In this paper, the derivation of irreducible bases for a class of isotropic 2nth-order tensors having particular ‘minor symmetries’ is presented. The methodology used for obtaining these bases consists of extending the concept of deviatoric and spherical parts, commonly used for second-order tensors, to the case of an nth-order tensor. It is shown that these bases are useful for effecting the classical tensorial operations and especially the inversion of a 2nth-order tensor. Finally, the formalism introduced in this study is applied for obtaining the closed-form expression of the strain field within a spherical inclusion embedded in an infinite elastic matrix and subjected to linear or quadratic polynomial remote strain fields.
Some specific problems in mechanics take the form of linear equations between two tensors having an order higher than n=2. For instance, the theories of generalized continuum (Toupin 1962; Mindlin 1964; Mindlin & Eshel 1968; Suiker & Chang 2000) introduce higher order gradients of the displacement for the description of the continuum. The generalized gradient elastic constitutive equation introduces tensors of order 6,8,… More recently, such considerations were extended to nonlinear elasticity (Dell’Isola et al. 2009). From another point of view, the problem of an elastic inclusion embedded in an infinite elastic medium and subjected to a polynomial remote strain field has been studied by Asaro & Barnett (1975) and Mura (1987). It has been shown that the complete solution requires a linear system involving the inversion of tensors of order 6,8,… to be solved. However, there are no explicit known closed forms for the inverses of such higher order tensors.
To summarize, in all the problems quoted above, two nth-order tensors a and b are related through a linear relation that takes the form 1.1 where is a tensor of order 2n. We assume that the components of a and b are symmetric according to their two first indices aijp,…,q=ajip,…,q, bijp,…,q=bjip,…,q and are invariant by any permutation of their (n−2) last indices p,…,q. For instance, in the case n=3,4,5, one has 1.2
In relation (1.1), ⊙n denotes the nth contraction between and a such that bijp,…,q=Aijp,…,qklr,…,saklr,…,s. Owing to the symmetries of a and b, Aijp,…,qklr,…,s is invariant by any permutation of indices (i,j), (k,l), (p,…,q) and (r,…,s). Throughout this paper, these symmetries are called ‘minor symmetries’. Note that does not necessarily possess the ‘major symmetry’, namely Aijp,…,qklr,…,s≠Aklr,…,sijp,…,q.
In the present study, we assume that tensor is isotropic and we denote by E2n the space of isotropic 2nth-order tensors having minor symmetries. We propose to build the inverse of , namely the tensor such that 1.3
In the case n=2, is an isotropic tensor of fourth order. Owing to the symmetries mentioned above, depends on two independent coefficients a1 and a2 and can be expressed as . Tensors and are defined by Jijkl=δijδkl/3, Kijkl=Iijkl−Jijkl and Iijkl=(δikδjl+δilδjk)/2, where δij is the Kronecker symbol. and are two projectors that define an irreducible basis for isotropic tensors having minor symmetries. The use of these two tensors easily produces the inversion of , since .
We aim at extending this basis in the case of higher order isotropic tensors. The paper is organized as follows. In §2, we first state the case n=3. An irreducible basis, constituted of six independent tensors, is obtained and appears to be convenient for effecting the classical tensorial operations and especially the inversion. The methodology applied for obtaining this basis is clearly depicted in this section. It consists of generalizing the concept of deviatoric and spherical parts, commonly used for second-order tensors, to the case of a tensor of order 3. This approach is then applied to the case of an eighth-order tensor. Its generalization to the case of a tensor of order up to 2n=8 is addressed in §4. Specifically, the application of the methodology to the inclusion problem is performed.
