L}_{\epsilon }\mathbf{u}^{\epsilon }=\mathbf{F}\quad \mbox{in } \varOmega , $$

(95)

where $\varOmega$ is a bounded Lipschitz domain in $\mathbb{R}^{d}$. By the Lax–Milgram theorem, we have

$$ \bigl\| \mathbf{u}^{\epsilon }\bigr\| _{H_{0}^{1}(\varOmega )}\leq C\|\mathbf{F}\| _{H^{-1}(\varOmega )}, $$

where the constant $C$ independent of $\epsilon$. Note that $\mathbf{u}^{\epsilon }\in (H_{0}^{1}(\varOmega ) )^{d}$ is a weak solution of (95) if for all $\boldsymbol{\varphi }\in (H _{0}^{1}(\varOmega ) )^{d}$, we have

$$\int _{\varOmega } \biggl(\mathbf{A} \biggl(\dfrac{x}{\epsilon }\biggr)\nabla \mathbf{u}^{\epsilon } \biggr):\nabla \boldsymbol{\varphi }dx= \langle \mathbf{F}, \boldsymbol{\varphi } \rangle _{H^{-1}(\varOmega )\times H_{0}^{1}( \varOmega )}. $$

Next, we want to derive the homogenized equation by using the following asymptotic analysis. We consider $\mathbf{u}^{\epsilon }$ to be the perturbation of $\mathbf{u}^{{\scriptscriptstyle (0)}}$ with respect to $\epsilon$-parameter. Moreover, by observing the elliptic operator $\mathcal{L}_{\epsilon }$, we introduce the famous two-scale homogenization method in the homogenization theory: Let us regard $x=x$, and $y=\frac{x}{\epsilon }$ as two independent parameters. Let

$$\mathbf{u}^{\epsilon }:=\mathbf{u}^{{\scriptscriptstyle (0)}}+\epsilon \mathbf{u}^{(1)}+ \epsilon ^{2}\mathbf{u}^{{\scriptscriptstyle (2)}}+\cdots $$

be the asymptotic expansion of $u_{\epsilon }$, where

$$\mathbf{u}^{{\scriptscriptstyle (j)}}:=\mathbf{u}^{{\scriptscriptstyle (j)}}(x,y)=\mathbf{u}^{{\scriptscriptstyle (j)}} \biggl(x,\dfrac{x}{ \epsilon }\biggr). $$

In addition,

$$\nabla \mathbf{u}^{{\scriptscriptstyle (j)}}=\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (j)}}(x,y)+\cfrac{1}{ \epsilon }\nabla _{y}\mathbf{u}^{{\scriptscriptstyle (j)}}(x,y),\quad \mbox{as }y=\dfrac{x}{ \epsilon }, $$

which means under our two-scaled method, the operator $\nabla =\nabla _{x}+\frac{1}{\epsilon }\nabla _{y}$. Therefore, (95) will become

$$ - \biggl(\nabla _{x}+\dfrac{1}{\epsilon }\nabla _{y} \biggr)\cdot \biggl\{ \mathbf{A}(y) \biggl[ \biggl(\nabla _{x}+\cfrac{1}{ \epsilon }\nabla _{y} \biggr) \bigl(\mathbf{u}^{{\scriptscriptstyle (0)}}+\epsilon \mathbf{u}^{{\scriptscriptstyle (1)}}+\epsilon ^{2}\mathbf{u}^{{\scriptscriptstyle (2)}}+\cdots \bigr) \biggr] \biggr\} =\mathbf{F}(x)\quad \mbox{in }\varOmega . $$

(96)

We point out that the derivation of the homogenized equation did not need to take care of the boundary condition of certain equations. Expand (96) and compare it with the same $\epsilon ^{N}$-orders (for $N=0,-1,-2$), so we get

