Exact determination of the volume of an inclusion in a body having constant shear modulus
aa r X i v : . [ m a t h . A P ] J un Exact determination of the volume of an inclusionin a body having constant shear modulus
Andrew E Thaler and Graeme W Milton
Department of Mathematics, University of Utah, Salt Lake City, UT 84112, USAE-mail: [email protected], [email protected]
Abstract.
We derive an exact formula for the volume fraction of an inclusion in abody when the inclusion and the body are linearly elastic materials with the same shearmodulus. Our formula depends on an appropriate measurement of the displacementand traction around the boundary of the body. In particular, the boundary conditionsaround the boundary of the body must be such that they mimic the body being placedin an infinite medium with an appropriate displacement applied at infinity.AMS classification scheme numbers: 74B05, 35Q74
Keywords : volume fraction, Dirichlet-to-Neumann map, artificial boundary conditions
Submitted to:
Inverse Problems xact determination of the volume of an inclusion
1. Introduction
A fundamental and interesting problem in the study of materials is the estimation of thevolume fraction occupied by an inclusion D in a body Ω. Although the volume fractioncould be determined by weighing the body, the densities of the materials may be close orunknown or weighing the body may be impractical. Because of this, many methods havebeen developed which utilize measurements of certain fields around ∂ Ω to derive boundson the volume fraction | D | / | Ω | , where | U | is the Lebesgue measure of the set U [1–23].In this paper, we show that under certain circumstances the volume fraction | D | / | Ω | canbe computed exactly from a single appropriate boundary measurement around ∂ Ω. Wenote that many of the results in the literature (and our results in this paper) can alsobe applied when Ω contains a two-phase composite with microstructure much smallerthan the dimensions of Ω.We consider an inclusion D in a body Ω (or a two-phase composite inside Ω), whereΩ is a subset of R d ( d = 2 or 3). We assume that the inclusion and body are filled withlinearly elastic materials with the same shear modulus µ and Lam´e moduli λ and λ ,respectively. Our goal is to determine the volume fraction occupied by the inclusion,namely | D | / | Ω | , in terms of a measurement of the displacement and traction around ∂ Ω.The boundary conditions around ∂ Ω are taken to be such that they mimic the body Ωbeing placed in an infinite medium with a suitable field at infinity. The starting pointfor our result is based on an exact relation due to Hill [24], which we now describe.One of the most important problems in the study of composite materials is thedetermination of effective moduli given information about the local moduli — see,e.g., the work by Hashin [25] and the book by Milton [26] (Chapters 1 and 2 inparticular). In general, it is extremely difficult (if not impossible) to determine effectiveparameters exactly, even if the microgeometry of the composite is known and relativelyuncomplicated. However, many useful approximation techniques and bounds on effectiveproperties of composites have been derived in the literature — see the book by Milton[26] and the references therein for a vast collection of such results. Surprisingly, there areseveral circumstances in which exact links between effective moduli (or exact formulas forthe moduli themselves) can be derived regardless of the complexity of the microstructure;such links are known as exact relations .Exact relations exist for a variety of problems including elasticity and coupledproblems such as thermoelasticity, thermoelectricity, piezoelectricity, thermo-piezo-electricity, and others — see the review articles by Dvorak and Benveniste [27] andMilton [28], the work by Grabovsky, Milton, and Sage [29], the works by Hegg [30, 31],and the references therein for summaries of numerous previous and current results onexact