2. A basis for sixth-order isotropic tensors
In this section, we first consider the case n=3 in relation (1.1), a and b are then two third-order tensors, while is a sixth-order isotropic tensor. has the symmetries: Aijkpqr=Ajikpqr, Aijkpqr=Aijkqpr. Tensor is invariant under the orthogonal group O3, consequently 2.1 where Qip,…,Qjq are the orthogonal matrices of the group O3, which satisfy QikQjk=δij and . Every isotropic tensor of order 2n (n being an integer) can be expressed in terms of the Kronecker symbol. Particularly, a sixth-order isotropic tensor can be read as a linear combination of 2.2 For a tensor having minor symmetries, only six tensors are needed. They are denoted , and their components are 2.3 where it is recalled that Iijkl=(δikδjl+δilδjk)/2.
The triple contractions between two tensors taken from are given in table 1. constitute a basis for all tensors . However, this basis is not convenient for the inversion of sixth-order tensors since it leads to a complex linear system of dimension 6.
In order to provide a simplified basis, we first introduce the spherical part, S(a), and the deviatoric part, D(a), of a third-order tensor a as follows: 2.4 S(a) has the properties (S(a))ipp=aipp and (S(a))ppi=appi. Consequently, all contractions of indices of (D(a)) are null: (D(a))ipp=(D(a))ppi=0. It is, therefore, natural to consider D(a) as the generalization to third-order tensors of the deviatoric part, which is well known for second-order tensors. This decomposition suggests introducing the sixth-order tensors , and given by 2.5 These tensors are such that , and . Here, is the identity for the triple contraction ⊙3 and and produce the deviatoric and spherical parts of a. From another point of view, every sixth-order isotropic tensor having minor symmetries is defined by six independent coefficients. Owing to the previous relations, it is natural to introduce the decomposition , where and . It is easy to show that tensor is defined by four independent coefficients. It suggests, therefore, that there exist four tensors with n=1,2,3,4, such that and whatever the value of n. In other words, we search defined by , such that and . For its part, tensor is defined by two independent coefficients. It suggests that there exist two tensors and such that and , whenever n=1,2.
The following expressions were found for and : 2.6 Note that and . In addition, the triple contractions between the different tensors and are given in table 2.
These results lead to the following remarks:
— It can be observed that define a monoid (an algebraic structure with a single associative binary operation and an identity element). is the identity for the composition ⊙3 and is defined by equation (2.5). The six elements for n=1,2 and m=1,2,3,4 constitute an irreducible basis for . Every tensor can be read as 2.7
— Introducing K6, the space of isotropic sixth-order tensors given by , it can also be shown from table 2 that define a sub-monoid. Tensor is the unit element of K6 for the composition ⊙3.
— Introducing J6, the space of isotropic sixth-order tensors given by , it can be also shown from table 1, that define a sub-monoid. is the unit element of J6 for the composition ⊙3.
For a given third-order tensor, the contractions provide four spherical tensors, which can be named partial spherical parts of a, 2.8 Operators Sn have the properties Sn(Sn(a))=Sn(a) for n=1 and 4, but Sn(Sn(a))=0 for n=2 and 3. On the other hand, the deviatoric part of a can be decomposed into the partial deviatoric parts of a, D1(a) and D2(a), which are defined by 2.9
These partial spherical and deviatoric parts have the properties Dn(Sm(a))=Sm(Dn(a))=0 whatever n=1,2 and m=1,2,3,4. As a consequence, every third-order tensor a can be decomposed into 2.10
A decomposition of a third-order symmetric tensor (called SFH decomposition) was introduced by Smyshlyaev & Fleck (1996), formalized by Fleck & Hutchinson (1997) and used more recently in the context of gradient plasticity by Gurtin & Anand (2005). The SFH decomposition of a third-order tensor reads a=a(1)+a(2)+a(3), where the expressions of a(n) for n=1,2,3 are given in appendix A. The concept of spherical and deviatoric parts of a third-order tensor has not been used by the authors. There are close relationships with our approach because it can be shown that a(1)=D1(a), a(2)=D2(a) and a(3)=S(a). However, the SFH decomposition uses only three terms, whereas in the present study, a is decomposed into four terms: two spherical parts and two deviatoric parts, as shown in equation (2.10).
The third-order tensor can be decomposed into its partial spherical and deviatoric parts, which are related to those of a by 2.11 in which a1,…,a6 are the components of on the basis , as defined in equation (2.7).