$$\begin{aligned} O\biggl(\dfrac{1}{\epsilon ^{2}}\biggr):\quad -\nabla _{y}\cdot \bigl(\mathbf{A}(y) \nabla _{y} \mathbf{u}^{{\scriptscriptstyle (0)}}(x,y)\bigr) & =0, \\ O\biggl(\dfrac{1}{\epsilon }\biggr):\quad -\nabla _{y}\cdot \bigl( \mathbf{A}(y)\nabla _{y}\mathbf{u}^{{\scriptscriptstyle (1)}}(x,y)\bigr) & =\nabla _{y}\cdot \bigl(\mathbf{A}(y) \nabla _{x} \mathbf{u}^{{\scriptscriptstyle (1)}}\bigr)+\nabla _{x}\cdot \bigl(\mathbf{A}(y) \nabla _{y}\mathbf{u}^{{\scriptscriptstyle (0)}}\bigr), \\ O(1):\quad -\nabla _{y}\cdot \bigl(\mathbf{A}(y)\nabla _{y}\mathbf{u}^{{\scriptscriptstyle (2)}}(x,y)\bigr) & =\nabla _{y} \cdot \bigl(\mathbf{A}(y)\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (1)}}\bigr)+ \nabla _{x}\cdot \bigl(\mathbf{A}(y)\nabla _{y} \mathbf{u}^{{\scriptscriptstyle (1)}}\bigr) \\ &\quad {} +\nabla _{x}\cdot \bigl(\mathbf{A}(y)\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (0)}}\bigr)+ \mathbf{F}(x). \end{aligned}$$

(97)

Recall that for the periodic elliptic equation

$$-\nabla \cdot \bigl(\mathbf{A}(y)\nabla \mathbf{v}(y)\bigr)=\mathbf{h}(y),\quad \mbox{whenever }\mathbf{A}(y)\mbox{ is Y-periodic,} $$

then we have

$$\int _{Y}\mathbf{h}(y)dy=0, $$

by using the divergence theorem. For $O(\frac{1}{\epsilon ^{2}})$ term, this equation is solvable because the right hand side is zero. In further, we multiply $\mathbf{u}^{{\scriptscriptstyle (0)}}(x,y)$ on both sides and integrate by parts, which will imply

$$0=\int _{Y} \bigl( \mathbf{A}(y)\nabla _{y}\mathbf{u}^{{\scriptscriptstyle (0)}} \bigr): \nabla _{y}\mathbf{u}^{{\scriptscriptstyle (0)}}\geq \mu \int _{Y} \bigl|\nabla _{y}\mathbf{u} ^{{\scriptscriptstyle (0)}}(x,y)\bigr|^{2}dy\geq 0, $$

which gives us the information that

$$\mathbf{u}^{{\scriptscriptstyle (0)}}(x,y)\equiv \mathbf{u}^{{\scriptscriptstyle (0)}}(x) $$

and we know that $\mathbf{u}_{0}$ is independent of $y$.

Now, for the second term $O(\frac{1}{\epsilon })$, the second term on the right hand side should be zero since $\nabla _{y}\mathbf{u}^{{\scriptscriptstyle (0)}}(x)=0$. Solve the equation

$$\begin{aligned} -\nabla _{y}\cdot \bigl(\mathbf{A}(y)\nabla _{y} \mathbf{u}^{{\scriptscriptstyle (1)}}(x,y)\bigr) = \nabla _{y}\cdot \bigl( \mathbf{A}(y)\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (0)}}\bigr)=\bigl(\nabla _{y}\cdot \mathbf{A}(y)\bigr) \bigl(\nabla _{x} \mathbf{u}^{{\scriptscriptstyle (0)}}\bigr) \end{aligned}$$

formally. Note that since $\mathbf{A}(y)$ is $Y$-periodic, then the equation is solvable for $\mathbf{u}^{{\scriptscriptstyle (1)}}$ if

$$\int _{Y} \bigl(\nabla _{y}\cdot \mathbf{A}(y)\bigr)\cdot \bigl(\nabla _{x}\mathbf{u} ^{{\scriptscriptstyle (0)}}\bigr)dy=\int _{\partial Y} \bigl(\mathbf{A}(y)\nabla _{x}\mathbf{u} ^{{\scriptscriptstyle (0)}}\bigr)\cdot \boldsymbol{\nu }(y)dS(y)=0. $$

