高阶多尺度方法理论推导

符号约定

\(n\) 维欧几里得空间定义为 \(\mathbb{E}^{n}\), 如无特别说明, 方程或函数的定义域 \(\Omega \subset \mathbb{E}^{3}\).

欧式空间中的元素使用粗体小写字母 \(\boldsymbol{x}\) 或者 \(\boldsymbol{y}\) 表示.

一般的 \(n\) 阶张量空间表示为 \(\mathbb{T}^{n}\), 特别的, 二阶对称张量空间记作 \(\mathbb{S}\). 二阶张量一般使用斜粗体 \(\boldsymbol{T}\) 表示, 四阶张量使用黑板粗体 \(\mathbb{T}\) 表示, 其它阶次的张量使用无衬线字体 \(\mathsf{T}\) 表示.

对于高阶张量, 点积的意义会与缩并指标的顺序相关, 点积 \(\cdot\) 定义为 \[ \boldsymbol{A} \cdot \boldsymbol{B} \triangleq A_{...i} B_{i...} \] 点积 \(:\) 定义为 \[ \boldsymbol{A} : \boldsymbol{B} \triangleq A_{...ij} B_{ij...} \] 点积 \(\therefore\) 定义为 \[ \boldsymbol{A} \therefore \boldsymbol{B} \triangleq A_{...ijk} B_{ijk...} \]

函数空间 \(L(\Omega)\) 中的元素 \(f: \Omega \mapsto \mathbb{T}^{n}\) 是定义在 \(\Omega\) 上的函数, 其值域是 \(n\) 阶张量.

周期性函数空间 \(L(\mathbb{E}^{3})\) 可以看作是定义在单胞区域 \(\Theta\) 的函数到全空间 \(\mathbb{E}^{3}\) 的延拓.

对于张量场 \(\boldsymbol{T}: \Omega \mapsto \mathbb{T}^{n}\), 它的梯度定义为 \[ \mathrm{grad}\ \boldsymbol{T} \triangleq T_{...,i} \ \underbrace{\boldsymbol{e}_{i_1} \otimes \cdots \otimes \boldsymbol{e}_{i_n} \otimes \boldsymbol{e}_{i}}_{n+1} \] 散度定义为 \[ \mathrm{div}\ \boldsymbol{T} \triangleq T_{..i_{n},i_{n}} \ \underbrace{\boldsymbol{e}_{i_1} \otimes \cdots \otimes \boldsymbol{e}_{n-1}}_{n-1} \] 方程算子 \(\mathcal{A}: L \mapsto L\) 使用花体表示. 特别的, 使用 \(\nabla\) 表示偏导数算子, 由此可将张量场的梯度和散度重写为 \[ \mathrm{grad}\ \boldsymbol{T} \leftrightarrow \boldsymbol{T} \nabla, \quad \mathrm{div}\ \boldsymbol{T} \leftrightarrow \boldsymbol{T} \cdot \nabla \] 定义 \[ \nabla^{s} \boldsymbol{T} \triangleq \frac{1}{2} ( T_{i \ldots,j} + T_{j \ldots,i} ) \] 特别的, 对于一阶张量有 \[ \nabla^{s} \boldsymbol{u} \triangleq \frac{1}{2} ( \boldsymbol{u} \nabla + \nabla \boldsymbol{u} ) \]

