Efficient fixed point and Newton–Krylov solvers for FFT-based homogenization of elasticity at large deformations

Abstract

In recent years the FFT-based homogenization method of Moulinec and Suquet has been established as a fast, accurate and robust tool for obtaining effective properties in linear elasticity and conductivity problems. In this work we discuss FFT-based homogenization for elastic problems at large deformations, with a focus on the following improvements. Firstly, we exhibit the fixed point method introduced by Moulinec and Suquet for small deformations as a gradient descent method. Secondly, we propose a Newton–Krylov method for large deformations. We give an example for which this methods needs approximately 20 times less iterations than Newton’s method using linear fixed point solvers and roughly $100$ times less iterations than the nonlinear fixed point method. However, the Newton–Krylov method requires 4 times more storage than the nonlinear fixed point scheme. Exploiting the special structure we introduce a memory-efficient version with 40 % memory saving. Thirdly, we give an analytical solution for the micromechanical solution field of a two-phase isotropic St.Venant–Kirchhoff laminate. We use this solution for comparison and validation, but it is of independent interest. As an example for a microstructure relevant in engineering we discuss finally the application of the FFT-based method to glass fiber reinforced polymer structures.

Green’s operator for large deformations in Fourier space

An efficient implementation of the basic scheme (Algorithm 1) needs two main ingredients. A fast implementation of the discrete Fourier transformations as provided for example by the FFTW⁶ library [20] and an explicit formula for the Fourier coefficients $\hat{\varGamma }^0$ of Green’s operator.

The operators $\nabla ,$ $\mathrm{Div}$ and $G^0$ are understood in Fourier space via

$$\begin{aligned}&\left[ \widehat{\nabla d}(\xi )\right] _{kl} = i \xi _l \hat{d}_k(\xi ),\end{aligned}$$

(54)

$$\begin{aligned}&\left[ \widehat{\mathrm{Div}D}(\xi )\right] _{k} = i \xi _l \hat{D}_{kl}(\xi ),\end{aligned}$$

(55)

$$\begin{aligned}&\left[ \widehat{G^0 d}(\xi )\right] _{k} = \hat{G}^0(\xi )_{kl}\hat{d}_{l}(\xi ), \end{aligned}$$

(56)

for $k,l = 1,\ldots ,d$, where $\xi = 2\pi z/L$, $z \in {\mathbb {Z}}^d$, denotes a wave vector with the short notation $z/L=(z_1/L_1,\ldots z_d/L_d)$. Therefore, in Fourier space the constitutive Eq. (4) and the equilibrium condition (1) have the form

$$\begin{aligned}&\hat{P}_{kL}(\xi ) = i {\mathbb {C}}^0_{kLmN} \xi _N \hat{u}_m(\xi ) + \hat{\tau }_{kL}(\xi ), \end{aligned}$$

(57)

$$\begin{aligned}&i \hat{P}_{kL}(\xi ) \xi _L = 0, \end{aligned}$$

(58)

Eliminating $\hat{P}_{kL}$ yields

$$\begin{aligned} {\mathbb {C}}^0_{kLmN} \xi _L \xi _N \hat{u}_m(\xi ) = i \hat{\tau }_{kL}(\xi ) \xi _L. \end{aligned}$$

(59)

Due to the definition of the solution operator $G^0$ this implies

$$\begin{aligned}&(\hat{G}^0)^{-1}_{km}(\xi ) = {\mathbb {C}}^0_{kLmN} \xi _L \xi _N, \end{aligned}$$

(60)

$$\begin{aligned}&\hat{u}_m(\xi ) = \hat{G}^0_{mk}(\xi ) i \hat{\tau }_{kL}\xi _L. \end{aligned}$$

(61)

Using (9) yields additionally

$$\begin{aligned} \hat{\varGamma }^0_{kLmN}(\xi ) = \xi _L \xi _N \hat{G}^0_{km}(\xi ). \end{aligned}$$

(62)

In the case of an isotropic reference material ${\mathbb {C}}^0$ with Lamé moduli $\lambda _0$ and $\mu _0$, i.e. ${\mathbb {C}}^0=\lambda _0 I \otimes I + 2 \mu _0 {\mathbb {I}}^S$ or ${\mathbb {C}}^0_{kLmN} = \lambda _0 \delta _{kL}\delta _{mN} + \mu _0 (\delta _{km}\delta _{LN} + \delta _{kN}\delta _{Lm})$ for $k,L,m,N = 1,\ldots ,d$, the Fourier coefficients $\hat{G}^0$ of the solution operator read (cf. [33])

$$\begin{aligned}&(\hat{G}^0)^{-1}_{km}(\xi ) \!=\! \lambda _0 \delta _{kL}\delta _{mN} \xi _L \xi _N \!+\! \mu _0 (\delta _{km}\delta _{LN} \!+\! \delta _{kN}\delta _{Lm})\xi _L \xi _N, \end{aligned}$$

