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On coated inclusions neutral to bulk strain fields in two dimensions∗ Hyeonbae Kang† Abstract 6 1 Theneutralinclusionproblemintwodimensionalisotropicelasticityisconsidered. 0 The neutral inclusion, when inserted in a matrix having a uniform applied field, does 2 notdisturbthe fieldoutside the inclusion. The inclusionconsistsofthe coreandshell n of arbitrary shapes, and their elasticity tensors are isotropic. We show that if the a J coated inclusion is neutral to a uniform bulk field, then the core and shell must be 8 concentric disks, provided that the shear and bulk moduli satisfy certain conditions. 2 Key words. Elastic neutral inclusion, bulk strain field, concentric disk ] h p 1 Introduction - h t a Some inclusions, when inserted in a matrix having a uniform field, do not disturb the field m outside the inclusion. Such inclusions are called neutral inclusions (to the given field). A [ typical neutral inclusion consists of a core coated by a shell having the material property 2 different from that of the core. v Itiseasytoconstructneutralinclusionsofcircularshapesinthecontextofconductivity 7 2 (or anti-plane elasticity). Let D = {|x| < r1} and Ω = {|x| < r2} (r1 < r2) so that D is 8 the core and Ω\D is the shell. The conductivity is σc in the core, σs in the shell, and σm 4 in the matrix (R2\Ω). So the conductivity distribution is given by 0 . 1 σ = σ χ(D)+σ χ(Ω\D)+σ χ(R2\Ω) c s m 0 6 where χ is the characteristic function. If σ , σ and σ satisfy the relation 1 c s m : v 2 2 r (σ +σ )(σ −σ )−r (σ −σ )(σ +σ )= 0, (1.1) i 2 s c m s 1 s c m s X r then Ω is neutral to uniform fields. In other words, for any constant vector a, the solution a u to the problem ∇·σ∇u= 0 in R2, u(x)−a·x= O(|x|−1) as |x|→ ∞ (cid:26) satisfies u(x)−a·x =0 in R2\Ω. ∗This work is supported by the A3 Foresight Program among China, Japan, and Korea through NRF grant NRF-2014K2A2A6000567. †Department of Mathematics, InhaUniversity,Incheon 22212, S. Korea ([email protected]). 1 Much interest in neutral inclusions was aroused by the work of Hashin [6, 7], where it is shown that since insertion of neutral inclusions does not perturb the outside uniform field, the effective conductivity of the assemblage filled with neutral inclusions of many different scales is σ satisfying (1.1). It is also proved that this effective conductivity is m a bound of the Hashin-Shtrikman bounds on the effective conductivity of arbitrary two phase composites. We refer to a book of Milton [13] for development on neutral inclusions in relation to theory of composites. Anotherinterestinneutralinclusionshasarousedinrelationtotheinvisibilitycloaking by transformation optics. In this regard, we first observe that in general the solution u to ∇·σ∇u= 0 in R2, (1.2) u(x)−h(x) = O(|x|−1) as |x| → ∞ (cid:26) for a given harmonic function h satisfies u(x)−h(x) = O(|x|−1) as |x| → ∞. But, if the inclusion is neutral to all uniform fields, then the linear part of h is unperturbed