On the Orientation Average Based on Central Orientation Density Functions for Polycrystalline Materials

Abstract

The present work continues the investigation first started by Lobos et al. (J. Elast. 128(1):17–60, 2017) concerning the orientation average of tensorial quantities connected to single-crystal physical quantities distributed in polycrystals. In Lobos et al. (J. Elast. 128(1):17–60, 2017), central orientation density functions were considered in the orientation average for fourth-order tensors with certain index symmetries belonging to single-crystal quantities. The present work generalizes the results of Lobos et al. (J. Elast. 128(1):17–60, 2017) for the orientation average of tensors of arbitrary order by presenting a clear connection to the Fourier expansion of central orientation density functions and of the general orientation density function in terms of tensorial texture coefficients. The closed form of the orientation average based on a central orientation density function is represented in terms of the Fourier coefficients (referred to as texture eigenvalues) and the central orientation of the central orientation density function. The given representation requires the computation of specific isotropic tensors. A pragmatic algorithm for the automated generation of a basis of isotropic tensors is given. Applications and examples are presented to show that the representation of the orientation average offers a low-dimensional parametrization with major benefits for optimization problems in materials science. A simple implementation in Python 3 for the reproduction of all examples is offered through the GitHub repository https://github.com/mauricio-fernandez-l/centralODF-average.

Appendices

Appendix A: Generation of Bases for Isotropic Tensors

Consider the isotropic tensor

$$ \mathbb{D}_{\langle n \rangle } = \int _{\mathsf{SO(3)}}D(\omega ) \boldsymbol{Q}^{\star (n/2)} { \,\mathrm{dQ}} $$

(91)

for some Dirichlet kernel $D(\omega )$, as considered in the present work. The isotropic tensor (91) can be computed with a given basis for the space of isotropic $n$-th-order tensors $\mathsf{T}^{ \mathrm{iso}}_{n}$ with dimension $d^{\mathrm{iso}}_{n}$, given in terms of the Motzkin sum numbers $a_{n}$

$$ d^{\mathrm{iso}}_{n} = a_{n} , \qquad a_{i} = \frac{i-1}{i+1}(2 a _{i-1} + 3 a_{i-2}) , \quad a_{0} = 1, \ a_{1} = 0 , $$

(92)

see, e.g., http://oeis.org/A005043. The generation of basis of $\mathsf{T}^{\mathrm{iso}}_{n}$ is non-trivial for high $n$, see, [4, 21] or [3]. This section offers a pragmatic computational algorithm for the automated generation of a basis, which is easily implemented in Python 3, see [12].

The fundamental isotropic tensor

$$ \mathbb{B}_{{\langle n \rangle }} = \textstyle\begin{cases} \boldsymbol{I}^{\otimes (n/2)} & n \ \text{even} \\ \boldsymbol{\epsilon }\otimes \boldsymbol{I}^{\otimes ((3-n)/2)} & n\ \text{odd} \end{cases} $$

(93)

and the list of permutations $L_{P}$ with a total of $n_{P}$ permutations of indices $(i_{1},i_{2},\dots ,i_{n})$

$$ L_{P} = \{p_{1},p_{2},\dots ,p_{n_{P}}\} , \quad n_{P} = n! $$

(94)

are considered for the generation of isotropic tensors and a basis $B_{n}$. It is shortly noted that, of course, $n_{P}$ is much larger than $d^{\mathrm{iso}}_{n}$, as given in Table 3.

Table 3 Numbers $n_{P}$ and $d^{\mathrm{iso}}_{n}$ for $n = 4, \dots ,12$

The application $\rho$ of a permutation $p \in L_{P}$ on $\mathbb{B} _{{\langle n \rangle }}$ yields, in general, a different but still isotropic tensor

$$ \rho (\mathbb{B}_{\langle n \rangle },p) \in \mathsf{T}^{\mathrm{iso}} _{n} \quad \forall p \in L_{P} . $$

(95)

We say that two permutations $\hat{p}_{1}$ and $\hat{p}_{2}$ are linear independent, if the generated isotropic tensors $\rho (\mathbb{B}_{ \langle n \rangle },\hat{p}_{1})$ and $\rho (\mathbb{B}_{\langle n \rangle },\hat{p}_{2})$ are linear independent. Correspondingly, we say that a permutation list $\hat{L}_{P} \subset L_{P}$ is a permutation basis, if it contains exactly $d^{\mathrm{iso}}_{n}$ linear independent permutations.