By using the separation of variables, we put the ansatz

$$\mathbf{u}^{{\scriptscriptstyle (1)}}(x,y)=\boldsymbol{\chi }(y)\cdot \bigl(\nabla _{x} \mathbf{u}^{{\scriptscriptstyle (0)}}(x)\bigr) $$

with $\mathbf{u}^{{\scriptscriptstyle (1)}}=(u^{{\scriptscriptstyle (1)}}_{\alpha })_{1\leq \alpha \leq d}$ such that

$$u^{{\scriptscriptstyle (1)}}_{\alpha }(x,y)=\chi _{\alpha j \beta }(y)\dfrac{\partial u^{{\scriptscriptstyle (0)}}_{\beta }}{\partial x_{j}}(x). $$

Moreover, the corrector $\chi _{\alpha j \beta }$ is $Y$-periodic and solves the cell problem

$$\textstyle\begin{cases} \dfrac{\partial }{\partial y_{i}} \biggl(a_{\mathit{ijmn}}-a_{\mathit{ijk}\ell }\dfrac{ \partial }{\partial y_{k}} \chi _{\ell m n} \biggr)=0 \quad \mbox{in } Y, \\ \displaystyle\int _{Y}\chi _{\ell mn}(y)dy=0, \end{cases} $$

and plug $\mathbf{u}^{{\scriptscriptstyle (1)}}$ to the $O(\frac{1}{\epsilon })$ equation (97) to obtain

$$-\nabla \cdot \bigl(\mathbf{A}(y)\nabla _{y}\boldsymbol{\chi }(y)\bigr)\bigl(\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (0)}}\bigr)=\bigl(\nabla _{y}\cdot \mathbf{A}(y)\bigr)\bigl(\nabla _{x} \mathbf{u}^{{\scriptscriptstyle (0)}}\bigr). $$

Finally plug $\mathbf{u}^{{\scriptscriptstyle (1)}}(x,y)=\boldsymbol{\chi }(y)\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (0)}}$ into the $O(1)$ equation and examine the solvability condition for $\mathbf{u}^{{\scriptscriptstyle (2)}}(x,y)$, we have

$$\begin{aligned} 0 & = \int _{Y} \bigl[\nabla _{y}\cdot \bigl( \mathbf{A}(y)\nabla _{x} \mathbf{u}^{{\scriptscriptstyle (1)}}\bigr)+\nabla _{x}\cdot \bigl(\mathbf{A}(y)\nabla _{y} \mathbf{u}^{{\scriptscriptstyle (1)}}\bigr)+\nabla _{x}\cdot \bigl(\mathbf{A}(y) \nabla _{x} \mathbf{u}^{{\scriptscriptstyle (0)}}\bigr)+\mathbf{F}(x) \bigr]dy \\ & =\nabla _{x}\cdot \biggl\{ \biggl[ \int _{Y}\mathbf{A}(y) \bigl(\nabla _{y} \boldsymbol{\chi }(y)\bigr)dy \biggr]\nabla _{x}\mathbf{u}^{{\scriptscriptstyle (0)}} \biggr\} + \nabla _{x}\cdot \biggl\{ \biggl[ \int _{Y}\mathbf{A}(y)dy \biggr]\nabla _{x} \mathbf{u}^{{\scriptscriptstyle (0)}} \biggr\} +\mathbf{F}(x), \end{aligned}$$

where the first term vanishes by the periodicity of $\mathbf{A}$ and $\boldsymbol{\chi }$. Thus, we can obtain that $\mathbf{u}^{{\scriptscriptstyle (0)}} \in ( H_{0}^{1}(\varOmega ) )^{d}$ is a solution of

$$ \overline{\mathcal{L}}\mathbf{u}^{{\scriptscriptstyle (0)}}:=-\nabla \cdot \bigl(\overline{ \mathbf{A}}\nabla \mathbf{u}^{{\scriptscriptstyle (0)}}\bigr)=\mathbf{F}(x)\quad \mbox{in } \varOmega , $$

(98)

where

$$\overline{\mathbf{A}}=\int _{Y} \bigl\{ \mathbf{A}(y)+\mathbf{A}(y) \bigl( \nabla _{y}\boldsymbol{\chi }(y)\bigr) \bigr\} dy, $$

where $\overline{\mathbf{A}}$ is the (constant) homogenized operator and we call (98) to be the homogenized equation. In addition, $\overline{\mathbf{A}}=(\overline{a}_{\mathit{ijk}\ell })_{1\leq i,j,k, \ell \leq d}$ and

$$ \overline{a}_{\mathit{ijk}\ell }= \int _{Y} \biggl(a_{\mathit{ijk}\ell }-a_{\mathit{ijmn}} \dfrac{ \partial }{\partial y_{m}} \chi _{nk\ell } \biggr)dy. $$

(99)

For the rigorous derivation of the homogenized equation, we need to use a famous result, which is called the Div-Curl lemma. We skip the rigorous analysis here and refer readers to the lecture note [26] for more details.

