Multi-Scale Asymptotic Expansion for a Small Inclusion in Elastic Media

Abstract

The aim of this paper is to present an asymptotic expansion of the influence of a small inclusion of different stiffness in an elastic media. The applicative interest of this study is to provide tools which take into account this influence and correct the deformation without inclusion by additive terms that can be precalculated and which depend only on the shape of the inclusion. We treat two problems: an anti-plane linearized elasticity problem and a plane strain problem. On every expansion order we provide corrective terms modeling the influence of the inclusion using techniques of scaling and multi-scale asymptotic expansions. The resulting expansion is validated by comparing it to a test case obtained by solving the Poisson transmission problem in the case of an inclusion of circular shape using the separation of variables method. Proofs of existence and uniqueness of our fields on unbounded domains are also adapted to the bidimensional Poisson problem and the linear elasticity problem.

Appendix:  Poincaré-Type Inequality

We establish the existence and uniqueness of the profile functions $V^{(k)}$ introduced previously in the asymptotic expansion at order $K$ given by (26). The functions $V^{(k)}$ are defined on $\mathbb{R}^{2}$ (two-dimensional geometry) and are solutions to the family of Problems (7a)–(7e). For that, we will introduce weighted Sobolev spaces $\mathcal{W}_{\alpha ,\beta }^{m,p}$ ($\mathcal{W}_{0,0} ^{1,2}$ in our case) similar to classical ones $\mathcal{W}^{m,p}$ ($H ^{1}$ in our case) but with weights that describe the growth or the decay of functions at infinity. The idea of using weighted spaces arises naturally from Hardy’s inequalities and will allow us to establish a Poincaré inequality relating to norms of functions and to that of their derivatives.

The use of these weights is necessary to obtain a Poincaré-type inequality and eliminates the drawbacks of spaces defined by the closure of $\mathcal{D}(\mathbb{R}^{2})$ for the Dirichlet norm and which are not always spaces of distributions. The appropriate weights arise naturally from Hardy’s inequality or from a generalized Hardy’s inequality (see [2]) and the classical weights are of the form $\rho =(1+ \vert x \vert^{2})^{\frac{1}{2}}$, but there is appearance of a logarithmic factor in our case.

1.1 A.1 Notations and Functional Setting

All functions and distributions here are defined on the two-dimensional real Euclidean space $\mathbb{R}^{2}$.

Let $r=\vert x\vert =(x_{1}^{2}+x_{2}^{2})^{\frac{1}{2}}$ be the distance of a point $x=(x_{1},x_{2})$ to the origin. Recall that $\mathcal{D}(\mathbb{R}^{2})$ denotes the space of infinitely differentiable functions with compact support and $\mathcal{D}'( \mathbb{R}^{2})$ its dual space called space of distributions. With $\lambda =(\lambda_{1},\ldots ,\lambda_{n}) \in \mathbb{N}^{n}$ a multi-index $D^{\lambda }=D_{1}^{\lambda_{1}} \cdots D_{n}^{\lambda _{n}}$ is the differential operator of order $\vert \lambda \vert = \lambda_{1}+ \cdots +\lambda_{n}$. $L^{2}(\mathbb{R}^{2})$ is the classical space of measurable functions for which $(\int_{\mathbb{R} ^{2}} \vert u \vert^{2}\, \mathrm{d}\boldsymbol{x}) < \infty$. It is a Banach space for the norm $\Vert u \Vert =(\int_{\mathbb{R}^{2}} \vert u \vert^{2}\, \mathrm{d}\boldsymbol{x})^{\frac{1}{2}}$.

With $\rho =(1+r^{2})^{\frac{1}{2}}$ we introduce the weighted Sobolev space, which is appropriate to our case, by:

$$ \mathcal{W}_{0,0}^{1,2}\bigl(\mathbb{R}^{2}\bigr) = \bigl\{ u \in \mathcal{D}'\bigl( \mathbb{R}^{2}\bigr), \rho^{-1}(\lg \rho )^{-1} u \in L^{2}\bigl( \mathbb{R}^{2}\bigr), \nabla u \in \bigl(L^{2}\bigl( \mathbb{R}^{2}\bigr)\bigr)^{2} \bigr\} , $$

which is a reflexive Banach space equipped with its natural norm:

$$ \Vert u \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})} = \bigl(\bigl\Vert \rho ^{-1}(\lg \rho )^{-1} u \bigr\Vert _{L^{2}(\mathbb{R}^{2})}^{2}+\Vert \nabla u \Vert_{L^{2}(\mathbb{R}^{2})}^{2}\bigr)^{\frac{1}{2}}. $$

We also define the semi-norm:

$$ \vert u \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}=\Vert \nabla u \Vert_{L^{2}(\mathbb{R}^{2})}. $$

With $r=\vert x \vert$ we set $\lg (r)=\ln (2+r^{2})$, $B_{R}=B(0,R)$ the open ball with center 0 and radius $R$ and $B'_{R}=( \overline{B}_{R})^{c}$ the exterior of $\overline{B}_{R}$. Finally, we define $P_{0}$ as the space of constant functions.

