Uniform convergence for linear elastostatic systems with periodic high contrast inclusions

Abstract

We consider the Lamé system of linear elasticity with periodically distributed inclusions whose elastic parameters have high contrast compared to the background media. We develop a unified method based on layer potential techniques to quantify three convergence results when some parameters of the elastic inclusions are sent to extreme values. More precisely, we study the incompressible inclusions limit where the bulk modulus of the inclusions tends to infinity, the soft inclusions limit where both the bulk modulus and the shear modulus tend to zero, and the hard inclusions limit where the shear modulus tends to infinity. Our method yields convergence rates that are independent of the periodicity of the inclusions array, and are sharper than some earlier results of this type. A key ingredient of the proof is the establishment of uniform spectra gaps for the elastic Neumann-Poincaré operator associated to the collection of periodic inclusions that are independent of the periodicity.

Some useful facts and technical tools

In this appendix we record some important facts that are used frequently in this paper.

1.1 The second Korn’s inequality

Lemma A.1

(The second Korn’s inequality) Let $\Omega$ be a bounded Lipschitz domain in ${\mathbb {R}}^d$, and let V be a closed subspace of vector valued functions in $H^1(\Omega )$ such that $V\cap {\textbf{R}} =\{0\}$, where ${\textbf{R}}$ is the rigid displacements space. Then every ${\textbf{v}}\in V$ satisfies

$$\begin{aligned} \Vert {\textbf{v}}\Vert _{H^1(\Omega )}\le C\Vert \mathbb {D}({\textbf{v}})\Vert _{L^2(\Omega )}, \end{aligned}$$

where the constant C only depends on $\Omega$.

For the proof we refer to [31].

1.2 A coercive lemma

The following version of Lax-Milgram theorem allowed us to prove that certain bounded linear operators are invertible by showing they are coercive.

Lemma A.2

Let H be a Hilbert space, and let $A \in {\mathcal {L}}(H)$ be a bounded linear operator. Suppose that there exists $\gamma >0$ such that

$$\begin{aligned} (Ax,x)_{H} \ge \gamma \Vert x\Vert _{H}^2 \quad \mathrm {for \ any\ }x\in H. \end{aligned}$$

(A.1)

Then A has a bounded inverse, and $\Vert A^{-1}\Vert _{{\mathcal {L}}(H)}\le \gamma ^{-1}$.

We then have the following result as a direct consequence.

Lemma A.3

Let H be a Hilbert space and $H^*$ its dual space. Suppose $A: H \rightarrow H^*$ is a bounded linear operator, and, moreover, there exists $\gamma >0$ such that

$$\begin{aligned} \langle x,Ax \rangle _{H,H^*} \ge \gamma \Vert x\Vert _{H}^2 \quad \mathrm {for \ any\ }x\in H. \end{aligned}$$

(A.2)

Then A has a bounded inverse, and $\Vert A^{-1}\Vert _{{\mathcal {L}}(H^*,H)}\le \gamma ^{-1}$.

Proof

Let $\iota : H^* \rightarrow H$ be the canonical dual isometry, i.e. for any $\phi \in H^*$ and for any $y\in H$,

$$\begin{aligned} (\iota (\phi ), y)_{H}=\langle y,\phi \rangle _{H,H^*}. \end{aligned}$$

It follows that $\iota A$ satisfies the conditions of Lemma A.2; indeed,

$$\begin{aligned} (\iota (Ax),x)_{H} =\langle x,Ax\rangle _{H,H^*} \ge \gamma \Vert x\Vert _{H}^2. \end{aligned}$$

It follows that $\iota A$ has an inverse with the desired bound. Since $\iota$ is an isometry, the desired result for A also follows. $\square$

1.3 Proof of Lemma 4.1

The uniqueness of ${\textbf{u}}$ is clear, so it suffices to construct a solution. Let ${\textbf{v}}\in H^1(\Omega )$ be the unique solution of

$$\begin{aligned} {\mathcal {L}}_{\lambda ,\mu }{\textbf{v}}=0 \ \ \textrm{in}\ \Omega _{\varepsilon }, \quad {\textbf{v}}|_{D_{\varepsilon }}=0, \quad \left. \frac{\partial {\textbf{v}}}{\partial \nu _{(\lambda ,\mu )}}\right| _{\partial \Omega }={\textbf{g}}. \end{aligned}$$