and one can show using multi-polar expansions that u(x)−h(x) = O(|x|−2) as |x| → ∞ for any h (not necessarily linear). Recently, Ammari et al [2] extend the idea of neutral inclusions to construct multi-coated circular structures which are neutral not only to uniform fields but also to fields of higher order, so that the solution u to (1.2) satisfies u(x)−h(x) = O(|x|−N)as|x| → ∞foranygivenN andanyh(suchstructuresarecalledGPTvanishing structures). Such structures have a strong connection to the cloaking by transformation optics. The transformation optics proposed by Pendry et al [16] transforms a punctured disk (or a sphere)to an annulus to achieve perfect cloaking. Thesame transform was used to show non-uniqueness of the Calderón’s problem by Greenleaf et al [5]. Kohn et al [12] showedthatifonetransformsadiskwithsmallhole,thenonecanavoidsingularities ofthe conductivity which occur on the inner boundary of the annulus and achieve near-cloaking instead of the perfect cloaking. In [2] it is shown that if we coat the hole by multiple layers so that the structure becomes neutral to fields of higher order (and transform the structure), then the near-cloaking effect is dramatically improved. All above mentioned neutral inclusions have circular shapes and it is of interest to consider neutral inclusions of arbitrary shapes. For a given core of arbitrary shape, the shape of the outer boundary of the shell has been constructed by Milton & Serkov [14] so that the coated inclusion is neutral to a single uniform field. This is done when the conductivity σ of the core is either 0 or ∞. See [9] for an extension to the case when c σ is finite. It is also proved in [14] that if an inclusion is neutral to all uniform field (or c equivalently, to two linearly independent uniform fields), then the inclusion is concentric disks (confocal ellipses if the conductivity of the matrix is anisotropic), when σ is 0 or c ∞. In recent paper [10], Kang and Lee proved that this is the case even when σ is finite. c See also [11] for an extension to three dimensions. In this paper the problem of neutral inclusions in two dimensional linear isotropic elasticity is considered. Let the shear and bulk moduli of the core, the shell, and the matrix be (µ ,κ ), (µ ,κ ), and (µ ,κ ), respectively, and let µ and κ denote their c c s s m m distributions in R2. Define the elasticity tensor C = (C ) by ijkl C = (κ−µ)δ δ +µ(δ δ +δ δ ), i,j,k,l = 1,2, (1.3) ijkl ij kℓ ik jℓ iℓ jk whereδ is theKronecker’s delta. Leth(x) = x,whosegradientrepresentsthebulkstrain ij 2 field, and consider the following interface problem: divC∇u = 0 in R2, (1.4) ( u(x)−h(x) = O(|x|−1) as |x| → ∞, b where ∇u is the symmetric gradient (or the strain tensor), i.e., 1 b ∇u := (∇u+(∇u)T) (T for transpose). 