relations.Perhaps even more surprising than the existence of exact relations is the existenceof a general mathematical theory of exact relations, developed by Grabovsky, Milton,and Sage [29, 32–35], that allows us to determine all of the above mentioned exactrelations and many more. For example, Hegg applied this general theory to the study xact determination of the volume of an inclusion µ but differentLam´e moduli λ and λ . Hill proved that such a composite is macroscopically elasticallyisotropic with shear modulus µ and effective Lam´e modulus λ ∗ ; he also derived anexact formula for λ ∗ that holds regardless of the complexity of the microgeometry. Hisderivation of this formula [24, 36] provides the starting point of our work in this paper.We begin by assuming that the body Ω is embedded in an infinite medium withLam´e modulus λ E and shear modulus µ (we take λ E = λ for simplicity) and that adisplacement u = ∇ g is applied at infinity. Using a method similar to Hill’s derivation of λ ∗ , we derive a formula for | D | / | Ω | in terms of a measurement of the displacement around ∂ Ω, the (known) parameters λ , λ , and µ , and the (known) function g . In order to makethe situation more practical, we derive a certain nonlocal boundary condition that canbe applied to ∂ Ω that forces the body to behave as if it actually were embedded in aninfinite medium with Lam´e modulus λ , shear modulus µ , and an applied displacement u = ∇ g at infinity. This nonlocal boundary condition couples the measurements of thetraction and displacement around ∂ Ω.Nonlocal boundary conditions which mimic infinite media similar to the onementioned above are common tools used in the numerical solution of PDEs (and ODEs)in infinite domains — see, e.g., the review article by Givoli [37] and references thereinfor examples specific to scattering problems, the work by Han and Wu on the Laplaceequation [38] and elasticity equations [39], the work by Lee, Caflisch, and Lee on theelasticity equations [40], and references therein.As discussed above, our formula for the volume fraction | D | / | Ω | holds as long asthe body Ω is embedded in an infinite medium with an applied displacement u = ∇ g atinfinity. In Section 4 we derive a nonlocal boundary condition such that if this boundarycondition is applied to ∂ Ω the solution inside Ω will be equal to the restriction to Ω ofthe solution to the infinite problem. Our boundary condition depends on the function g and on the exterior Dirichlet-to-Neumann map on ∂ Ω (which, when the body Ω isabsent, maps the displacement on ∂ Ω to the traction on ∂ Ω when no fields are applied atinfinity). Thus it is closely related to the boundary condition of Han and Wu [38, 39] andBonnaillie-No¨el, Dambrine, H´erau, and Vial [41] — see Section 4 for complete details.The rest of this paper is organized as follows. In Section 2 we briefly review thelinear elasticity equations and relevant results from homogenization theory. Next, inSection 3 we derive a formula that gives the exact volume fraction of an inclusion in abody when the inclusion and the body have the same shear modulus µ and the bodyis embedded in an infinite medium with shear modulus µ . We discuss the nonlocalboundary condition relevant to our problem in Section 4 so we can focus on a (morerealistic) finite domain. Finally, in Section 5 we present the analytical expression ofthe nonlocal boundary condition in the particular case when Ω is a disk in R — thisexpression was first derived by Han and Wu [38, 39]. A complete derivation of our xact determination of the volume of an inclusion
2. Elasticity