预备定理

对于定义在单胞区域 \(\Theta\) 的方程 \[ - \nabla \cdot [A(\nabla u)] = f \] 方程的弱形式可以描述为 \[ \mathrm{find} \ u \in \mathcal{W}_{\mathrm{per}}(\Theta) \ \mathrm{s.t.} \\ \int_{\Theta} \nabla v \cdot (A \nabla u) \ \mathrm{d} y = \int_{\Theta} f v \ \mathrm{d} y, \quad \forall v \in \mathcal{W}_{\mathrm{per}}(\Theta) \] 其中, 函数空间 \(\mathcal{W}_{\mathrm{per}}(\Theta)\) 定义为 \[ \mathcal{W}_{\mathrm{per}}(\Theta) \triangleq \{ u \in {H}_{\mathrm{per}}^{1}(\Theta) \mid \int_{\Theta} u \ \mathrm{d}y = 0 \} \] 上述弱形式有唯一解 (在等价类意义下) 的充分必要条件是方程右端项在单胞上的积分等于 0: \[ \boxed{\int_{\Theta} f \ \mathrm{d}y = 0} \tag{1} \] 若函数 \(g\) 是 \(\Theta\)-周期性函数, 那么该函数的偏导数在单胞域内的积分总是等于 0: \[ \boxed{\int_{\Theta} \frac{ \partial g }{ \partial y_{i} } \ \mathrm{d} y = 0} \tag{2} \]

控制方程

控制方程的指标形式为 \[ \begin{aligned} &\begin{array}{ll} \sigma_{i j, j}^\zeta(\boldsymbol{x})+f_i^\zeta(\boldsymbol{x})=0, & \boldsymbol{x} \in \Omega \\ \sigma_{i j}^\zeta(\boldsymbol{x})=L_{i j k l}^\zeta(\boldsymbol{x} / \zeta)\left(\varepsilon_{k l}^\zeta(\boldsymbol{x})-\mu_{k l}^\zeta(\boldsymbol{x})\right), & \boldsymbol{x} \in \Omega \\ \varepsilon_{k l}^\zeta(\boldsymbol{x})=u_{\left(k, l\right)}^\zeta(\boldsymbol{x}) \equiv \frac{1}{2}\left(u_{k, l}^\zeta(\boldsymbol{x})+u_{l, k}^\zeta(\boldsymbol{x})\right), & \boldsymbol{x} \in \Omega \end{array}\\ &\begin{array}{ll} u_i^\zeta=\bar{u}_i(\boldsymbol{x}), & \boldsymbol{x} \in \Gamma_u \\ n_i \sigma_{i j}^\zeta(\boldsymbol{x})=\bar{t}_i(\boldsymbol{x}), & \boldsymbol{x} \in \Gamma_t \end{array} \end{aligned} \] 将其中的微分方程写成张量形式 \[ - (\mathbb{L}^{\zeta} : \nabla^{s} \boldsymbol{u}^{\zeta} ) \cdot \nabla = \boldsymbol{f}^{\zeta} - (\mathbb{L}^{\zeta} : \boldsymbol{\mu}^{\zeta}) \cdot \nabla \] 定义如下算子: \[ \mathcal{A}^{\zeta} \square \triangleq - (\mathbb{L}^{\zeta} : \nabla^{s} \square) \cdot \nabla, \quad \mathcal{B}^{\zeta} \square \triangleq - (\mathbb{L}^{\zeta} : \square) \cdot \nabla \] 因此微分方程写成算子的形式为 \[ \mathcal{A}^{\zeta} \boldsymbol{u}^{\zeta} = \boldsymbol{f}^{\zeta} + \mathcal{B}^{\zeta} \boldsymbol{\mu}^{\zeta} \] 方程系数 \(\mathbb{L}\) 满足 \[ \boxed{\mathbb{L}^{\zeta}(\boldsymbol{x}) = \mathbb{L}(\frac{\boldsymbol{x}}{\zeta}), \quad \boldsymbol{x} \in \Omega} \] 式中, \(\mathbb{L}\) 是 \(\Theta\)-周期性函数.