Note that $\overline{\mathcal{L}}:=-\nabla \cdot (\overline{ \mathbf{A}}\nabla )$ is the homogenized second order elliptic operator with respect to $\mathbf{A}$ and we want to prove $\overline{ \mathcal{L}}$ is an elliptic operator with constant coefficients.

Theorem 5

The homogenized operator $\overline{\mathcal{L}}$ satisfies that

1. $\overline{\mathcal{L}}$is an elliptic operator, which means

$$ \mu _{1}\sum_{i,j=1}^{d}| \varepsilon _{ij}|^{2}\leq \overline{a}_{\mathit{ijk} \ell }(y) \varepsilon _{ij}\varepsilon _{k\ell }\leq \dfrac{1}{\mu _{1}} \sum _{i,j=1}^{d}|\varepsilon _{ij}|^{2}, $$

(100)

for some constant$\mu _{1}>0$.

2. The effective coefficient $\overline{a}_{\mathit{ijk}\ell }$ is major and minor symmetric provided $a_{\alpha \beta \gamma \delta }$ is major and minor symmetric.

Proof

It is easy to see that $|\overline{a}_{\mathit{ijk}\ell }|\leq C$ by using (99) and the ellipticity of $A(y)$, for some constant $C>0$. It remains to show $\overline{a}_{\mathit{ijk}\ell } \varepsilon _{ij}\varepsilon _{k\ell }\geq \mu _{1}\sum_{i,j=1}^{d}| \varepsilon _{ij}|^{2}$ for some constant $\mu _{1}>0$. We can rewrite (99) as

$$\overline{a}_{\mathit{ijk}\ell }=\int _{Y}\dfrac{\partial }{\partial y_{\alpha }} \{ \delta _{\beta j}y_{i}+\chi _{\beta ij} \} \cdot a_{ \alpha \beta \gamma \delta }\cdot \dfrac{\partial }{\partial y_{ \gamma }} \{ \delta _{\delta \ell }y_{k}+\chi _{\delta k \ell } \} dy, $$

where $\delta _{s\alpha }$ is the standard Kronecker delta (i.e., $\delta _{s\alpha }=1$ if $s=\alpha$, and $\delta _{s\alpha }=0$ otherwise). Hence, for $\varepsilon =(\varepsilon _{ij})\in \mathbb{R} ^{d\times d}$, we have

$$\begin{aligned} \overline{a}_{\mathit{ijk}\ell }\varepsilon _{ij}\varepsilon _{k\ell } &= \int _{Y}\dfrac{ \partial }{\partial y_{\alpha }} \{ \delta _{\beta j}y_{i} \varepsilon _{ij}+\chi _{\beta ij}\varepsilon _{ij} \} \cdot a_{ \alpha \beta \gamma \delta }\cdot \dfrac{\partial }{\partial y_{ \gamma }} \{ \delta _{\delta \ell }y_{k} \varepsilon _{k\ell }+ \chi _{\delta k \ell }\varepsilon _{k\ell } \} dy \\ & \geq \mu \sum_{\beta =1}^{d} \int _{Y}|\nabla (y_{i}\varepsilon _{i \beta }+\chi _{\beta ij}\varepsilon _{ij})|^{2}dy \geq 0. \end{aligned}$$

If $\overline{a}_{\mathit{ijk}\ell }\varepsilon _{ij}\varepsilon _{k\ell }=0$ for some $\varepsilon =(\varepsilon _{ij})\in \mathbb{R}^{d\times d}$, then $y_{i}\varepsilon _{i\beta }+\chi _{\beta ij}$ must be a constant. Recall that $\chi _{\beta ij}(y)$ is $Y$-periodic, so this implies that $\varepsilon =0$. This means that there exists $\mu _{1}>0$ such that (100) holds. □

1.3 5.3 Tools and Estimates

In the last part, for the completeness of this paper, we provide some elliptic estimate where we have utilized in previous sections. The following theorem was proved in [6, Theorem 5.7] for the scalar case. It will hold for the vector case. For completeness, we provide the theorem and its proof as follows.