We begin by recalling a few definitions from Hegg’s work [30, 31]. Let d = 2 or 3 be thedimension under consideration; then Sym( R d ) is the set of all symmetric linear mappingsfrom R d to itself, i.e.,Sym( R d ) ≡ (cid:8) B ∈ R d ⊗ R d | B = B T (cid:9) . Similarly, the set Sym(Sym( R d )) is defined as the set of symmetric linear mappings fromSym( R d ) to itself. If A ∈ Sym(Sym( R d )) and B ∈ Sym( R d ), then A : B ∈ Sym( R d )with elements ( A : B ) ij = A ijkl B kl , (1)where here and throughout the paper we use the Einstein summation convention thatrepeated indices are summed from 1 to d .Consider a linearly elastic body which is either a periodic composite material withunit cell Ω ⊂ R d or which occupies an open, bounded set Ω ⊂ R d . Let u ( x ), ε ( x ),and σ ( x ) denote the displacement, linearized strain tensor, and Cauchy stress tensor,respectively, at the point x ∈ Ω. Then u ∈ R d while ε and σ belong to Sym( R d ). ByHooke’s Law, the stress and strain tensor are related through the linear constitutiverelation σ ( x ) = C ( x ) : ε ( x ) , (2)where C ∈ Sym(Sym( R d )) is the elasticity (or stiffness) tensor. We assume C is ellipticfor all x ∈ Ω, i.e., there are positive constants a and b such that B : [ C ( x ) : B ′ ] ≤ a k B kk B ′ k and B : [ C ( x ) : B ] ≥ b k B k for all B , B ′ ∈ Sym( R d ) and where k B k = ( B : B ). If there are no body forcespresent, then at equilibrium the elasticity equations are ∇ · σ ( x ) = 0 , ε ( x ) = 12 (cid:2) ∇ u ( x ) + ∇ u ( x ) T (cid:3) , and ε ( x ) = C ( x ) : σ ( x ); (3)see, e.g., Chapter 2 of the book by Milton [26].If the composite is locally isotropic, the local elasticity tensor takes the form C ( x ) = λ ( x ) I ⊗ I + 2 µ ( x ) I , where λ is the Lam´e modulus, µ is the shear modulus, I ∈ Sym( R d ) is the second-orderidentity tensor with elements I ij = δ ij (where δ ij is the Kronecker delta which is 1 if i = j and 0 otherwise), and I ∈ Sym(Sym( R d )) is the fourth-order identity tensor whichmaps an element in Sym( R d ) to itself under contraction [26]. In this case, Hooke’s Law(2) reduces to S u ( x ) ≡ σ ( x ) = λ ( x ) Tr [ ε ( x )] I + 2 µ ( x ) ε ( x ) (4)= λ ( x ) [ ∇ · u ( x )] I + µ ( x ) (cid:2) ∇ u ( x ) + ∇ u ( x ) T (cid:3) , (5) xact determination of the volume of an inclusion S : R d → Sym( R d ) is the linear stress operator that maps the displacement u tothe stress σ (note that S itself depends on x through λ ( x ) and µ ( x )).The bulk modulus, Young’s modulus, and the Poisson ratio are related to λ and µ by κ = λ + 2 µd , E = 2 µ ( dλ + 2 µ )( d − λ + 2 µ , and ν = λ ( d − λ + 2 µ , respectively [26]. Throughout this paper we assume [42] µ ( x ) > dλ ( x ) + 2 µ ( x ) > . (6)We define the average of a tensor-valued function M ( x ) over a set M ⊂ R d by h M i M ≡ |M| Z M M ( x ) d x , (7)where |M| denotes the Lebesgue measure of the set M . The effective elasticity tensor C ∗ is defined at sample points x ∈ Ω through h σ i Ω ′ ( x ) = C ∗ ( x ) h ε i Ω ′ ( x ) , (8)where Ω ′ ( x ) is a suitably chosen representative volume element centered at x .
3. Exact Volume Fraction
In this section we derive a formula that gives the exact volume fraction occupied byan inclusion in a body, where our formula depends on a boundary measurement of thedisplacement. Let D and Ω be open, bounded sets in R d with D ⊂ Ω. Suppose R d isfilled with a linearly elastic, locally isotropic material with constant shear modulus µ and Lam´e modulus λ ( x ) = λ χ D ( x ) + λ χ R d \ D ( x ) . (9)Since the material is locally isotropic, the elasticity tensor is C ( x ) = λ ( x ) I ⊗ I + 2 µ I . (10)We can write S u ( x ) from (5) as S u ( x ) = ( S u ( x ) ≡ λ [ ∇ · u ( x )] I + µ (cid:2) ∇ u ( x ) + ∇ u ( x ) T (cid:3) for x ∈ D , S u ( x ) ≡ λ [ ∇ · u ( x )] I + µ (cid:2) ∇ u ( x ) + ∇ u ( x ) T (cid:3) for x ∈ R d \ D . (11)According to the elasticity equations in (3), the displacement u satisfies L u = 0 in D , L u = 0 in R d \ D , u , σ · n D = ( S u ) · n D continuous across ∂D , u − f = O ( | x | − d ) as | x | → ∞ , (12)where L j u = − ( λ j + µ ) ∇ ( ∇ · u ) − µ ∆ u (for j = 1, 2) is the Lam´e operator, n D is theoutward unit normal vector to ∂D , σ = S u is the stress tensor associated with u , andthe function f = ∇ g is given and satisfies L f = L ∇ g = 0 for all x ∈ R d . To avoid xact determination of the volume of an inclusion g