渐进展开式

假设有如下独立的渐进展开式: \[ \boldsymbol{u}^{\zeta}(\boldsymbol{x}) = \boldsymbol{u}^{(0)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \zeta \boldsymbol{u}^{(1)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \zeta^2 \boldsymbol{u}^{(2)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \cdots \\ \boldsymbol{\mu}^{\zeta}(\boldsymbol{x}) = \boldsymbol{\mu}^{(0)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \zeta \boldsymbol{\mu}^{(1)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \zeta^2 \boldsymbol{\mu}^{(2)}(\boldsymbol{x},\frac{\boldsymbol{x}}{\zeta}) + \cdots \] \(\boldsymbol{u}^{(i)}:\Omega \times \Theta \mapsto \mathbb{T}^1\), \(\boldsymbol{\mu}^{(i)}:\Omega \times \Theta \mapsto \mathbb{S}\), 并对坐标 \(y\in\Theta\) 满足 \(\Theta\)-周期性. 若函数 \(\hat{f}(\boldsymbol{x}): \Omega \mapsto \mathbb{R}\) 定义为 \[ \hat{f}(\boldsymbol{x}) \triangleq f(\boldsymbol{x}, \frac{\boldsymbol{x}}{\zeta}), \quad f: \Omega \times \Theta \mapsto \mathbb{R} \] 那么根据链式法则, 函数 \(\hat{f}\) 对 \(\boldsymbol{x}\) 的偏导数为 \[ \frac{ \partial \hat{f} }{ \partial x_{i} } = \frac{ \partial f }{ \partial x_{i} } + \frac{1}{\zeta} \frac{ \partial f }{ \partial y_{i} } \] 记成算子形式为 \[ \nabla = \nabla_{x} + \frac{1}{\zeta} \nabla_{y} \] 所以, 将渐进展开式代入之后, 算子 \(\mathcal{A}^{\zeta}\) 可以重新表示为 \[ \boxed{\mathcal{A}^{\zeta} = \frac{1}{\zeta^2} \mathcal{A}_{0} + \frac{1}{\zeta} \mathcal{A}_{1} + \mathcal{A}_{0}, \quad \mathcal{B}^{\zeta} = \frac{1}{\zeta} \mathcal{B}_{0} + \mathcal{B}_{1}} \] 式中, \[ \begin{aligned} \mathcal{A}_{0} \square &\triangleq - (\mathbb{L} : \nabla_{y}^{s} \square) \cdot \nabla_{y} \\ \mathcal{A}_{1} \square &\triangleq - (\mathbb{L} : \nabla_{y}^{s} \square) \cdot \nabla_{x} - (\mathbb{L} : \nabla_{x}^{s} \square) \cdot \nabla_{y}\\ \mathcal{A}_{2} \square &\triangleq - (\mathbb{L} : \nabla_{x}^{s} \square) \cdot \nabla_{y} \\ \mathcal{B}_{0} \square &\triangleq - (\mathbb{L} : \square) \cdot \nabla_{y} \\ \mathcal{B}_{1} \square &\triangleq - (\mathbb{L} : \square) \cdot \nabla_{x} \end{aligned} \]

将渐进展开式代入到微分方程中得到 \[ \begin{aligned} \mathcal{O}(\zeta^{-2}) &: \mathcal{A}_{0} \boldsymbol{u}^{(0)} = \boldsymbol{0} \\ \mathcal{O}(\zeta^{-1}) &: \mathcal{A}_{0} \boldsymbol{u}^{(1)} = -\mathcal{A}_{1} \boldsymbol{u}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(0)} \\ \mathcal{O}(1) &: \mathcal{A}_{0} \boldsymbol{u}^{(2)} = -\mathcal{A}_{2} \boldsymbol{u}^{(0)} -\mathcal{A}_{1} \boldsymbol{u}^{(1)} + \mathcal{B}_{1} \boldsymbol{\mu}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(1)} + \boldsymbol{f} \end{aligned} \]

不同阶次的控制方程求解

\(\mathcal{O}(\zeta^{-2})\) 阶方程

\[ \mathcal{A}_{0} \boldsymbol{u}^{(0)} = \boldsymbol{0} \]

方程解的唯一性

方程右端项积分等于 0, 因此方程有唯一解.