1.2 A.2 An Intermediate Result

The following result is an intermediate result to prove the equivalence of the norm and the semi-norm.

Lemma 3

For any large enough real number $R$, there exists a constant $C_{R}$ such that:

$$ \forall \phi \in \mathcal{D}\bigl(B'_{R}\bigr), \quad \Vert \varphi \Vert_{\mathcal{W}_{0,0}^{1,2}(B'_{R})} \leq C_{R}\vert \varphi \vert_{\mathcal{W}_{0,0}^{1,2}(B'_{R})}. $$

Proof

Let $\varphi$ belong to $\mathcal{D}'_{R}$. First, observe that, owing to the support of $\varphi$, all integrals in the norm and semi-norm are taken on $B'_{R}$ instead of $\mathbb{R}^{2}$. Hence, since the origin is in the interior of $B_{R}$, we can use $r$ and $\ln r$ instead of $\rho (r)$ and $\lg (r)$ in the expression of the norm and seminorm. Then using $\frac{\partial \varphi }{\partial r}= \nabla \varphi \cdot \frac{\boldsymbol{r}}{r}$ we can write:

$$ \biggl\vert \frac{\partial \varphi }{\partial r} \biggr\vert ^{2} \leq 2 \sum_{i= 1}^{2}{ \biggl\vert \frac{\partial \varphi }{\partial x _{i}} \biggr\vert ^{2}}. $$

(94)

Let $\theta$ be the angular variable, then we have:

$$ \varphi (r,\theta )= \int_{R}^{r} \varphi (t,\theta ) \, \mathrm{d}t. $$

Now, assuming that $R$ is large enough, we apply the generalized Hardy’s inequality (see [2]) (with $\gamma =-2$) to the function $r\rightarrow \varphi (r,\theta )$. Integrating with respect to $\theta$ and applying (94) we obtain:

$$ \bigl\Vert r^{-1}\ln r^{-1} \varphi \bigr\Vert _{L^{2}(B'_{R})}\leq C \Vert D \varphi \Vert_{L^{2}(B'_{R})}. $$

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The needed result can now be proven.

1.3 A.3 A Poincaré-Type Inequality

Theorem 3

The semi-norm $\vert \cdot \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R} ^{2})}$ defines on $\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})/P_{0}$ a norm which is equivalent to the quotient norm.

Proof

It is clear that $\vert \cdot \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}$ is a norm on $\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})/P_{0}$, and that:

$$ \forall u \in \mathcal{W}_{0,0}^{1,2}\bigl( \mathbb{R}^{2}\bigr), \quad \vert u \vert_{\mathcal{W}_{0,0}^{1,2}}\bigl( \mathbb{R}^{2}\bigr) \leq \Vert u \Vert_{\mathcal{W}_{0,0}^{1,2}}\bigl( \mathbb{R}^{2}\bigr). $$

(95)

Thus, we only have to prove that there exits $c > 0$ such that:

$$ \forall \dot{u} \in \mathcal{W}_{0,0}^{1,2} \bigl(\mathbb{R}^{2}\bigr)/P_{0}, \quad \Vert \dot{u} \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})/P_{0}} \leq c\vert u \vert_{\mathcal{W}_{0,0}^{1,2}}\bigl( \mathbb{R}^{2}\bigr). $$

(96)

The proof proceeds in two steps. The first step consists in eliminating the quotient norm by choosing an adequate representative of the class of $\dot{u}$. To this end, we fix a bounded open domain of $\mathbb{R}^{2}$, with positive measure, say $O$, and we choose the representative $U$ of $\dot{u}$ in $\mathcal{W}_{0,0}^{1,2}( \mathbb{R}^{2})$ that satisfies the system of equations:

$$ \forall \mu \in P_{0}, \int_{O} U \mu \mathrm{d}\boldsymbol{x}=0. $$

(97)

It is easy to see that (97) determines $U$ uniquely and that:

$$ \Vert \dot{u} \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})/P_{0}} \leq \Vert U \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}. $$

(98)

Therefore, the second step consists in proving that there exists a constant $C$ such that the following bound holds for all $U$ in $\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})$ satisfying (97):

$$ \Vert U \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})} \leq C\vert U \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}. $$

(99)

We shall prove it by contradiction. If (99) is not true, there exists a sequence $(U_{\nu })$ of elements of $\mathcal{W}_{0,0}^{1,2}( \mathbb{R}^{2})$ satisfying (98) and such that:

$$ \Vert U_{\nu } \Vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})} = 1 \quad \text{and}\quad \vert U_{\nu } \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})} \rightarrow 0. $$

(100)

Hence the sequence $(U_{\nu })$ is bounded in $\mathcal{W}_{0,0}^{1,2}( \mathbb{R}^{2})$ and since this is a reflexive Banach space, we can extract a subsequence, still denoted by $(U_{\nu })$, that converges weakly to an element $U_{*}$ of $\mathcal{W}_{0,0}^{1,2}(\mathbb{R} ^{2})$ and it is easy to check from this weak convergence that $U_{*}$ also satisfies (98). But since $\vert u \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}$ tends to 0, the lower semi-continuity of the norm implies that $\vert u \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}=0$. Thus, $U_{*}$ is a constant (polynomial of $P_{0}$) and the fact that $U_{*}$ satisfies (98) implies that $U_{*}=0$.