(A.3)

For each ${\textbf{m}}\in \Pi _{\varepsilon }$ and $l\in \{1,\ldots ,\frac{d(d+1)}{2}\}$, let ${\textbf{v}}_{{\textbf{m}}}^l\in H^1(\Omega )$ be the solution of

$$\begin{aligned} {\mathcal {L}}_{\lambda ,\mu } {\textbf{v}}_{{\textbf{m}}}^l=0 \ \ \textrm{in}\ \Omega _{\varepsilon }, \quad {\textbf{v}}_{{\textbf{m}}}^l|_{\omega _{{\textbf{n}}}^{\varepsilon }}=\delta _{{\textbf{m}}{\textbf{n}}}{\textbf{r}}_l\quad \mathrm {for\ all}\ {\textbf{n}}\in \Pi _{\varepsilon },\quad \left. \frac{\partial {\textbf{v}}_{{\textbf{m}}}^l}{\partial \nu _{(\lambda ,\mu )}}\right| _{\partial \Omega }=0. \end{aligned}$$

(A.4)

The existence and uniqueness, both for ${\textbf{v}}$ and ${\textbf{v}}_{{\textbf{m}}}^l$, follow from a standard practice of weak formulation and an application of Lax-Milgram theorem. Moreover, the functions $\{{\textbf{v}}^l_{{\textbf{m}}}\}$ are independent as elements of ${\mathcal {E}}$.

For the solution to (4.5), we consider a function ${\textbf{u}}$ of the form

$$\begin{aligned} {\textbf{u}}:={\textbf{v}}+\sum _{{\textbf{m}}\in \Pi _{\varepsilon }}\sum _{l=1}^{\frac{d(d+1)}{2}} a_{{\textbf{m}}}^l {\textbf{v}}_{{\textbf{m}}}^l, \end{aligned}$$

(A.5)

where $\{a_{{\textbf{m}}}^l\}\subset {\mathbb {R}}$ are constants to be chosen. Clearly, ${\textbf{u}} \in H^1(\Omega )$ already satisfies ${\textbf{u}} \in {\textbf{R}}$ in each component of $D_\varepsilon$, and

$$\begin{aligned} {\mathcal {L}}_{\lambda ,\mu }{\textbf{u}} =0 \ \ \textrm{in} \ \Omega _{\varepsilon },\quad \left. \frac{\partial {\textbf{u}}}{\partial \nu _{(\lambda ,\mu )}}\right| _{\partial \Omega }={\textbf{g}}. \end{aligned}$$

We choose $a_{{\textbf{m}}}^l$’s so that ${\textbf{u}}$ also satisfies the remaining equation in (4.5), namely,

$$\begin{aligned} \sum _{{\textbf{m}}\in \Pi _{\varepsilon }}\sum _{l=1}^{\frac{d(d+1)}{2}} a_{{\textbf{m}}}^l \int _{\partial \omega ^{{\textbf{n}}}_{\varepsilon }} \left. \frac{\partial {\textbf{v}}_{{\textbf{m}}}^l}{\partial \nu _{(\lambda ,\mu )}} \right| _+\cdot {\textbf{r}}_j = b^l_{{\textbf{n}}}:= c_{{\textbf{n}}}^j -\int _{\partial \omega ^{{\textbf{n}}}_{\varepsilon }} \left. \frac{\partial {\textbf{v}}}{\partial \nu _{(\lambda ,\mu )}} \right| _+\cdot {\textbf{r}}_j. \end{aligned}$$

(A.6)

The equation above can be viewed as a linear system of the form $A {\textbf{a}} = {\textbf{b}}$ where the unknown is ${\textbf{a}} = (a^l_{{\textbf{m}}}) \in \mathbb {R}^{|\Pi _\varepsilon |d(d+1)/2}$, the right hand side vector ${\textbf{b}} = (b^j_{{\textbf{n}}})$ is defined above and the coefficient matrix $A = (A^{jl}_{{\textbf{n}}{\textbf{m}}})$ is defined by

$$\begin{aligned} A^{jl}_{{\textbf{n}}{\textbf{m}}}= \int _{\partial \omega ^{{\textbf{n}}}_{\varepsilon }} \Big (\left. \frac{\partial {\textbf{v}}_{{\textbf{m}}}^l}{\partial \nu _{(\lambda ,\mu )}} \right| _+\cdot {\textbf{r}}_j \Big ). \end{aligned}$$

(A.7)

We need to show that this linear system has a solution.