is at least three times continuouslydifferentiable in R d . The function f represents the “displacement at infinity”; perhapsthe simplest example of such a function is f ( x ) = x , in which case g = ( x · x )+constant.As shown in Chapters 9 and 10 of the book by Ammari and Kang [42], there exists aunique solution u to (12) if D is a Lipschitz domain.Following Hill’s work [24, 36], we assume there is a continuously differentiablepotential φ such that u = ∇ φ . In particular, we assume φ and ∇ φ are continuousacross ∂D (by (12), u = ∇ φ must be continuous across ∂D ). Also, for i , j = 1 , . . . , d ,we have ( ∇∇ φ ) ij = ∂ φ∂x i ∂x j ; (13)note that the matrix ∇∇ φ is symmetric in each phase. We only assume that φ and ∇ φ are continuous across ∂D (indeed, as shown by Hill [43], ∂ φ/∂x i ∂x j is discontinuousacross ∂D ). Then from (3) we have ε = 12 (cid:0) ∇ u + ∇ u T (cid:1) = 12 (cid:2) ∇∇ φ + ( ∇∇ φ ) T (cid:3) = ∇∇ φ. (14)From (13) and (14) we have Tr( ε ) = Tr( ∇∇ φ ) = ∆ φ , where ∆ = ∇ · ∇ = ∂ /∂x i ∂x i isthe Laplacian. Then (4) and (14) imply σ ( x ) = C ( x ) : ε ( x ) = λ ( x )∆ φ I + 2 µ ∇∇ φ. (15)Finally, for j = 1 and j = 2 we have L j u = − ( λ j + µ ) ∇ ( ∇ · ∇ φ ) − µ ∆ ( ∇ φ )= − ( λ j + µ ) ∇ (∆ φ ) − µ ∇ (∆ φ )= − ( λ j + 2 µ ) ∇ (∆ φ ) . (16)By assumption, we have0 = L f = − ( λ + 2 µ ) ∇ (∆ g )for all x ∈ R d , so ∇ (∆ g ) = 0 for all x ∈ R d . Then we must have ∆ g = C g = 0 forall x ∈ R d , where C g is a constant. (The constant C g is known since g is known; wewill see later why we must take C g = 0.) Thus the function g must be chosen so that g = ( C g / x · x + g h , where g h is harmonic in R d . This implies that g is infinitelydifferentiable in R d [44].Recalling that u = ∇ φ and f = ∇ g , we see that (16) implies that (12) becomes ∇ (∆ φ ) = 0 in D and R d \ D , ∇ φ, σ · n D = ( S∇ φ ) · n D continuous across ∂D , ∇ φ − ∇ g = O ( | x | − d ) as | x | → ∞ , (17)where σ = S∇ φ is given in (15). xact determination of the volume of an inclusion ∆ φ In this section we study the behavior of ∆ φ . Recall that we assume φ to be at leastcontinuously differentiable in R d ; in particular, this implies that φ and u = ∇ φ arecontinuous across ∂D (see (17)). If f is smooth the solution u to (12) is smooth in R d \ D and in D [42, equation (10.2)] (although it is only continuous across ∂D ).Since φ is smooth in D and R d \ D , (17) implies that ∆ φ is constant in each phase,i.e., ∆ φ = ( C in D , C in R d \ D . (18)Recall from (3) that ∇ · σ = 0 in R d . By (15), this becomes0 = ∇ · σ ( x )= ∇ · [ λ ( x )∆ φ ( x ) I + 2 µ ∇∇ φ ( x )]= ∇ [ λ ( x )∆ φ ( x )] + 2 µ ∇ · ∇∇ φ ( x )= ∇ [ λ ( x )∆ φ ( x )] + 2 µ ∆ [ ∇ φ ( x )]= ∇ { [ λ ( x ) + 2 µ ] ∆ φ ( x ) } . (19)This implies that[ λ ( x ) + 2 µ ] ∆ φ ( x ) = C ⇔ ∆ φ ( x ) = Cλ ( x ) + 2 µ (20)almost everywhere in R d , where C is a constant [26, Chapter 5]. Then, due to (18) and(20), we have ∆ φ = C = Cλ + 2 µ in D , C = Cλ + 2 µ in R d \ D . (21)By (17), we have ∇ φ − ∇ g → | x | → ∞ ; thus ∆ φ − ∆ g → | x | → ∞ . Since∆ g = C g , ∆ φ → C g as | x | → ∞ . Since λ ( x ) = λ for large enough x , we take the limitof (20) and find that C = lim | x |→∞ { [ λ ( x ) + 2 µ ] ∆ φ ( x ) } = ( λ + 2 µ ) C g . (22)Finally, (9), (20), and (22) imply that∆ φ = ∇ · u = C = (cid:18) λ + 2 µλ + 2 µ (cid:19) C g in D , C g in R d \ D . (23) The divergence theorem and (23) imply Z ∂ Ω u · n Ω d S = Z D ∇ · u d x + Z Ω \ D ∇ · u d x = ( C − C ) | D | + C | Ω | . xact determination of the volume of an inclusion | D || Ω | = 1 C − C (cid:18) | Ω | Z ∂ Ω u · n Ω d S − C (cid:19) , (24)where C and C are related to C g by (23), respectively. Since we are assuming we havecomplete knowledge of u around ∂ Ω from our measurement, and since C g = ∆ g is given,(24) allows us to exactly determine | D | / | Ω | . Note also that we must take C g = 0. If C g = 0, then (23) implies that C = C = 0, which makes the formula in (24) undefined.We have thus proved the following theorem. Theorem 3.1.