2 The inclusion is neutralbto the (strain) field ∇h if the solution u to (1.4) satisfies u(x)− h(x) = 0 in R2\Ω. Inclusions neutral to the bulk field was found using the exact effective bulkmodulusoftheassemblageofcoateddiskswhichwasderivedbyHashinandRosen[8]. The purpose of this paper is to prove that concentric disks are the only coated inclusions neutral to bulk fields under some conditions on the shear and bulk moduli. The following is the main theorem of this paper. Theorem 1.1 Let Ω and D be bounded simply connected domains in R2 with Lipschitz boundaries such that D ⊂ Ω. Suppose that µ 6= µ , κ 6= κ , and κ < 2κ +µ . (1.5) c s m s c s s If (Ω,D) isneutral to the bulk field, or equivalently, if the solution u to (1.4)with h(x) = x satisfies u(x)−x = 0 in R2\Ω, then D and Ω are concentric disks. The conditions in (1.5) are required to show that the solution is linear in the core. The first two conditions seem natural because the elasticity properties of the core, the shell, and the matrix must be different. However, we don’t know if the third condition is necessary. Itis worth mentioning that inclusions consisting of theconcentric disks arenot neutral to shear fields: for example, if h(x) = (y,x)T, then u(x)−h(x) has a term of order |x|−1 and a term of order |x|−3 as |x| → ∞, and it is not possible to make both terms vanish. Christensen and Lo [3] constructed circular inclusions such that the term of order |x|−1 vanishes and derived an effective transverse shear modulus of the assemblage of coated disks. It is interesting to construct coated inclusions neutral to shear fields or to prove non-existence of such inclusions. Therestofthepaperisorganized asfollows: Inthenextsection weshow thatif(Ω,D) is neutral to the bulk field, then ∇u is symmetric and divu is constant in the shell. The main theorem is proved in section 3 by showing that u is linear in the core. To do so we use a complex representation of the displacement vector. 2 Properties of the solution in the shell Inthis section weprove the following proposition. We emphasizethat (1.5)is not required for this proposition. Proposition 2.1 Let Ω and D be bounded simply connected domains in R2 with Lipschitz boundaries such that D ⊂ Ω. If (Ω,D) is neutral to the bulk field, then the solution u to (1.4) satisfies the following: 3 (i) ∆u = 0, or equivalently divu = constant in Ω\D. (ii) There is a function χ such that u = ∇χ in R2\D. In particular, ∇u is symmetric in Ω\D. To prove Proposition 2.1, we need some preparartion. The Kelvin matrix Γ(x) = (Γ (x))2 of the fundamental solution to the Lamé operator divC∇ in two dimensions ij i,j=1 is given by Γ (x) := α1δ log|x|− α2 xixj, x 6= 0, b (2.1) ij 2π ij 2π |x|2 where 1 1 1 1 1 1 α1 = + and α2 = − . (2.2) 2 µ µ+κ 2 µ µ+κ (cid:18) (cid:19) (cid:18) (cid:19) A straight-forward computation shows that α2−α1 1 div Γ(x−y) = ∇ log|x−y| = − ∇ log|x−y|. (2.3) y x x 2π 2π(µ+κ) In particular, we have 1 div Γ(x−y)dy = − ∇ log|x−y|dy. (2.4) y 2π(µ+κ) Ω Ω Z Z Since 1 1 if x∈ Ω, ∆ log|x−y|dy = 2π Ω (0 if x∈ R2\Ω, Z we have 1 − if x∈ Ω, div div Γ(x−y)dy = µ+κ (2.5) y ZΩ 0 if x∈ R2\Ω. We also have  rot div Γ(x−y)dy = 0. (2.6) y Ω Z Proof of Proposition 2.1. Suppose that (Ω,D) is neutral to the bulk field. Then