Consider two sixth-order tensors . We denote by a1,…,a6 and b1,…,b6 their components within the basis . The triple contraction between and leads to 2.12
It is now possible to look for an inverse of . Let be the inverse of defined by . Note that and, consequently, the inverse of is the sum of , the inverse of , and , the inverse of . Finally, the components of are given by 2.13 with ΔJ=a3a6−a4a5. As a consequence, the condition for having an inverse is a1a2ΔJ≠0. The tensors that comply with this condition constitute a sub-monoid that has the properties of a group. The production of the inverse, if it exists, of any sixth-order isotropic tensor having minor symmetries is a clear advantage of the basis , compared with the basis . The following section is devoted to the construction of a similar basis for eighth-order isotropic tensors.
3. A basis for eighth-order isotropic tensors
We now consider in relation (1.1) the case n=4. Consequently, a and b are now two fourth-order tensors, while is an eighth-order tensor. In the general case of an isotropic eighth-order tensor having no symmetries, it can be decomposed into a linear combination of 105 isotropic tensors whose components are obtained by the permutation according to indices i,j,k,l,p,q,r,s of δijδklδpqδrs. In fact, 91 independent tensors are needed (Kearsley & Fong 1975). Now, tensors of components Aijklpqrs are assumed to be symmetric according to indices (i,j), (k,l), (p,q) and (r,s) (called minor symmetries). So, among the 105 isotropic tensors quoted above, we can define 17 isotropic tensors having these four minor symmetries, which are given by 3.1 Note that a more refined analysis of the 17 tensors defined above show that they do not constitute an irreducible basis for the eighth-order tensor. More precisely, those tensors comply with the following relation: 3.2 All eighth-order isotropic tensors having minor symmetries are defined by 16 independent coefficients and can then be decomposed as a linear combination of 16 tensors chosen among those of equation (3.1). As for the case of a sixth-order tensor, a basis made up of tensors is not useful for doing the classical tensorial operations and especially the inversion since , whenever n,m=1,…,17.
The methodology used is the same as the one applied in the previous section. The first step consists of splitting a fourth-order tensor a into its deviatoric and spherical parts, 3.3 where D(a) is the deviatoric part of a, such that [D(a)]ijpp=[D(a)]ppij=[D(a)]ipjp=0. Now, for a fourth-order tensor, it is possible to find tensors for which the contraction over two indices, which defines the deviatoric part, is not zero, but for which the double contraction over the indices is null. Let us introduce the spherical part S1(a) of tensor a, which complies with [S1(a)]ppqq=[S1(a)]pqpq=0. Now, a can be decomposed as 3.4 where the second spherical part S2(a)=S(a)−S1(a) has been introduced. The components of D(a), S1(a) and S2(a) are given by 3.5 with 3.6 where α, β, γ and η are traceless. S1(a) and S2(a) have the properties 3.7
Similarly, a definition of the deviatoric part of a fourth-order tensor was proposed by Lubarda & Krajcinovic (1993). However, the definition introduced by these authors can be used only for a tensor that is invariant by any permutation of its indices, while, in our paper, the considered fourth-order tensor is symmetric only according to its two first and two last indices.
We introduce , , , and , such that , , and . These tensors can be expressed on the basis of as follows: 3.8 Let us decompose as follows: , where , and . As for the case of the sixth order, tensor is defined by four independent coefficients. This suggests that there exist four tensors , for n=1,2,3,4, such that , and , whenever the value of n=1,2,3,4. These tensors read 3.9 Note that . The quadruple contractions between the different tensors , for n=1,2,3,4, are given in table 3.
It can be observed that the structure of the composition of all has the same properties as the one obtained for in the last section.
Now it is possible to show that tensors constitute a vector space having dimension 8. Therefore, has 10 independent coefficients, which suggests the existence of 10 tensors , such that , and , whenever the value of n=1,…,10. These tensors read 3.10 The contractions between the different tensors , for n=1,…,10, are given in table 4. Note that .