方程求解

观察到与细观尺度 \(y\) 无关的常数是方程的一个解, 所以 \[ \boxed{\boldsymbol{u}^{(0)}(\boldsymbol{x},\boldsymbol{y}) = \boldsymbol{u}^{(0)}(\boldsymbol{x})} \tag{3} \] ### \(\mathcal{O}(\zeta^{-1})\) 阶方程

\[ \mathcal{A}_{0} \boldsymbol{u}^{(1)} = -\mathcal{A}_{1} \boldsymbol{u}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(0)} \]

方程解的唯一性

根据定理 (1), 保证该方程有唯一解的充要条件是: \[ \int_{\Theta} -\mathcal{A}_{1} \boldsymbol{u}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(0)} \ \mathrm{d}y = 0, \quad \forall x \]

也即 \[ \int_{\Theta} (\mathbb{L} : \nabla_{x}^{s} \boldsymbol{u}^{(0)}) \cdot \nabla_{y} \ \mathrm{d}y - \int_{\Theta} (\mathbb{L} : \boldsymbol{\mu}^{(0)}) \cdot \nabla_{y} \ \mathrm{d}y = 0 \]

将 \(x\) 视作参数, 又因为 \(\nabla_{x}^{s} \boldsymbol{u}^{(0)}\) 与 \(y\) 无关, 而 \(\boldsymbol{\mu}^{(0)}(\cdot,y)\) 是 \(\Theta\)​-周期函数, 根据定理 (2), 上式恒等于 0, 所以方程有唯一解的条件满足.

方程求解

定义宏观应变 \(\boldsymbol{\varepsilon}^{c} \in \mathbb{S}\): \[ \boldsymbol{\varepsilon}^{c} \triangleq \nabla_{x}^{s} \boldsymbol{u}^{(0)} \] 假设方程解 \(\boldsymbol{u}^{(1)}\) 具有如下形式: \[ \boxed{\boldsymbol{u}^{(1)}(x,y) = \mathsf{H} : \boldsymbol{\varepsilon}^{c} + \int_{\Theta} \mathsf{h} : \boldsymbol{\mu}^{(0)} \ \mathrm{d} y} \] 式中, \(\mathsf{H}: \Theta \mapsto \mathbb{T}^3\) , \(\mathsf{h}: \Theta \times \Theta \mapsto \mathbb{T}^3\). 张量 \(\mathsf{H}\) 和 \(\mathsf{h}\) 具有如下对称性 \[ H_{imn} = H_{inm}, \quad h_{imn} = h_{inm} \] 单胞上的本征应变采用分片常值形函数进行插值近似: \[ \boldsymbol{\mu}^{(0)}(x,y) \approx \sum_{\alpha=1}^{N} \mathbb{I}^{(\alpha)}(y) : \boldsymbol{\mu}^{(0,\alpha)}(x) \] 所以 \[ \boxed{\boldsymbol{u}^{(1)}(x,y) \approx \mathsf{H} : \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y : \boldsymbol{\mu}^{(0,\alpha)} } \] 代入方程中有 \[ \mathcal{A}_{0} \left( \mathsf{H} : \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y : \boldsymbol{\mu}^{(0,\alpha)} \right) = -\mathcal{A}_{1} \boldsymbol{u}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(0)} \] 