the following over-determined problem admits a solution: ∇·(C∇u) = 0 in Ω, (2.7) ( u(x) = x, (Cs∇u)n = (CmI)n on ∂Ω. b Here and throughout this paper, n denobtes the outward normal to ∂Ω (and ∂D). Let uc and u denote the solution on D and Ω\D, respectively. Then the transmission condition s along ∂D is given by u = u and (C ∇u )n = (C ∇u )n on ∂D. (2.8) c s c c s s Let v be a smooth vector field in Ω. Then we have b b (C ∇u)n·vdσ = C∇u : ∇vdy. s Z∂Ω ZΩ b b b 4 Here and afterwards, A : B denotes the contraction of two matrices A and B, i.e., A : B = a b = tr(ATB). On the other hand, we have from the Neumann boundary ij ij condition in (2.7) P (C∇u)n·vdσ = C I :∇vdy. m Z∂Ω ZΩ So, we have b b C ∇u : ∇vdy+ C ∇u :∇vdy = C I :∇vdy. (2.9) s c m ZΩ\D ZD ZΩ Using the Dirichlet bboundbary condition ibn (2.7b) we have for anybelasticity tensor C0 u·(C0∇v)ndσ = C0∇u :∇vdy+ u·div(C0∇v)dy Z∂Ω ZΩ ZΩ and b b b b u·(C0∇v)ndσ = y·(C0∇v)ndσ = C0I :∇vdy+ y·div(C0∇v)dy. Z∂Ω Z∂Ω ZΩ ZΩ Thus we haveb b b b C0∇u : ∇vdy+ u·div(C0∇v)dy = C0I : ∇vdy+ y·div(C0∇v)dy. (2.10) Ω Ω Ω Ω Z Z Z Z Subtracbting (b2.10) with C0 = Cs frbom (2.9) we obtainb b (C −C )∇u :∇vdy− u·div(C ∇v)dy c s s ZD ZΩ = (C −Cb )I :b∇vdy− y·div(Cb∇v)dy. (2.11) m s s Ω Ω Z Z b b Let Γs and Γc be the Kelvin matrices for divC ∇ and divC ∇, respectively. For s c x ∈ Ω, let v(y) be a column of Γs(x− y). Then we may apply the same argument of integration by parts (over Ω with an ǫ ball around x dbeleted) as abbove and obtain from (2.11) the following representation of the solution: u(x) =x+ (C −C )∇u(y) : ∇Γs(x−y)dy c s ZD + (C −C )I :b∇Γs(x−b y)dy, x ∈ Ω. (2.12) s m Ω Z b Since (C −C )I :∇v = 2(κ −κ )divv, the identity (2.11) takes the form m s m s b(C −C )∇u : ∇vdy− u·div(C ∇v)dy c s s ZD ZΩ = 2(κ −κ )b dbivvdy− y·div(Cb∇v)dy, (2.13) m s s Ω Ω Z Z b 5 One can also see from (2.4) that the representation formula (2.12) takes the form u(x) = x+ (C −C )∇u(y) : ∇ Γs(x−y)dy c s y ZD κ −κ − m s ∇ blog|x−by|dy, x∈ Ω. (2.14) π(µ +κ ) s s Ω Z Let Γs,j be the j-th column of Γs. Let x∈ R2\Ω. Substitute v (y) := ∂ Γs,j(x−y) j ∂xj for v in (2.13) and add the identities for j = 1,2. Note that div(C ∇v ) = 0 in Ω since s j x∈/ Ω. We infer from (2.5) that b 2 2 ∂ divv = div Γs,j(x−y)dy = 0. j y ∂x Ω j Ω j=1Z j=1 Z X X It then follows from (2.13) that 2 ∂ (C −C )∇u(y) : ∇Γs,j(x−y)dy = 0, x∈ R2\Ω. (2.15) c s ∂x j=1 j ZD X b b Observe that the left-hand side in the above is a real analytic function in R2\D. So, by unique continuation (2.15) holds for all x ∈ R2 \D. We then infer from (2.5) and (2.14) that divu =α in Ω\D, (2.16) where α is the constant given by 2(κ −κ ) m s α = 2− . (2.17) µ +κ s s Since div(C ∇u) = µ ∆u+κ ∇divu =0, we also have s s s b ∆u = 0 in Ω\D. (2.18) We now prove (ii). Let x ∈ R2 \Ω and substitute ∂ Γs,1(x−y) for v in (2.13) to ∂x2 obtain from (2.4) that ∂ ∂ (C −C )∇u(y) :∇Γs,1(x−y)dy =2(κ −κ ) div Γs,1(x−y)dy c s m s y ∂x2 ZD ∂x2 ZD (κ −κ ) ∂2 b b m s =− log|x−y|dy. π(µs+κs)∂x1∂x2 ZD By substituting ∂ Γs,2(x−y) for v in (2.13), we also obtain ∂x1 ∂ (κ −κ ) ∂2 (C −C )∇u(y) :∇Γs,2(x−y)dy =− m s log|x−y|dy. c s ∂x1 ZD π(µs+κs)∂x1∂x2 ZD b b 6 So, we have using unique continuation again that ∂ (C −C )∇u(y) :∇Γs,1(x−y)dy c s ∂x2 ZD ∂ = (C −Cb)∇u(y)b: ∇Γs,2(x−y)dy, x∈ R2\D. c s ∂x1 ZD Since D(Cc−Cs)∇u(y) : ∇Γs(x−by)dy →b0 as |x| → ∞, we infer that there is χ1 such that R b (Ccb−Cs)∇u(y) :∇Γs(x−y)dy =∇χ1(x). ZD Let b b 1 2(κ −κ ) χ(x) := |x|2+χ1(x)− m s log|x−y|dy. 2 µ +κ s s Ω Z Then (ii) is proved. (cid:3) 3 Neutral Inclusions to the bulk field 3.1 Complex representation of the solution and a lemma Let u = (u1,u2)T be the solution to (1.4). There are functions ϕ and ψ which are analytic in D, Ω\D, and C\D, separately, such that 1 u1+iu2 = kϕ(z)−zϕ′(z)−ψ(z) , (3.1) 2µ (cid:16) (cid:17) where 2µ k = 1+ . (3.2) κ See for example [1, 15]. Conversely, u = (u1,u2)T of the form (3.1) with k > 1 for a pair of analytic functions ϕ and ψ in D is a solution in D of the Lamé system determined by the shear modulus µ and the bulk modulus κ= 2µ/(k−1). We denote ϕ and ψ by ϕ and ψ in the core, ϕ and ψ in the shell, and ϕ and ψ c c s s m m in the matrix. Then the transmission conditions (2.7) and (2.8) along the interfaces ∂D and ∂Ω take the following forms: along ∂D, 1 1 k ϕ (z)−zϕ′(z)−ψ (z) = k ϕ (z)−zϕ′(z)−ψ (z) , 2µ s s s s 2µ c c c c s c (cid:18) (cid:19) (cid:18) (cid:19) d(ϕ (z)+zϕ′(z)+ψ (z)) = d(ϕ (z)+zϕ′(z)+ψ (z)), s s s c c c andsimilar conditions on ∂Ω, whered is theexterior differential. Thefirstcondition is the continuity of the displacement and the second one is that of the traction. Using complex notation dz = dx+idy and dz = dx−idy, the exterior differential is given by ∂f ∂f df = dz+ dz. ∂z ∂z It is convenient to use notation 1 U(z) := u1+iu2 = kϕ(z)−zϕ′(z)−ψ(z) , (3.3) 2µ (cid:16) (cid:17) 7 and DU(z) := d(ϕ+zϕ′+ψ) = (ϕ′+ϕ′)dz+(zϕ′′ +ψ′)dz. (3.4) Then the transmission conditions read U = U , DU = DU on ∂D, (3.5) c s c s and U = U , DU = DU on ∂Ω. (3.6) m s m s The proofs in the subsequent subsection use the following lemma, which may be well- known. We include a short proof for readers’ sake. Lemma 3.1 Let D be a simply connected bounded domain with the Lipschitz boundary, and let g be a square integrable function on ∂Ω. If g(z)f′(z)dz = 0 (3.7) Z∂D for any function f analytic in a neighborhood of D, then there is an analytic function G in D such that G = g on ∂D. Proof. Define the Cauchy transform by 1 g(z) C[g](w) := dz, w ∈ C\∂D. 2πi z−w Z∂D Then by Plemelj’s jump formula (see [15]), we have g(w) = C[g]|−(w)−C[g]|+(w), w ∈ ∂D, whereC[g]|− andC[g]|+ denotethelimitsfrominsideandoutsideofD,respectively. Since D is simply connected, f(z) = log(z −w) is well-defined and analytic in a neighborhood of D if w ∈/ D. So, C[g](w) = 0 if w ∈/ D by (3.7). Thus, we have g(w) = C[g]| (w), w ∈ ∂D. − So, G := C[g] in D is the desired analytic function. (cid:3) 3.2 Proof of Theorem 1.1 Let us prove the following proposition first. Proposition 3.2 Let Ω and D be bounded simply connected domains in