Theorem 6

Trace Theorem

Let$\mathbf{A}=(a_{\mathit{ijk}\ell })_{1\leq i,j,k,\ell \leq d}$be a four tensor satisfying the ellipticity condition (100) and$\varOmega \subset \mathbb{R}^{d}$be a bounded domain with a$C^{\infty }$-smooth boundary, for$d\geq 2$. The (conormal) mapping$\mathit{Tr}:\mathbf{u}\to \frac{\partial \mathbf{u}}{ \partial \boldsymbol{\nu }_{\mathbf{A}}}:=(\mathbf{A}\nabla \mathbf{u})\cdot \boldsymbol{\nu }$defined in$C^{\infty }(\overline{ \varOmega })$can be continuously extended to a linearly continuous mapping (still denote by$\mathit{Tr}$) from$H^{1}(\varOmega ,\mathbf{A})$to$H^{-1/2}( \partial \varOmega )$, where$H^{1}(\varOmega ,\mathbf{A})$is the space equipped with the graph norm

$$ \|\mathbf{u} \|_{H^{1}(\varOmega ,\mathbf{A})}^{2}:=\|\mathbf{u}\|_{H ^{1}(\varOmega )}^{2} + \bigl\| \nabla \cdot (\mathbf{A}\nabla \mathbf{u})\bigr\| _{L^{2}(\varOmega )}^{2}. $$

Proof

Let $\boldsymbol{\varphi }\in (C^{\infty }(\overline{\varOmega }) )^{d}$ be a test function and $\mathbf{u}\in C^{\infty }(\overline{ \varOmega };\mathbb{R}^{d})$. The integration by parts formula gives

$$\begin{aligned} \int _{\partial \varOmega }(\mathbf{A}\nabla \mathbf{u}\cdot \boldsymbol{\nu }) \cdot \boldsymbol{\varphi }\, dS= \int _{\varOmega }( \mathbf{A}\nabla \mathbf{u}):\nabla \boldsymbol{ \varphi }dx+ \int _{ \varOmega }\nabla \cdot (\mathbf{A}\nabla \mathbf{u})\cdot \boldsymbol{\varphi }\,dx. \end{aligned}$$

By the standard density arguments, the above equation holds for $\boldsymbol{\varphi }\in (H^{1}(\varOmega ) )^{d}$ so that

$$\begin{aligned} \biggl\vert \int _{\partial \varOmega }(\mathbf{A}\nabla \mathbf{u}\cdot \boldsymbol{\nu }) \cdot \boldsymbol{\varphi }dS \biggr\vert \leq C \| \mathbf{u} \|_{H^{1}(\varOmega ,\mathbf{A})}\|\boldsymbol{\varphi }\|_{H ^{1}(\varOmega )}, \end{aligned}$$

(101)

for any $\boldsymbol{\varphi }\in (H^{1}(\varOmega ) )^{d}$, $\mathbf{u}\in (C^{\infty }(\overline{\varOmega }) )^{d}$, where constant $C>0$ is a constant independent of $\boldsymbol{\varphi }$ and $\mathbf{u}$. Let $\mathbf{g}\in (H^{1/2}(\partial D) )^{d}$, by using the trace theorem, then there exists a function $\boldsymbol{\varphi }\in (H^{1}(\varOmega ) )^{d}$ such that $\gamma _{\partial \varOmega }\boldsymbol{\varphi }=\mathbf{f}$, where $\gamma _{\partial \varOmega }$ stands for the trace operator. Continuing the inequality (101) and the trace theorem,

$$\begin{aligned} \biggl\vert \int _{\partial \varOmega }(\mathbf{A}\nabla u\cdot \boldsymbol{\nu }) \cdot \mathbf{f}\, dS \biggr\vert \leq C \|\mathbf{u}\| _{H^{1}(\varOmega ,\mathbf{A})}\|\mathbf{f} \|_{H^{1/2}(\partial \varOmega )}, \end{aligned}$$

for any $\mathbf{f} \in (H^{1/2}(\partial \mathbf{)} )^{d}$, $\mathbf{u}\in (C^{\infty }(\overline{\varOmega }) )^{d}$.