Let D and Ω be open, bounded sets in R d ( d = 2 or ) such that D ⊂ Ω and ∂D , ∂ Ω are smooth. Suppose R d is filled with a material described bythe local elasticity tensor given by (10) and (9). Also suppose f = ∇ g is given and L f = − ( λ + µ ) ∇ ( ∇ · f ) − µ ∆ f = 0 ( ⇔ ∆ g = C g = 0 ) for all x ∈ R d . Assume that u · n Ω is known around ∂ Ω . Then the volume fraction of the inclusion D is given by(24).
4. Finite Medium
Consider again the linear elasticity problem from Section 3, namely that of an inclusion D in a body Ω which in turn is embedded in an infinite medium R d \ Ω. The isotropicand homogeneous materials in D and R d \ D have Lam´e moduli λ and λ , respectively;we also assume that both materials have the same shear modulus µ . If a displacement f = ∇ g is applied at infinity, then the displacement u = ∇ φ satisfies (12) (so φ satisfies(17)). Recall that we require L f = 0 in R d , which implies ∆ g = C g in R d . Since g and f = ∇ g are smooth in R d , g , f , and S f are continuous up to ∂D from outside D ; inother words, the limits g | ∂D + , f | ∂D + , and ( S f ) | ∂D + exist and are finite at each point of ∂D , where h | ∂D + and h | ∂D − denote the restriction of the function h to ∂D from outsideand inside D , respectively.We now derive a boundary condition P so that the solution u ′ to L u ′ = 0 for x ∈ D , L u ′ = 0 for x ∈ Ω \ D , u ′ , σ ′ · n D = ( S u ′ ) · n D continuous across ∂D , P ( u ′ , t ′ , f , F ) = 0 on ∂ Ω (25)is equal to the solution u to (12) restricted to Ω, i.e., u ′ = u | Ω ; we have defined u ′ ≡ u ′ | ∂ Ω − , t ′ ≡ ( σ ′ | ∂ Ω − ) · n Ω , f ≡ f | ∂ Ω + , and F ≡ [( S f ) | ∂ Ω + ] · n Ω . (26)This allows us to apply our formula (24) to the problem (25), which is posed on thefinite domain Ω. For details on a related problem (including proofs of the well-posednessof problems similar to (25)), see the papers of Han and Wu [38, 39]. xact determination of the volume of an inclusion P , we begin by considering the following exteriorproblem: L E e u E = 0 for x ∈ R d \ Ω, e u E = e u on ∂ Ω, e u E → | x | → ∞ , (27)where L E u = − ( λ E + µ ) ∇ ( ∇ · u ) − µ ∆ u , λ E = λ , and e u is a given displacement on ∂ Ω. Ultimately we wish to find the normal stress distribution ( e σ E | ∂ Ω + ) · n Ω around ∂ Ωgiven e u — this mapping from the displacement on the boundary to the traction on theboundary is defined as the exterior Dirichlet-to-Neumann map. Definition 4.1.
The exterior Dirichlet-to-Neumann (DtN) map Λ E is defined byΛ E ( e u E | ∂ Ω + ) = Λ E ( e u ) ≡ ( e σ E | ∂ Ω + ) · n Ω = [( S e u E ) | ∂ Ω + ] · n Ω , (28)where e u E solves (27) and e σ E and S e u E are given by (5) (with λ ( x ) = λ E ). We now return to the problem (12), which has a unique solution u . We introduceexterior fields u E ( x ) ≡ u | R d \ Ω and σ E ( x ) ≡ σ | R d \ Ω ; (29)we also introduce interior fields u I ( x ) ≡ u | Ω and σ I ( x ) ≡ σ | Ω . (30)Recall that λ E = λ . Lemma 4.2.
Define e u E ≡ u E − f where u E is defined in (29) and f = ∇ g satisfies L E f = 0 in R d . Then e u E solves (27) with e u = ( u I | ∂ Ω − ) − f , where f = f | ∂ Ω + is definedin (26).Proof. First, since L E u E = 0 in R d \ Ω (by (12)) and L E f = 0 in R d \ Ω, we have L E e u E = L E ( u E − f ) = L E u E − L E f = 0in R d \ Ω as well. Second, recall from (27) that e u ≡ e u E | ∂ Ω + ≡ ( u E | ∂ Ω + ) − f .Since λ E = λ , u must be continuous across ∂ Ω, i.e., u E | ∂ Ω + = u I | ∂ Ω − . Hence e u = ( u I | ∂ Ω − ) − f . Finally, e u E = u E − f → | x | → ∞ by (12). Thus e u E solves (27)with e u = ( u I | ∂ Ω − ) − f . Theorem 4.3.