Now, we need a strong convergence to conclude by contradiction, but we cannot use a standard compactness argument on an unbounded domain. Instead, we shall derive a strong convergence via an adequate partition of unity that will enable us to consider separately a bounded domain where the topologies of $\mathcal{W}_{0,0}^{1,2}$ and $H^{1}$ coincide and the exterior of a ball, where Lemma 3 can be applied.

Let $R$ denote a real number, large enough to apply the generalized Hardy’s inequality. Let $\varphi$ and $\psi$ be two functions of $C^{\infty }(\mathbb{R}^{2})$ such that:

$$\begin{aligned} \forall \boldsymbol{x} \in \mathbb{R}^{2},\quad (\varphi + \psi ) (\boldsymbol{x})=1, \quad 0 \leq \varphi (\boldsymbol{x}) \leq 1,\quad \operatorname{supp}(\phi )\subset \overline{B}_{R+1},\quad \operatorname{supp}(\psi ) \subset B'_{R}. \end{aligned}$$

(101)

Since for fixed $R$, $\mathcal{W}_{0,0}^{1,2}(B_{R+1})$ is isomorphic to $R$, $H^{1}(B_{R+1})$, we have that $U_{\mu }$ converges weakly to 0 in $H^{1}(B_{R+1})$. Since $H^{1}(B_{R+1})$ is compactly embedded into $L^{2}(B_{R+1})$, it follows that

$$ U_{\nu }\rightarrow 0\quad \mbox{strongly in } L^{2}(B_{R+1}). $$

(102)

In addition, as $\vert U_{\nu } \vert_{\mathcal{W}_{0,0}^{1,2}(\mathbb{R}^{2})}$ tends to 0, it follows that

$$ U_{\nu }\rightarrow 0\quad \mbox{strongly in } \mathcal{W}^{1,2}(B_{R+1}), $$

(103)

so that

$$ \varphi U_{\nu }\rightarrow 0 \quad \mbox{strongly in } \mathcal{W}^{1,2} _{0,0}(B_{R+1}). $$

(104)

Now, let us examine the behavior of $\psi U_{\nu }$. For fixed $\nu$, let $(\theta_{j})$ be a sequence of functions of $\mathcal{D}( \mathbb{R}^{2})$ that tends to $U_{\nu }$ in $\mathcal{W}^{1,2}_{0,0}( \mathbb{R}^{2})$. Then, $\psi \theta_{j}$ belongs to $\mathcal{D}(B '_{R})$ and we can apply to it Lemma 3

$$ \Vert \psi \theta_{j} \Vert_{\mathcal{W}^{1,2}_{0,0}(\mathbb{R}^{2})} \leq C_{R} \vert \psi \theta_{j} \Vert_{\mathcal{W}^{1,2}_{0,0}(B'_{R})}. $$

(105)

Then, letting $j$ tend to infinity and using the fact that $\psi$ is identically one outside $B_{R+1}$, we obtain:

$$ \Vert \psi U_{\nu } \Vert_{\mathcal{W}^{1,2}_{0,0}(\mathbb{R}^{2})} \leq C_{R} \vert \psi U_{\nu } \vert_{\mathcal{W}^{1,2}_{0,0}(B' _{R})} \leq C_{R} \bigl( \vert \psi U_{\nu } \vert^{p}_{\mathcal{W} ^{1,2}_{0,0}(B'_{R} \cap B_{R+1})}+ \vert U_{\nu } \vert^{p}_{ \mathcal{W}^{1,2}_{0,0}(B'_{R+1})} \bigr) ^{\frac{1}{p}}. $$

(106)

Next, let $\nu$ tend to infinity and observe that $\psi U_{\nu }$ tends to zero strongly in $\mathcal{W}^{1,2}_{0,0}(B'_{R} \cap B_{R+1})$ because $B'_{R} \cap B_{R+1}$ is bounded and $\mathcal{W}^{1,2} _{0,0}(B'_{R} \cap B_{R+1})$ is isomorphic to $\mathcal{W}^{1,2}(B '_{R} \cap B_{R+1})$; we derive that:

$$ \psi U_{\nu }\rightarrow 0\quad \mbox{strongly in } \mathcal{W}^{1,2} _{0,0}\bigl(B'_{R} \bigr). $$

(107)

Since $U_{\nu }=\varphi U_{\mu }+\psi U_{\mu }$, we obtain:

$$ U_{\mu }\rightarrow 0\quad \mbox{strongly in } \mathcal{W}^{1,2}_{0,0}\bigl( \mathbb{R}^{2}\bigr), $$

(108)

which contradicts the assumption (100) that:

$$ \Vert U_{\mu } \Vert_{\mathcal{W}^{1,2}_{0,0}(\mathbb{R}^{2})}=1. $$

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