Hence, the mapping

$$ \mathbf{f} \to \int _{\partial \varOmega }(\mathbf{A}\nabla \mathbf{u} \cdot \boldsymbol{\nu })\cdot \mathbf{f} dS, \quad \text{for any } \mathbf{f}\in \bigl(H^{1/2}( \partial \varOmega ) \bigr)^{d} $$

defines a continuous linear operator and from the duality argument,

$$\begin{aligned} \bigl\| (\mathbf{A}\nabla )\cdot \boldsymbol{\nu }\bigr\| _{H^{-1/2}(\partial \varOmega )}\leq C\| \mathbf{u} \|_{H^{1}(\varOmega ,\mathbf{A})}. \end{aligned}$$

Therefore, the linear mapping $\mathit{Tr}: \mathbf{u} \to (\mathbf{A}\nabla u) \cdot \boldsymbol{\nu }$ defined on $(C^{\infty }(\overline{ \varOmega }) )^{d}$ is continuous under the norm $H^{1}(\varOmega , \mathbf{A})$. Thus, the assertion follows from the density arguments. □

Let $\mathbf{C}=(C_{\mathit{ijk}\ell })$ be an anisotropic elastic four tensor and $\mathbf{C}_{0}$ be a constant isotropic elastic tensor defined by (2), which satisfy all the conditions given in Sect. 1. Next, we provide the stability estimate for the following transmission problem. The scalar case was demonstrated in [6, Sect. 5] and here we generalize the result to a system version.

Theorem 7

Let$\varOmega \subset \mathbb{R}^{d}$be a bounded$C^{\infty }$-smooth domain. Given$\mathbf{f}\in (H^{1/2}(\partial \varOmega ) )^{d}$and$\mathbf{g}\in (H^{-1/2}(\partial \varOmega ) )^{d}$. Let$\mathbf{u}\in (H^{1}(\varOmega ) )^{d}$and$\mathbf{v}\in (H^{1}_{\mathit{loc}}(\mathbb{R}^{d}\setminus \overline{\varOmega }) )^{d}$be the solutions of the following transmission problem

$$\begin{aligned} \textstyle\begin{cases} \nabla \cdot (\mathbf{C}\nabla \mathbf{u} )+\omega ^{2} \rho \mathbf{u}=0 &\textit{in}\ \varOmega , \\ \Delta ^{*}\mathbf{v}+\omega ^{2} \mathbf{v}=0 &\textit{in}\ \mathbb{R} ^{d} \setminus \overline{\varOmega }, \\ \mathbf{u}-\mathbf{v}=\mathbf{f} &\textit{on}\ \partial \varOmega , \\ T_{\boldsymbol{\nu }} \mathbf{u}-(\mathbf{C}\nabla \mathbf{u})\cdot \boldsymbol{\nu }=\mathbf{g} &\textit{on}\ \partial \varOmega , \end{cases}\displaystyle \end{aligned}$$

(102)

where$T_{\boldsymbol{\nu }}$is the boundary traction operator given by (3), $\omega \in \mathbb{R}$is not an eigenvalue of the transmission problem (102) and$v$satisfies the Kupradze radiation condition (4). Then for any ball$B_{R}$with$\varOmega \subset B_{R}$, there exists a constant$C_{R}>0$such that

$$\begin{aligned} \|\mathbf{u}\|_{H^{1}(\varOmega )}+\|\mathbf{v}\|_{H^{1}(B_{R} \setminus \overline{\varOmega })} \leq C_{R} \bigl\{ \|\mathbf{f}\|_{H^{1/2}(\partial \varOmega )}+\|\mathbf{g} \|_{H^{-1/2}(\partial \varOmega )} \bigr\} . \end{aligned}$$

(103)