Finally, is within a vector space having dimension 2. So we introduce , such that , and , whenever the value of n=1,…,2. These tensors read 3.11 Note that .
The contractions between the different tensors , for n=1,…,2, are given in table 5. Note that the table of products is the same as for the case of sixth-order tensors.
All eighth-order tensors having minor symmetries can be decomposed by using the irreducible basis . Appendix B produces the relations allowing the components of any eighth-order tensor within the basis to be obtained from its components in the basis . As for the case of a sixth-order tensor, define a monoid for the composition ⊙4, the unit tensor for ⊙4 being .
The following decomposition of the space is used where defines the sub-space of isotropic eighth-order tensors that can be decomposed on the basis of tensors , for n=1,…,4, the sub-space and K8 being, respectively, associated with , for n=1,…,10, and , for n=1,…,2. It can be observed that define a commutative sub-monoid, while and define two sub-monoids.
Let us decompose by using the new basis 3.12 Let us do the same with a second tensor , its components within the new basis being denoted by bi for i=1,…,16. Defining by , its components within the new basis are given by 3.13 We now look for the inverse of an eighth-order tensor . The components of , solution of the equations , are 3.14 with 3.15 The condition for having an inverse is a1a2a12ΔJ1ΔJ2≠0.
4. The case of higher order isotropic tensors
We aim at generalizing the methodology proposed in the previous sections to the case of a 2nth-order tensor for n≥5. To this aim, consider an nth order tensor a, its components being denoted by aijk,…,l. This tensor is assumed symmetric according to its two first indices aijk,…,l=ajik,…,l and also according to its (n−2) last indices k,…,l. Tensor a can be decomposed as follows: 4.1 with n=2p if n is an even number but n=2p+1 if n is an odd number. D(a) is the deviatoric part of a, such that . In the expression above, Sp(a) denotes the pth spherical part of a. Tensors Sp(a) have the properties 4.2 Consequently, we introduce tensors , for p=1,2,3,… and , such that , and . We can define the independent sub-spaces K2n, for p=1,2,3,… associated with and , used for the decomposition of . The second step is to apply the decomposition , . Consequently, , is decomposed as 4.3 The numbers of irreducible tensors , for 2≤n≤6 are given in table 6.
5. Higher order inhomogeneity problem: spherical inclusion in an infinite matrix subjected to a polynomial remote strain field
Eshelby’s ‘inhomogeneity problem’ (Eshelby 1957) is well known for the case of a given constant strain field at infinity: it gives the strain field inside an ellipsoidal inclusion having elastic properties that are different from the material outside the inclusion. This problem uses the solution of the ‘inclusion problem’ for which a constant free deformation is given within an ellipsoidal part of an homogeneous material. The inclusion problem can be extended to the case of free deformations that have a polynomial form, but the solution of the ‘inhomogeneity problem’ for the case of polynomial strain fields at infinity needs the inversion of higher order tensors. In this section, the method used for obtaining the inverse of higher order tensors is used for solving the ‘inhomogeneity problem’ in the case of spherical inhomogeneities made up of an isotropic material and located within an infinite isotropic medium. Let us consider a spherical inclusion located at xi=0 made up of an isotropic elastic material of rigidity Cijkl embedded in an infinite isotropic elastic matrix whose rigidity is . We denote by λ, μ, ν (respectively λ0, μ0, ν0) the Lamé moduli and the Poisson ratio of the inclusion (respectively of the matrix). The inclusion is subjected to a polynomial remote strain field . It has been proved (see Mura (1987) in the case of an infinite isotropic medium and the work of Asaro & Barnett (1975) in an anisotropic context) that the strain field within the inclusion is also a polynomial and reads 5.1 In the following, a series for aij(x) at the second order is considered, aij, aijk and aijkl are solutions of 5.2 with 5.3
In the expression above, and are the components of the Hill tensor (Hill 1975), which are obtained from the components of Eshelby’s tensor and from the components of the inverse Sklmn of the elasticity tensor by . This tensor depends only on the elastic properties of the infinite medium. , and are the components of higher order Hill-type tensors that are introduced by aijk and aijkl. As for the classical Hill’s tensor, they are built from the inverse of the elasticity tensor and from higher order Eshelby’s tensors that can be found in Mura (1987). These tensors can be derived within the basis composed of for both cases of a sixth- and eighth-order tensor and translated into the basis composed of ) by using the base change relations given in appendix B. Note that, once aijkl is determined by solving the last equation in (5.2), one can compute cij for obtaining aij from the first equation in (5.2).