(63)

$$\begin{aligned}&(\hat{G}^0)^{-1}(\xi ) = (\lambda _0 + \mu _0) \xi \otimes \xi + \mu _0 |\xi |^2 I,\end{aligned}$$

(64)

$$\begin{aligned}&\hat{G}^0(\xi ) = \frac{(\lambda _0 + 2 \mu _0) |\xi |^2 I - (\lambda _0 + \mu _0) \xi \otimes \xi }{\mu _0(\lambda _0+2\mu _0)|\xi |^4}. \end{aligned}$$

(65)

By applying (62) we arrive at an explicit formula for the Fourier coefficients of the Green’s operator

$$\begin{aligned}&\hat{\varGamma }^0(\xi )\hat{\tau } = \frac{\hat{\tau }\xi \otimes \xi }{\mu _0 |\xi |^2} -\frac{\lambda _0+\mu _0}{\mu _0\left( \lambda _0+2\mu _0\right) } \frac{\hat{\tau }\xi \cdot \xi }{|\xi |^4} \xi \otimes \xi , \end{aligned}$$

(66)

$$\begin{aligned}&\hat{\varGamma }^0_{kLmN}(\xi ) = \frac{\delta _{km}\xi _L\xi _N}{\mu _0 |\xi |^2} - \frac{\lambda _0+\mu _0}{\mu _0\left( \lambda _0+2\mu _0\right) } \frac{\xi _k \xi _L \xi _m \xi _N}{|\xi |^4}. \end{aligned}$$

(67)

Symmetrizing $\hat{\varGamma }^0$ gives the Fourier coefficients of Green’s operator for linear elasticity.

If the isotropic reference material ${\mathbb {C}}^0$ with Lamé moduli $\lambda _0$ and $\mu _0$ is not symmetrized (cf. Sects. 3.1 and 3.2.5), i.e. ${\mathbb {C}}^0=\lambda _0 I \otimes I + 2 \mu _0 {\mathbb {I}}$ or ${\mathbb {C}}^0_{kLmN} = \lambda _0 \delta _{kL}\delta _{mN} + 2\mu _0 \delta _{km}\delta _{LN}$ for $k,L,m,N = 1,\ldots ,d$, the Fourier coefficients $\hat{G}^0$ of the solution operator read

$$\begin{aligned}&(\hat{G}^0)^{-1}_{km}(\xi ) = \lambda _0 \delta _{kL}\delta _{mN} \xi _L \xi _N + 2 \mu _0 \delta _{km}\delta _{LN}\xi _L \xi _N, \end{aligned}$$

(68)

$$\begin{aligned}&(\hat{G}^0)^{-1}(\xi ) = \lambda _0 \xi \otimes \xi + 2\mu _0 |\xi |^2 I,\end{aligned}$$

(69)

$$\begin{aligned}&\hat{G}^0(\xi ) = \frac{(\lambda _0 + 2 \mu _0) |\xi |^2 I - \lambda _0 \xi \otimes \xi }{2\mu _0(\lambda _0+2\mu _0)|\xi |^4}. \end{aligned}$$

(70)

By applying (62) we arrive at an explicit formula for the Fourier coefficients of the Green’s operator

$$\begin{aligned}&\hat{\varGamma }^0(\xi )\hat{\tau } = \frac{\hat{\tau }\xi \otimes \xi }{2\mu _0 |\xi |^2} - \frac{\lambda _0}{2\mu _0\left( \lambda _0+2\mu _0\right) } \frac{\hat{\tau }\xi \cdot \xi }{|\xi |^4} \xi \otimes \xi , \end{aligned}$$

(71)

$$\begin{aligned}&\hat{\varGamma }^0_{kLmN}(\xi ) = \frac{\delta _{km}\xi _L\xi _N}{2\mu _0 |\xi |^2} - \frac{\lambda _0}{2\mu _0\left( \lambda _0+2\mu _0\right) } \frac{\xi _k \xi _L \xi _m \xi _N}{|\xi |^4}. \end{aligned}$$

(72)