Suppose u solves (12) with f = ∇ g and g = ( C g / x · x + g h where C g = 0 is an arbitrary constant and ∆ g h = 0 in R d . Define u E , σ E and u I , σ I as in(29) and (30), respectively. Finally, define e u E = u E − f . Then u I satisfies L u I = 0 in D , L u I = 0 in Ω \ D , u I , σ I · n D = ( S u I ) · n D continuous across ∂D , P ( u I | ∂ Ω − , ( σ I | ∂ Ω − ) · n Ω , f , F ) = 0 on ∂ Ω, (31) xact determination of the volume of an inclusion where P ( u I | ∂ Ω − , ( σ I | ∂ Ω − ) · n Ω , f , F ) ≡ ( σ I | ∂ Ω − ) · n Ω − Λ E (( u I | ∂ Ω − ) − f ) − F (32) and f and F are defined in (26).Proof. By definition (see (12) and (30)), u I satisfies the differential equations andcontinuity conditions in (31). By Lemma 4.2, e u E = u E − f solves (27) with e u =( u I | ∂ Ω − ) − f | ∂ Ω+ . By (28), then, we have( e σ E | ∂ Ω + ) · n Ω = Λ E ( e u ) = Λ E (( u I | ∂ Ω − ) − f ) . (33)Since S E is linear, we have( e σ E | ∂ Ω + ) · n Ω = [( S E e u E ) | ∂ Ω + ] · n Ω = [( S E u E ) | ∂ Ω + ] · n Ω − F . (34)Then (33) and (34) imply[( S E u E ) | ∂ Ω + ] · n Ω = Λ E (( u I | ∂ Ω − ) − f ) + F . (35)Since λ E = λ , the traction across ∂ Ω must be continuous, i.e.,[( S E u E ) | ∂ Ω + ] · n Ω = ( σ E | ∂ Ω + ) · n Ω = ( σ I | ∂ Ω − ) · n Ω . Inserting this into (35) givesΛ E (( u I | ∂ Ω − ) − f ) + F = ( σ I | ∂ Ω − ) · n Ω . (36)We define P ( u I | ∂ Ω − , ( σ I | ∂ Ω − ) · n Ω , f , F ) as in (32). Then, due to (36), the interiorpart of the solution u , namely u I , satisfies (31).We can thus identify the solution u ′ of (25) with u I which solves (31), i.e., u ′ = u I = u | Ω . In other words, the solution to (25) in the finite domain Ω will beexactly the same as if Ω were placed in an infinite medium with Lam´e parameters λ E = λ and µ and a displacement ∇ g were applied at infinity. Therefore, if we applythe boundary condition P ( u ′ , t ′ , f , F ) = t ′ − Λ E ( u ′ − f ) − F = 0 (37)on ∂ Ω, where u ′ , t ′ , f , and F are defined in (26), we can use the measurement of u ′ · n Ω around ∂ Ω (i.e., u ′ · n Ω ) along with (24) (with u replaced by u ′ ) to find thevolume fraction occupied by D . Remark . Since the geometry inside the body Ω is unknown, we cannot write σ ′ · n Ω interms of u ′ (since we would not know whether or not to use λ or λ in (5)). Practically,we would typically apply a displacement u ′ around ∂ Ω with a known f and measurethe resulting traction t ′ around ∂ Ω. The displacement u ′ and traction t ′ around ∂ Ωmust be tailored so that P ( u ′ , t ′ , f , F ) = 0 — see (37). xact determination of the volume of an inclusion
5. Two Dimensional Example
The results presented here were first derived in a slightly different form by Han and Wu[38, 39]. We consider the case when d = 2 and Ω is a disk of radius R centered at theorigin, denoted B R . In this geometry, it is possible to determine Λ E exactly by firstsolving (27) for the displacement e u E in terms of e u = e u E | ∂B R and then computing thecorresponding traction around ∂B R , namely( e σ E | ∂B + R ) · n B R = (cid:16) e σ E · x R (cid:17)(cid:12)(cid:12)(cid:12) ∂B + R = h S E ( e u E ) · x R i(cid:12)(cid:12)(cid:12) ∂B + R . We state the main results here; the complete calculations are given in work by one ofthe authors [23]. For more general regions, Λ E may have to be computed numerically. We denote the polar components of e u E by e u E,r and e u E,θ . It is convenient to write e u E ( r, θ ) = e u E,r ( r, θ ) + i e u E,θ ( r, θ ) , where i = √−