Let ${\mathcal {X}} \subset {\mathbb {R}}^{ |\Pi _{\varepsilon }| d(d+1)/2}$ be the subspace defined by

$$\begin{aligned} {\mathcal {X}}:=\left\{ (x_{{\textbf{n}}}^j ) \in {\mathbb {R}}^{|\Pi _{\varepsilon }|\times \frac{d(d+1)}{2}}:\sum _{{\textbf{n}}\in \Pi _{\varepsilon }} x_{{\textbf{n}}}^j =0\quad \mathrm {for \ all\ }1\le j \le \frac{d(d+1)}{2} \right\} , \end{aligned}$$

(A.8)

Clearly, $\textrm{dim}\,{\mathcal {X}}=\frac{d(d+1)}{2}|\Pi _{\varepsilon }|-\frac{d(d+1)}{2}$. By the definition of ${\textbf{v}}$ and the condition (4.4), we see ${\textbf{b}} \in {\mathcal {X}}$. It suffices to show that the range of A contains (actually, is) ${\mathcal {X}}$. First, we can check directly that the range of A is contained in ${\mathcal {X}}$. Indeed, for any ${\textbf{p}} = (p^l_{{\textbf{m}}})$, we compute and get

$$\begin{aligned} \sum _{{\textbf{n}}\in \Pi _\varepsilon } (A{\textbf{p}})^j_{{\textbf{n}}} = \sum _{{\textbf{n}} \in \Pi _\varepsilon } \int _{\partial \omega ^{{\textbf{n}}}_\varepsilon } \Big (\left. \frac{\partial (p^l_{{\textbf{m}}} {\textbf{v}}_{{\textbf{m}}}^l)}{\partial \nu _{(\lambda ,\mu )}} \right| _+\cdot {\textbf{r}}_j \Big ) = -J^{\Omega _\varepsilon }(\mathbf {{\textbf{w}}},r_j) = 0, \end{aligned}$$

where we defined ${\textbf{w}}:= p^l_{{\textbf{m}}} {\textbf{v}}^l_{{\textbf{m}}}$ and the summation convention is used. We also used the fact that ${\textbf{w}}$ solves the homogeneous Lamé system in $\Omega _\varepsilon$ with $\partial {\textbf{w}}/\partial \nu |_{\partial \Omega } = 0$ and ${\textbf{w}}|_{\omega ^{{\textbf{n}}}} = p^l_{{\textbf{n}}} {\textbf{r}}_l$ (in particular, ${\textbf{w}} \in {\textbf{R}}$ in each component of $D_\varepsilon$).

For our purpose, it remains to show that the kernel of A has dimension $d(d+1)/2$. Suppose ${\textbf{p}} = (p^l_{{\textbf{m}}})$ satisfies $A{\textbf{p}} = 0$. Then the function ${\textbf{w}} = p^l_{{\textbf{m}}}{\textbf{v}}^l_{{\textbf{m}}}$ has the property discussed above and further satisfies $\int _{\partial \omega ^{{\textbf{n}}}_\varepsilon } \partial {\textbf{w}}/\partial \nu |_+ \cdot {\textbf{r}}_j = 0$ for all j and ${\textbf{n}}$. From this we get $J^{\Omega _\varepsilon }(\omega ) = 0$. Hence ${\textbf{w}} \in {\textbf{R}}$ in $\Omega _\varepsilon$ and then ${\textbf{w}} \in {\textbf{R}}$ in $\Omega$. To summarize, we proved that $A{\textbf{p}} = 0$ implies ${\textbf{w}}({\textbf{p}}):= p^l_{{\textbf{m}}}{\textbf{v}}^l_{{\textbf{m}}} \in {\textbf{R}}$. It is easier to show that the reverse implication also holds, and that ${\textbf{p}} \rightarrow {\textbf{w}}({\textbf{p}})$ is an isomorphsim from $\mathbb {R}^{|\Pi _\varepsilon |d(d+1)/2}$ to $\textrm{span}({\textbf{v}}^l_{{\textbf{m}}})$. Moreover ${\textbf{R}}\subset \textrm{span}\{{\textbf{v}}^l_{{\textbf{m}}}\}$ is a $d(d+1)/2$ dimensional subspace. The proof is hence complete.