The objective of this appendix is to generate a permutation basis computationally and pragmatically. For a straight-forward implementation with vector and matrix operations, each $\hat{\mathbb{B}}_{{\langle n \rangle }i} = \rho (\mathbb{B}_{\langle n \rangle },\hat{p}_{i}), \hat{p}_{i} \in \hat{L}_{P}$ from a candidate list $\hat{L}_{P}$ of length $\hat{n} \geq d^{\mathrm{iso}}_{n}$ can be reshaped into a vector $\hat{\underline{b}}_{i}$ delivering a matrix $\hat{\underline{\underline{B}}}$, defined as

$$ \hat{\underline{b}}_{i} = \mathrm{vec}(\hat{\mathbb{B}}_{{\langle n \rangle }i}) \in \mathsf{R}^{3^{n}}, \qquad \hat{\underline{\underline{B}}}= (\hat{ \underline{b}}_{1} \hat{\underline{b}}_{2} \dots \hat{ \underline{b}}_{\hat{n}}) \in \mathsf{R}^{3^{n} \times \hat{n}} $$

(96)

The candidate list $\hat{L}_{P}$ is a permutation basis if and only if $\mathrm{rank}(\hat{\underline{\underline{B}}}) = d^{\mathrm{iso}} _{n}$. This offers the following pragmatic strategy: start with a candidate list $\hat{L}_{P}$ and keep adding index permutations to it until $\mathrm{rank}(\hat{\underline{\underline{B}}}) = d^{ \mathrm{iso}}_{n}$ holds, then extract the index set of linearly independent columns of $\hat{\underline{\underline{B}}}$ (e.g., based on its QR factorization) and extract the corresponding permutation basis based on this index set. It remains to choose a generation of index permutations to be tested. Since for high $n$, the number of permutations $n_{P} = n!$ is large in comparison to $d^{\mathrm{iso}} _{n}$ and ordered index permutations are highly probable to be linear dependent due to the index symmetries of $\boldsymbol{I}$ and $\boldsymbol{\epsilon }$, the usage of random permutations is taken into account. The straight forward idea is then to generate some sensible/pragmatic number of additional random permutations $\hat{n}_{+} \leq d^{\mathrm{iso}}_{n}$, build the matrix $\hat{\underline{\underline{B}}}$, compute its rank and repeat until $\mathrm{rank}(\hat{\underline{\underline{B}}}) = d^{\mathrm{iso}} _{n}$ holds. This simple approach is given in Algorithm 1.

Algorithm 1

Generation of a permutation basis based on random permutations

Files containing permutation bases up to $n=12$ generated with Algorithm 1 are available at [12].

Now, we assume that a permutation basis $\hat{L}_{P}$ has been found. Define the resulting basis of isotropic tensor as

$$ \hat{B}_{n} = \{\hat{\mathbb{B}}_{{\langle n \rangle }1},\dots \}, \qquad \hat{\mathbb{B}}_{{\langle n \rangle }i} = \rho (\mathbb{B}_{ \langle n \rangle }, \hat{p}_{i}), \quad \hat{p}_{i} \in \hat{L} _{P}. $$

(97)

The isotropic tensor (91) has a representation in terms of the basis $\hat{B}_{n}$

$$ \mathbb{D}_{\langle n \rangle } = \sum_{i=1}^{d^{\mathrm{iso}}_{n}} \hat{c}_{i} \hat{\mathbb{B}}_{{\langle n \rangle }i} $$

(98)

with coefficients $\hat{c}_{i}$. These coefficients are determined through the corresponding linear system

$$ \hat{\underline{D}}= \tilde{\underline{\underline{B}}}\ \hat{\underline{c}}, \qquad \hat{D}_{i} = \mathbb{D}_{\langle n \rangle } \cdot \hat{ \mathbb{B}}_{{\langle n \rangle }i} = \int _{\mathsf{SO(3)}}D(\omega ) \bigl(\boldsymbol{Q}^{\star (n/2)} \cdot \hat{\mathbb{B}}_{{\langle n \rangle }i}\bigr) \,\mathrm{dQ}, \qquad \tilde{B}_{ij} = \hat{\mathbb{B}}_{{\langle n \rangle }i} \cdot \hat{ \mathbb{B}}_{{\langle n \rangle }j}. $$

(99)

The matrix $\tilde{\underline{\underline{B}}}$ is regular and symmetric, such that efficient decompositions can be considered for the solution of the linear system. Alternatively, based on the permutation basis and corresponding basis isotropic tensors, a new set of orthonormal isotropic tensors (or corresponding flattened vectors) may be computed with standard Gram-Schmidt or other orthogonalization approaches to reduced the computational effort and avoid badly conditioned $\tilde{\underline{\underline{B}}}$. Such an orthonormal basis

$$ B_{n} = \{\mathbb{B}_{{\langle n \rangle }1},\dots \} , \qquad \underline{b}_{i} = \mathrm{vec}(\mathbb{B}_{{\langle n \rangle }i}) , \qquad \mathbb{B}_{{\langle n \rangle }i} \cdot \mathbb{B}_{{\langle n \rangle }j} = \underline{b}_{i}^{ \mathrm{T}}\underline{b}_{j} = \delta _{ij} $$

(100)

would immediately deliver the results given in (49) (with $n = 2r$).