It can be observed that aij, aijk and aijkl are solutions of a linear equation having the form 5.4 for n=2,3,4. Obviously, a closed-form expression of the strain field within the inclusion requires the inversion of tensor , for which it will be convenient to use the formalism introduced in the last sections.
— In the case where a is a second-order tensor, , and are fourth-order isotropic tensors having minor symmetries. The components of b are given by bij=aij−cij. Solution of equation (5.4) is trivial, and can be found in Mura (1987), for instance.
— Consider now the case of a third-order tensor a. The components of the Hill-type tensor written in the basis given in §2, are 5.5
are the components of in the basis , as defined in equation (2.7). The components of are δCijpqδkr. The decomposition in the basis is given by 5.6
The computation of the inverse of , denoted leads to 5.7
— In the case n=4, the Hill-type tensor is decomposed within the basis given in §3. One has 5.8
are the components of in the basis , as defined in equation (3.12). δCijkl is replaced by an equivalent eighth-order tensor whose components are given by δCijklIpqrs, which reads, in the basis , 5.9 Components of are given by These results finalize the closed-form solution of the higher order ‘heterogeneity problems’.
The present study deals with the inversion of an isotropic 2nth-order tensor having particular symmetries (called ‘minor symmetries’). To reach this objective, irreducible bases for isotropic 2nth-order tensors have been provided in the paper. These bases extend the () basis used for isotropic fourth-order tensors. The particular case of sixth- and eighth-order tensors has been examined, and higher order cases have been addressed in §4. The methodology used consists of decomposing third- and fourth-order tensors into their deviatoric and spherical parts, as commonly used in the case of a tensor of order 2. The particularity with tensors of order n≥4 lies in the definition of spherical parts of order 1,2,3,…, while for tensors of order n≤3, only one definition of the spherical part is used (for instance, in the case of a second-order tensor, the first spherical part corresponds to the classical definition). This decomposition of an nth tensor appears to be useful for obtaining the irreducible bases ‘’ for isotropic 2nth-order tensors.
It is shown that the bases ‘’ are useful for effecting the tensorial operations and particularly for the inversion of a 2nth-order tensor. In order to show the relevance of this formalism, we derive the closed-form expression of the strain field within a spherical inclusion subjected to a polynomial remote strain field. This result is an extension of the well-known use of Eshelby’s tensor for obtaining the solution of the heterogeneity problem that is the base of numerous homogenization problems. It suggests that the results of higher order heterogeneity problems could be used for obtaining effective properties in the context of gradient elasticity. This will be developed in a forthcoming paper.
Appendix A. The SFH decomposition
The SFH decomposition of a third-order tensor a (symmetric according to its two first indices) has been introduced by Smyshlyaev & Fleck (1996) and formalized by Fleck & Hutchinson (1997). It reads A1 with A2 where εijk is the permutation symbol.
Appendix B. Base change relations
Let us denote by an, for n=1,…,6, the components of a sixth-order tensor in the basis for n=1,…,6. Let bn, for n=1,…,6, be the components of in the basis as defined in equation (2.7). The relationships giving bn as a function of an are B1 Let us denote by an, for n=1,…,16, the components of an eighth-order tensor on the basis , for n=1,…,16, and let bn, for n=1,…,16, be its components on the basis , as defined in equation (3.12). The relationships giving bn as a function of an are B2
- Received March 16, 2010.
- Accepted May 5, 2010.
- © 2010 The Royal Society