1.4 Proof of Lemma 4.2

Again, the uniqueness is clear and we only need to construct a solution. The proof is very similar to the proof of Lemma 4.1 presented in the previous section, so we omit some details.

Let ${\textbf{v}}\in H^1(\Omega )$ be the unique solution of

$$\begin{aligned} {\mathcal {L}}_{\lambda ,\mu }{\textbf{v}}=0 \ \ \textrm{in}\ \Omega _{\varepsilon }, \quad {\textbf{v}}|_{D_{\varepsilon }}=0, \quad {\textbf{v}}|_{\partial \Omega }={\textbf{f}}. \end{aligned}$$

For each ${\textbf{m}} \in \Pi _\varepsilon$ and $l = 1,2,\dots ,d(d+1)/2$, let ${\textbf{v}}_{{\textbf{m}}}^l\in H^1(\Omega )$ be the unique solution of

$$\begin{aligned} {\mathcal {L}}_{\lambda ,\mu } {\textbf{v}}_{{\textbf{m}}}^l=0 \ \ \textrm{in}\ \Omega _{\varepsilon }, \quad {\textbf{v}}_{{\textbf{m}}}^l|_{\omega _{{\textbf{n}}}^{\varepsilon }}=\delta _{{\textbf{m}}{\textbf{n}}}{\textbf{r}}_l \quad \mathrm {for\ all}\ {\textbf{n}}\in \Pi _{\varepsilon },\quad {\textbf{v}}_{{\textbf{m}}}^l|_{\partial \Omega }=0. \end{aligned}$$

We seek a solution to (4.6) of the form

$$\begin{aligned} {\textbf{u}}:={\textbf{v}}+\sum _{{\textbf{m}}\in \Pi _{\varepsilon }}\sum _{l=1}^{\frac{d(d+1)}{2}} a_{{\textbf{m}}}^l {\textbf{v}}_{{\textbf{m}}}^l. \end{aligned}$$

(A.9)

As in the previous section, it suffices to find the constant vector ${\textbf{a}} = (a^l_{{\textbf{m}}})$ that solve the linear system $A{\textbf{a}} = {\textbf{b}}$, where A and ${\textbf{b}}$ has the same forms as in (A.7) and (A.6), but with ${\textbf{v}}^l_{\textbf{m}}$’s and ${\textbf{v}}$ redefined in this section. We prove that A is invertible so the linear system has a unique solution. Suppose ${\textbf{p}} = (p^l_{{\textbf{m}}})$ satisfies $A{\textbf{p}} = 0$. Then ${\textbf{w}} = p^l_{\textbf{m}}{\textbf{v}}^l_{\textbf{m}}$ satisfies ${\textbf{w}} \in {\textbf{R}}$ in each component of $D_\varepsilon$, solves the homogeneous Lamé system in $\Omega _\varepsilon$, satisfies ${\textbf{w}}|_{\partial \Omega } = 0$, and, moreover, $\partial {\textbf{w}}/\partial \nu |_{+} = 0$ on $\partial D_\varepsilon$. It follows that

$$\begin{aligned} {\textbf{w}} \in \mathfrak {H}_{{\textbf{R}}}^{\textrm{D}}\cap {\mathcal {S}}^{\textrm{D}}\,\textrm{ker}\,\left( -\frac{{\mathbb {I}}}{2} +{\mathbb {K}}^{\textrm{D},*} \right) =\{0\}. \end{aligned}$$

Again, by the isometry of ${\textbf{p}} \mapsto \textrm{span}\{{\textbf{v}}^l_{{\textbf{m}}}\}$, we get ${\textbf{p}} = 0$. Hence, the kernel of the square matrix A is trivial. This completes the proof.