1; see the books by Muskhelishvili [45] and England [46] for more details.We begin by expanding e u ( θ ) = e u r ( θ ) + i e u θ ( θ ) in a Fourier series, namely e u r ( θ ) + i e u θ ( θ ) = ∞ X n = −∞ e u n e i nθ , where e u n = 12 π Z π [ e u r ( θ ′ ) + i e u θ ( θ ′ )] e − i nθ ′ d θ ′ . (38)Then it can be shown that [23] e u E,r ( r, θ ) + i e u E,θ ( r, θ ) = e u Rr − + ∞ X n =1 e u − n R n − r − ( n − e − i nθ + ∞ X n =1 (cid:20)e u n R n +1 r − ( n +1) + (cid:18) n − ρ E (cid:19) e u − n R n − r − ( n +1) (cid:0) r − R (cid:1)(cid:21) e i nθ (39)for r ≥ R and where ρ E ≡ ( λ E + 3 µ ) / ( λ + µ ).Next we recall from (28) that Λ E ( e u ) = ( e σ E | ∂B + R ) · n B R . In polar coordinates,the components of the traction around the boundary of the disk of radius r ≥ R are e σ E,rr ( r, θ ) + i e σ E,rθ ( r, θ ) (where e σ E,rr is the radial component of the traction and e σ E,rθ isthe angular component of the traction). In particular, the traction around ∂B R is givenby e σ E,rr ( R + , θ ) + i e σ E,rθ ( R + , θ ) = Λ E ( e u ) = ∞ X n = −∞ e σ n e i nθ , (40)where e σ E,rr ( R + , θ ) + i e σ E,rθ ( R + , θ ) = ( e σ E,rr + i e σ E,rθ ) | ∂B + R , e σ n = − µR ( n + 1) e u n ( n ≥ , e σ − n = − µRρ E ( n − e u − n ( n ≥ , (41)and the coefficients e u n are defined in (38). xact determination of the volume of an inclusion In this section, we derive an expression for the boundary condition P ( u ′ , t ′ , f , F ) = 0,where P is defined in (37). We begin by expanding f in a Fourier series around ∂B R ;we have( f r + i f θ ) | ∂B + R = f r ( R + , θ ) + i f θ ( R + , θ ) = f ,r ( θ ) + i f ,θ ( θ ) = ∞ X n = −∞ f ,n e i nθ , (42)where f ,n = 12 π Z π [ f ,r ( θ ′ ) + i f ,θ ( θ ′ )] e − i nθ ′ d θ. Next we define F ≡ S E f = S E ∇ g = λ E ∆ g I + 2 µ ∇∇ g, where the last equality holds by (11). Recall from (26) that F = ( F | ∂B + R ) · n B R . Incomplex notation, the normal components of F around the boundary of a disk of radius r > F rr ( r, θ ) + i F rθ ( r, θ ), where F rr is the radial component and F rθ isthe angular component. We can expand ( F rr + i F rθ ) | ∂B + R in a Fourier series as( F rr + i F rθ ) | ∂B + R = F rr ( R + , θ ) + i F rθ ( R + , θ ) = F ,r ( θ ) + i F ,θ ( θ ) = ∞ X n = −∞ F ,n e i nθ , (43)where F ,n = 12 π Z π [ F ,r ( θ ′ ) + i F ,θ ( θ ′ )] e − i nθ ′ d θ ′ . Next we expand g in a Fourier series around the disk of radius r > g ( r, θ ) = ∞ X n = −∞ g n ( r )e i nθ , where g n ( r ) = 12 π Z π g ( r, θ ′ )e − i nθ ′ d θ ′ . We can write the Fourier coefficients F ,n in terms of the coefficients g n as F ,n = ( λ E + 2 µ ) ∂ g n ( r ) ∂r (cid:12)(cid:12)(cid:12)(cid:12) r → R + + λ E (cid:20) R ∂g n ( r ) ∂r (cid:12)(cid:12)(cid:12)(cid:12) r → R + − n R g n ( R + ) (cid:21) +2 µ (cid:20) − nR ∂g n ( r ) ∂r (cid:12)(cid:12)(cid:12)(cid:12) r → R + + nR g n ( R + ) (cid:21) . (44)Returning to (32), recall that ( u ′ | ∂B − R ) − f | ∂B + R = u ′ − f = e u , where u ′ solves (25).Thus if we write( u ′ r + i u ′ θ ) | ∂B − R = u ′ r ( R − , θ ) + i u ′ θ ( R − , θ ) = u ′ ,r ( θ ) + i u ′ ,θ ( θ ) = ∞ X n = −∞ u ′ ,n e i nθ , where u ′ ,n = 12 π Z π (cid:2) u ′ ,r ( θ ′ ) + i u ′ ,θ ( θ ′ ) (cid:3) e i nθ ′ d θ ′ , xact determination of the volume of an inclusion u ′ ,n − f ,n = e u n — see (38). The components of the traction ( σ ′ | ∂B − R ) · n B R = t ′ canbe written in polar coordinates as t ′ ,r ( θ ) + i t ′ ,θ ( θ ). This can be expanded in a Fourierseries as well, namely( σ ′ rr + i σ ′ rθ ) | ∂B − R = σ ′ rr ( R − , θ ) + i σ ′ rθ ( R − , θ ) = t ′ ,r ( θ ) + i t ′ ,θ ( θ ) = ∞ X n = −∞ t ′ ,n e i nθ , where t ′ ,n = 12 π Z π (cid:2) t ′ ,r ( θ ′ ) + i t ′ ,θ ( θ ′ ) (cid:3) e − i nθ ′ d θ ′ . Recalling the Fourier expansions of f given in (42), F given in (43) (and (44)), andΛ E ( e u ) = Λ E ( u ′ − f ) given in (40)–(41), the condition P ( u ′ , t ′ , f , F ) = 0 is equivalentto t ′ − F − Λ E ( u ′ − f ) = 0 ⇔ ∞ X n = − (cid:20) t ′ ,n − F ,n + 2 µR ( n + 1) (cid:0) u ′ ,n − f ,n (cid:1)(cid:21) e i nθ + ∞ X n =2 (cid:20) t ′ , − n − F , − n + 2 µRρ E ( n − (cid:0) u ′− n − f , − n (cid:1)(cid:21) e − i nθ = 0 . (45)Therefore we have the following relationships between the Fourier coefficients of thepolar components of the displacement, traction, and applied stress around ∂B R : t ′ ,n − F ,n + 2 µR ( n + 1) ( u ′ n − f ,n ) = 0 ( n ≥ − ,t ′ , − n − F , − n + 2 µRρ E ( n − (cid:0) u ′− n − f , − n (cid:1) = 0 ( n ≥ . (46) Remark . Recall from (25) that t ′ is the traction around ∂B R due to the applieddisplacement u ′ . In practice, one could consider applying a displacement u ′ around ∂ Ωwith a known f and then measuring t ′ around ∂B R . The applied displacement u ′ andmeasured traction t ′ have to be such that (45) (and, hence, (46)) holds. Previously, Han and Wu also derived an expression for the exterior DtN map Λ E ( e u )[38, 39]. They found the solution e u E to (27) by a method slightly different from the onewe used; they then computed the Cartesian components of the traction e σ E · n Ω around ∂B R . In particular, if we denote the Cartesian components of e u E by e u E and e v E , theCartesian components of e u E | ∂B + R = e u by e u and e v , and the Cartesian components of thetraction ( e σ E | ∂B + R ) · n B R by e X and e Y , then Λ E ( e u ) = Λ E ( e u + i e v ) = e X + i e Y . In particular xact determination of the volume of an inclusion e X = 2 + 2 η η µπR ∞ X n =1 Z π d e u ( θ ′ )d θ ′ cos n ( θ − θ ′ ) n d θ ′ − η η µπR ∞ X n =1 Z π d e v ( θ ′ )d θ ′ sin n ( θ − θ ′ ) n d θ ′ ; e Y = 2 + 2 η η µπR ∞ X n =1 Z π d e v ( θ ′ )d θ ′ cos n ( θ − θ ′ ) n d θ ′ + 2 η η µπR ∞ X n =1 Z π d e u ( θ ′ )d θ ′ sin n ( θ − θ ′ ) n d θ ′ (47)where e u ( θ ′ ) = e u E ( R + , θ ′ ), e v ( θ ′ ) = e v E ( R + , θ ′ ), and η = µ/ ( λ E + µ ). Also see the books byMuskhelishvili [45, Section 83] and England [46, Section 4.2] for solutions to problemsrelated to (27) based on potential formulations. A proof that our formulas (40)–(41)agree with (47) as long as e u is smooth enough is given in the work by Thaler [23]. Acknowledgments
AET would like to thank Patrick Bardsley, Andrej Cherkaev, Elena Cherkaev, FernandoGuevara Vasquez, and Hyeonbae Kang for helpful discussions. The work of AET andGWM was supported by the National Science Foundation through grant DMS-1211359.
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