R2 with Lipschitz boundaries such that D ⊂ Ω, and assume that (1.5) holds. If (Ω,D) is neutral to the bulk field, then the solution u to (1.4) is linear in D and of the form u(x) = ax+b (3.8) for a constant a and a constant vector b. 8 Proof. Let u be the solution to (1.4) when h(x) = x, and U be defined by (3.3). Since (Ω,D) is neutral to the bulk field, u(x) = x in R2\Ω, and hence we have U (z) = z, ϕ (z) = κ z, ψ (z) = 0, DU (z) = 2κ dz. m m m m m m Moreover, Proposition 2.1 implies that ϕ (z) = βz+constant, z ∈ Ω\D, (3.9) s where β is a real constant. In fact, we see from Proposition 2.1 that ∂ 1 ∂ ∂ 1 ∂u2 ∂u1 α Us = −i (u1+iu2) = divu+i − = , ∂z 2 ∂x1 ∂x2 2 ∂x1 ∂x2 2 (cid:18) (cid:19) (cid:18) (cid:18) (cid:19)(cid:19) where α is the constant in (2.17). Thus we have α ∂ 1 = U (z) = k ϕ′(z)−ϕ′(z) , 2 ∂z s 2µ s s s s (cid:16) (cid:17) which implies that ϕ′(z) = β = κ α/2 by (3.2). One can see from (2.17) that s s (κ −κ )(2κ +µ ) m s s s κ −β = . (3.10) m κ +µ s s Let f and g be functions analytic on Ω, and let F(z) = f(z)+g(z). We have from the first identity in (3.6), Cauchy’s theorem and Stokes’ theorem that U dF = U dF = zg′dz = − g′dm, s m Z∂Ω Z∂Ω Z∂Ω ZΩ where dm := dz∧dz. We also have from Stokes’ theorem that ∂ ∂ U dF = d(UdF) = Uf′ − Ug′ dm s ∂z ∂z Z∂Ω ZΩ ZΩ(cid:20) (cid:21) 1 (cid:0) (cid:1) (cid:0) (cid:1) = − zϕ′′ +ψ′ f′+(kϕ′ −ϕ′)g′ dm. (3.11) 2µ Ω Z (cid:2)(cid:0) (cid:1) (cid:3) So we have 1 g′dm = zϕ′′+ψ′ f′+(k ϕ′ −ϕ′)g′ dm 2µ s s s s s ZΩ s ZΩ\D 1 (cid:2)(cid:0) (cid:1) (cid:3) + zϕ′′ +ψ′ f′+(k ϕ′ −ϕ′)g′ dm. 2µ c c c c c c ZD (cid:2)(cid:0) (cid:1) (cid:3) It then follows from (3.2) and (3.9) that β g′dm− g′dm κ ZΩ s ZΩ\D 1 1 1 = ψ′f′dm+ zϕ′′+ψ′ f′dm+ (k ϕ′ −ϕ′)g′dm. 2µ s 2µ c c 2µ c c c s ZΩ\D c ZD c ZD (cid:0) (cid:1) 9 Since f and g are arbitrary, we have 1 1 ψ′f′dm+ zϕ′′+ψ′ f′dm = 0, (3.12) µ s µ c c s ZΩ\D c ZD (cid:0) (cid:1) and 1 β (k ϕ′ −ϕ′)g′dm = g′dm− g′dm. (3.13) 2µ c c c κ c ZD ZΩ s ZΩ\D Similarly, we have from the second identity in (3.6) FDU = FDU = 2κ gdz = 2κ g′dm, s m m m Z∂Ω Z∂Ω Z∂Ω ZΩ and hence 2κ g′dm = d(FDU) m Ω Ω Z Z ∂ ∂ = (f +g)(ϕ′+ϕ′) − (f +g)(zϕ′′ +ψ′) dm ∂z ∂z Ω Z (cid:20) (cid:21) (cid:0) (cid:1) (cid:0) (cid:1) = g′(ϕ′+ϕ′)−f′ zϕ′′+ψ′ dm. Ω Z (cid:2) (cid:0) (cid:1)(cid:3) Since f and g are arbitrary and (3.9) holds, we obtain ψ′f′dm+ zϕ′′+ψ′ f′dm = 0, (3.14) s c c ZΩ\D ZD (cid:0) (cid:1) and (ϕ′ +ϕ′)g′dm = 2κ g′dm−2β g′dm. (3.15) c c m ZD ZΩ ZΩ\D Since µ 6= µ by the assumption (1.5), we infer from (3.12) and (3.14) that s c zϕ′′+ψ′ f′dm = 0, c c ZD (cid:0) (cid:1) or equivalently, zϕ′ +ψ f′dz = 0. (3.16) c c Z∂D (cid:0) (cid:1) Note that (3.16) holds for all functions f analytic in Ω. However, one can infer using Runge’s approximation theorem that it holds for all f analytic in a neighborhood of D. So, by Lemma 3.1, there is an analytic function η1 in D such that zϕ′c +ψc = η1 on ∂D. (3.17) On the other hand, (3.13) can be rewritten as 1 2βµ β (kϕ′ −ϕ′ − c)g′dm = 1− g′dm, (3.18) 2µ c c κ κ c ZD s (cid:18) s(cid:19)ZΩ 10