Proof

Firstly, by using similar arguments in [4, Sect. 2] and [6, Sect. 5], the elastic scattering problem (102) is equivalent to the following transmission problem: Let $\mathbf{u}\in (H^{1}(\varOmega ) )^{d}$ and $\mathbf{v}\in (H^{1}(B_{R}\setminus \overline{ \varOmega }) )^{d}$ be the solutions of

$$\begin{aligned} \textstyle\begin{cases} \nabla \cdot (\mathbf{C}\nabla \mathbf{u} )+\omega ^{2} \rho \mathbf{u}=0 &\text{in }\varOmega , \\ \Delta ^{*}\mathbf{v}+\omega ^{2} \mathbf{v}=0 &\text{in } B_{R} \setminus \overline{\varOmega }, \\ \mathbf{u}-\mathbf{v}=\mathbf{f} &\text{on }\partial \varOmega , \\ T_{\boldsymbol{\nu }} \mathbf{u}-(\mathbf{C}\nabla \mathbf{u})\cdot \boldsymbol{\nu }=\mathbf{g} &\text{on }\partial \varOmega , \\ T_{\boldsymbol{\nu }}\mathbf{v} = \varLambda \mathbf{v} &\text{on } \partial B_{R}, \end{cases}\displaystyle \end{aligned}$$

(104)

where $\varLambda$ is the DN map defined by (93) on $\partial B_{R}$. Furthermore, by using [4, Lemma 2.8], the DN map $\varLambda$ is a bounded operator and $\varLambda$ can decomposed into $\varLambda =\varLambda _{1}+ \varLambda _{2}$, where $-\varLambda _{1}$ is a positive operator and $\varLambda _{2}$ is a compact operator from $(H^{1/2}(\partial B_{R}) )^{d}$ to $(H^{-1/2}(\partial B_{R}) )^{d}$.

Next, let $\mathbf{v}_{\mathbf{f}}\in (H^{1}(B_{R}\setminus \overline{ \varOmega }) )^{d}$ be the unique solution of the Navier’s equation in the exterior domain

$$ \textstyle\begin{cases} \Delta ^{*}\mathbf{v}_{\mathbf{f}}+\omega ^{2}\mathbf{v}_{\mathbf{f}}=0 & \text{in }B_{R} \setminus \overline{\varOmega }, \\ \mathbf{v}_{\mathbf{f}}=\mathbf{f} &\text{on }\partial \varOmega , \\ \mathbf{v}_{\mathbf{f}} =0 &\text{on }\partial B_{R}. \end{cases} $$

By straight forward calculation, it is not hard to see that the variational formula of (104) can be written as follows: Find a function $w\in H^{1}(B_{R})$ such that

$$\begin{aligned} &\int _{\varOmega } \bigl((\mathbf{C}\nabla \mathbf{w}):\nabla \boldsymbol{\phi }-\omega ^{2}\rho \mathbf{w}\cdot \boldsymbol{\phi } \bigr) \,dx+ \int _{B_{R}\setminus \overline{\varOmega }} \bigl((\mathbf{C}_{0} \nabla \mathbf{w}):\nabla \boldsymbol{\phi }-\omega ^{2} \mathbf{w} \cdot \boldsymbol{\phi } \bigr)\,dx \\ &\qquad {} - \int _{\partial B_{R}}\boldsymbol{\phi }\cdot T_{\boldsymbol{\nu }} \mathbf{w} \, dS+ \int _{\partial B_{R}} \boldsymbol{\phi }\cdot T_{ \boldsymbol{\nu }} \mathbf{v}_{\mathbf{f}}\,dS \\ &\quad = \int _{\partial \varOmega }\mathbf{g}\cdot \boldsymbol{\phi }\, dS+ \int _{B_{R}\setminus \overline{\varOmega }} \bigl((\mathbf{C}_{0}\nabla \mathbf{v}_{\mathbf{f}}):\nabla \boldsymbol{\phi }-\omega ^{2} \mathbf{v}_{\mathbf{f}}\cdot \boldsymbol{\phi } \bigr)\,dx, \end{aligned}$$

(105)

for any test function $\boldsymbol{\phi }\in (H^{1}(B_{R}) )^{d}$, where $\mathbf{C} _{0}$ is a constant elastic tensor defined by (2). By using the integration by parts, one can easily see that $\mathbf{u}=\mathbf{w}|_{\varOmega }$ and $\mathbf{v}=\mathbf{w}|_{B_{R}\setminus \overline{\varOmega }}-\mathbf{v} _{\mathbf{f}}$ satisfy (104).