整理得到 \[ \mathcal{A}_{0} \mathsf{H} : \boldsymbol{\varepsilon}^{c} + \mathcal{A}_{1} \boldsymbol{u}^{(0)} + \sum_{\alpha=1}^{N} \left( \mathcal{A}_{0} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y - \mathcal{B}_{0} \mathbb{I}^{(\alpha)} \right) : \boldsymbol{\mu}^{(0,\alpha)} = 0 \] 由 \(\boldsymbol{\varepsilon}^{c}\) 的任意性, 得到一组关于 \(\mathsf{H}\) 的线性方程 \[ (\mathbb{L} : \nabla_{y}^{s} \mathsf{H} : \boldsymbol{\lambda}) \cdot \nabla_{y} + (\mathbb{L} : \boldsymbol{\lambda}) \cdot \nabla_{y} = \boldsymbol{0}, \quad \forall \boldsymbol{\lambda} \in \mathbb{S} \] 以及 \(\boldsymbol{\mu}^{(0,\alpha)}\) 的任意性, 得到关于 \(\int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y\) 的线性方程 \[ \left\{ \mathbb{L} : \nabla_{y}^{s} \left( \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) : \boldsymbol{\lambda} - \mathbb{L} : \mathbb{I}^{(\alpha)} : \boldsymbol{\lambda} \right\} \cdot \nabla_{y} = \boldsymbol{0}, \quad \forall \boldsymbol{\lambda} \in \mathbb{S} \] 定义 \(\boldsymbol{H}_{\lambda}, \boldsymbol{h}_{\lambda}^{(\alpha)}: \Theta \mapsto \mathbb{T}^1\) \[ \boldsymbol{H}_{\lambda} \triangleq \mathsf{H} : \boldsymbol{\lambda}, \quad \boldsymbol{h}_{\lambda}^{(\alpha)} \triangleq \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y : \boldsymbol{\lambda} \] 又因为方程是线性的, 所以若选取空间 \(\mathbb{S}\) 的一组规范正交基 \(\boldsymbol{e}_{i}\), 并定义 \[ \boldsymbol{H}_{i} \triangleq \mathsf{H} : \boldsymbol{e}_{i}, \quad \boldsymbol{h}_{i}^{(\alpha)} \triangleq \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y : \boldsymbol{e}_{i}, \quad i = 1,2, \ldots , \mathrm{dim}\ \mathbb{S} \] 那么对任意的 \(\boldsymbol{\lambda} = \lambda_{i} \boldsymbol{e}_{i} \in \mathbb{S}\) \[ \boldsymbol{H}_{\lambda} = \lambda_{i} \boldsymbol{H}_{i}, \quad \boldsymbol{h}_{\lambda}^{(\alpha)} = \lambda_{i} \boldsymbol{h}_{i}^{(\alpha)} \] ### \(\mathcal{O}(1)\) 阶方程