Appendix B: Simplification of Integrals for the Computation of $\mathbb{D}_{{\langle 2r \rangle }\alpha }$

As remarked after (49), the scalar

$$ q = \boldsymbol{Q}^{\star (n/2)} \cdot \mathbb{T}_{\langle n \rangle } , \quad \mathbb{T}_{\langle r \rangle } \in \mathsf{T}^{ \mathrm{iso}}_{n} $$

(101)

does not depend on the rotation axis $\boldsymbol{n}$ of $\boldsymbol{Q}= \boldsymbol{Q}(\boldsymbol{n},\omega )$, i.e., $q = q(\boldsymbol{n},\omega ) = q(\tilde{\boldsymbol{n}},\omega )$ for all normalized $\boldsymbol{n},\tilde{\boldsymbol{n}}\in T_{1}$. This can be shown by considering $\tilde{\boldsymbol{n}}= \boldsymbol{R} \boldsymbol{n}$ with an arbitrary rotation $\boldsymbol{R}$ and using the isotropy property of $\mathbb{T}_{\langle n \rangle }$. First, note that

$$ \boldsymbol{Q}(\boldsymbol{R}\boldsymbol{n},\omega ) = \boldsymbol{R} \boldsymbol{Q}(\boldsymbol{n},\omega )\boldsymbol{R}^{\mathrm{T}} $$

(102)

holds. Then, keeping in mind that any transposition of an isotropic tensor is just another isotropic tensor, we represent the Rayleigh power through the tensor power and the corresponding transposition ${\mathrm{T}}_{\star }$ and compute

$$ \boldsymbol{Q}^{\star (n/2)} \cdot \mathbb{T}_{\langle n \rangle } = \bigl( \boldsymbol{Q}^{\otimes (n/2)}\bigr)^{{\mathrm{T}}_{\star }} \cdot \mathbb{T}_{\langle n \rangle } = \boldsymbol{Q}^{\otimes (n/2)} \cdot \mathbb{T}^{*}_{\langle n \rangle } , $$

(103)

where $\mathbb{T}^{*}_{\langle n \rangle }$ denotes the corresponding transposed/permuted isotropic tensor of $\mathbb{T}_{\langle n \rangle }$. Inserting (102) into (103) yields

$$\begin{aligned} \bigl(\boldsymbol{R}\boldsymbol{Q}(\boldsymbol{n},\omega )\boldsymbol{R} ^{\mathrm{T}}\bigr)^{\otimes (n/2)} \cdot \mathbb{T}^{*}_{\langle n \rangle } &= \boldsymbol{Q}(\boldsymbol{n},\omega )^{\otimes (n/2)} \cdot \bigl( \boldsymbol{R}^{\mathrm{T}}\star \mathbb{T}^{*}_{\langle n \rangle } \bigr) \\ &= \boldsymbol{Q}(\boldsymbol{n},\omega )^{\otimes (n/2)} \cdot \mathbb{T}^{*}_{\langle n \rangle } \quad \forall \boldsymbol{R} \in \mathsf{SO(3)}, \end{aligned}$$

(104)

meaning that $q$, given in (101), is independent of $\boldsymbol{n}$, such that we could just use, e.g., $\boldsymbol{n}= \boldsymbol{e}_{1}$ for rapid computations. This, naturally, implies that, if we use the orthonormal basis (100) (with $n=2r$) to compute $\mathbb{D}_{{\langle 2r \rangle }\alpha }$, then the integrals $b_{i}^{\alpha }$ of (49) can be simplified to the one-dimensional integrals

$$ b_{i}^{\alpha }= \mathbb{D}_{{\langle 2r \rangle }\alpha }\cdot \mathbb{B}_{{\langle 2r \rangle }i} = \int _{0}^{\pi }(1+2\alpha ) D _{\alpha }( \omega ) \bigl(\boldsymbol{Q}(\boldsymbol{e}_{1},\omega )^{ \star r} \cdot \mathbb{B}_{{\langle 2r \rangle }i}\bigr) \ 4\pi \ m(\omega ) \,\mathrm{d}\omega , $$

(105)

which are easily treatable in computer algebra systems like Mathematica^®12 or even with numerical modules like Numpy for Python 3, see [12].