Now, let us consider two bilinear forms

$$\begin{aligned} b_{1}(\boldsymbol{\psi },\boldsymbol{\phi }):= & \int _{\varOmega } \bigl((\mathbf{C}\nabla \boldsymbol{\psi }):\nabla \boldsymbol{\phi }+ \boldsymbol{\psi }\cdot \boldsymbol{\phi } \bigr) \,dx+ \int _{B_{R}\setminus \overline{\varOmega }} \bigl((\boldsymbol{C} _{0}\nabla \mathbf{w}):\nabla \boldsymbol{\phi }+\mathbf{w}\cdot \boldsymbol{\phi } \bigr) \,dx \\ &- \int _{\partial B_{R}}\boldsymbol{\phi }\cdot (\varLambda _{1} \boldsymbol{\psi })\,dS, \\ b_{2}(\boldsymbol{\psi }, \boldsymbol{\phi }):= &- \int _{\varOmega } \bigl(\omega ^{2}\rho +1 \bigr) \boldsymbol{\phi }\cdot \boldsymbol{\psi }dx- \int _{B_{R}\setminus \overline{\varOmega }}\bigl(\omega ^{2}+1\bigr) \boldsymbol{\phi }\cdot \boldsymbol{\psi }dx \\ &- \int _{\partial B_{R}}\boldsymbol{\phi }\cdot (\varLambda _{2} \boldsymbol{\psi })\, dS, \quad \text{ for all }\boldsymbol{\phi }, \boldsymbol{ \psi }\in \bigl(H^{1}(B_{R}) \bigr)^{d}, \end{aligned}$$

and

$$\begin{aligned} F(\boldsymbol{\phi }):= \int _{\partial \varOmega }\mathbf{g}\cdot \boldsymbol{\phi }\, dS- \int _{\partial B_{R}}\boldsymbol{\phi }\cdot T_{\boldsymbol{\nu }} \mathbf{v}_{\mathbf{f}}\,dS+ \int _{B_{R}\setminus \overline{\varOmega }} \bigl((\mathbf{C}_{0}\nabla \mathbf{v}_{\mathbf{f}}):\nabla \boldsymbol{\phi }-\omega ^{2} \mathbf{v}_{\mathbf{f}}\cdot \boldsymbol{\phi } \bigr)\,dx. \end{aligned}$$

Then we can rewrite the problem (105) as finding a function $\mathbf{w}\in (H^{1}(B_{R}) )^{d}$ such that

$$\begin{aligned} b_{1}(\mathbf{w},\boldsymbol{\phi })+b_{2}(\mathbf{w}, \boldsymbol{\phi })=F(\boldsymbol{\phi }),\quad \text{for any } \boldsymbol{\phi }\in \bigl(H^{1}(B_{R}) \bigr)^{d}. \end{aligned}$$

Since $-\varLambda _{1}$ is a positive operator, one can conclude that $b_{1}(\cdot ,\cdot )$ is strictly coercive. Therefore, from the Lax–Milgram theorem, one can see that the operator $A: (H^{1}(B _{R}) )^{d} \to (H^{1}(B_{R}) )^{d}$ defined by $b_{1}( \mathbf{w},\boldsymbol{\phi })=(A\mathbf{w},\boldsymbol{\phi })_{H ^{1}(B_{R})}$ is invertible and has a bounded inverse. On the other hand, since $\varLambda _{2}$ is a compact operator from $(H^{1/2}( \partial B_{R}) )^{d}\to (H^{-1/2}(\partial B_{R}) )^{d}$ and $(H^{1}(B_{R}) )^{d} \to (L^{2}(B_{R}) )^{d}$ is a compact embedding, then it is not hard to see that the operator $B: (H^{1}(B_{R}) )^{d} \to (H^{1}(B_{R}) )^{d}$ defined by $b_{2}(\mathbf{w},\boldsymbol{\phi })=(B\mathbf{w}, \boldsymbol{\phi })_{H^{1}(B_{R})}$ is compact. Hence, by using [6, Theorem 5.16], one can derive that the existence of the transmission problem (104) from the uniqueness of (104) and the stability estimate (103) holds automatically. □