\[ \mathcal{A}_{0} \boldsymbol{u}^{(2)} = -\mathcal{A}_{2} \boldsymbol{u}^{(0)} -\mathcal{A}_{1} \boldsymbol{u}^{(1)} + \mathcal{B}_{1} \boldsymbol{\mu}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(1)} + \boldsymbol{f} \]

方程解的唯一性

方程有唯一解的条件为 \[ \langle -\mathcal{A}_{2} \boldsymbol{u}^{(0)} -\mathcal{A}_{1} \boldsymbol{u}^{(1)} + \mathcal{B}_{1} \boldsymbol{\mu}^{(0)} + \mathcal{B}_{0} \boldsymbol{\mu}^{(1)} + \boldsymbol{f} \rangle = 0 \] 而 \[ \begin{aligned} \langle -\mathcal{A}_{2} \boldsymbol{u}^{(0)} \rangle &= (\mathbb{L} : \boldsymbol{\varepsilon}^{c}) \cdot \nabla_{x} \\ \langle -\mathcal{A}_{1} \boldsymbol{u}^{(1)} \rangle &= \langle \mathbb{L} : \nabla_{y}^{s} \boldsymbol{u}^{(1)} \rangle \cdot \nabla_{x} , \\ &\langle \mathbb{L} : \nabla_{y} \boldsymbol{u}^{(1)} \rangle = \langle \mathbb{L} : \nabla_{y}^{s} \mathsf{H} \rangle : \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \langle \mathbb{L} : \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \rangle :\boldsymbol{\mu}^{(0,\alpha)} \\ \langle \mathcal{B}_{1} \boldsymbol{\mu}^{(0)} \rangle &= - \left( \sum_{\alpha=1}^{N} \langle \mathbb{L} : \mathbb{I}^{(\alpha)} \rangle : \boldsymbol{\mu}^{(0,\alpha)} \right) \cdot \nabla_{x} \\ \langle \mathcal{B}_{0} \boldsymbol{\mu}^{(1)} \rangle &= \boldsymbol{0} \\ \langle \boldsymbol{f} \rangle &= \boldsymbol{f} \end{aligned} \] 得到宏观尺度上, 考虑本征应变的控制方程. \[ \left\{ \left[ \mathbb{L} + \langle \mathbb{L} : \nabla_{y}^{s} \mathsf{H} \rangle \right] : \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \left\langle \mathbb{L} : \left( \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y - \mathbb{I}^{(\alpha)} \right) \right\rangle :\boldsymbol{\mu}^{(0,\alpha)} \right\} \cdot \nabla_{x} + \boldsymbol{f} = \boldsymbol{0} \] 定义 \[ \mathbb{E} \triangleq \nabla_{y}^{s} \mathsf{H} + \mathbb{I}, \quad \mathbb{P}^{(\alpha)} \triangleq \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y, \quad \mathbb{A}^{(\alpha)} \triangleq \mathbb{L} : (\mathbb{P}^{(\alpha)} - \mathbb{I}^{(\alpha)}) \] 就得到 \[ \left( \langle \mathbb{L} : \mathbb{E} \rangle : \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \langle \mathbb{A}^{(\alpha)} \rangle : \boldsymbol{\mu}^{(0,\alpha)} \right) \cdot \nabla_{x} + \boldsymbol{f} = 0 \]

所以, 方程有唯一解的条件为在宏观尺度下满足上述控制方程.

方程求解

整理方程右端项 \[ \begin{aligned} -\mathcal{A}_{2} \boldsymbol{u}^{(0)} &= \mathbb{L} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c}\\ -\mathcal{A}_{1} \boldsymbol{u}^{(1)} &= \left( \mathbb{L} : \nabla_{y}^{s} \mathsf{H} \right) \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \left( \mathbb{L} : \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \\ &+ \nabla_{y} \cdot ( \mathbb{L} \cdot \mathsf{H} ) \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \nabla_{y} \cdot \left( \mathbb{L} \cdot \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)}\\ \mathcal{B}_{1} \boldsymbol{\mu}^{(0)} &= - \sum_{\alpha=1}^{N} \left( \mathbb{L} : \mathbb{I}^{(\alpha)} \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \\ \mathcal{B}_{0} \boldsymbol{\mu}^{(1)} &= - \left( \mathbb{L} : \boldsymbol{\mu}^{(1)} \right) \cdot \nabla_{y} \\ \boldsymbol{f} &= - \mathbb{L}^{c} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} - \sum_{\alpha=1}^{N} \langle \mathbb{A}^{(\alpha)} \rangle \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \end{aligned} \] 由此得到 \[ \begin{aligned} \mathcal{A}_{0} \boldsymbol{u}^{(2)} &= \mathbb{L} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c}\\ &+ \left( \mathbb{L} : \nabla_{y}^{s} \mathsf{H} \right) \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \left( \mathbb{L} : \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \\ & + \nabla_{y} \cdot ( \mathbb{L} \cdot \mathsf{H} ) \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \nabla_{y} \cdot \left( \mathbb{L} \cdot \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \\ &- \sum_{\alpha=1}^{N} \left( \mathbb{L} : \mathbb{I}^{(\alpha)} \right) \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \\ &- \left( \mathbb{L} : \boldsymbol{\mu}^{(1)} \right) \cdot \nabla_{y} \\ &- \mathbb{L}^{c} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} - \sum_{\alpha=1}^{N} \langle \mathbb{A}^{(\alpha)} \rangle \therefore \nabla_{x} \boldsymbol{\mu}^{(0,\alpha)} \end{aligned} \] 或者 \[ \boxed{ \begin{aligned} \mathcal{A}_{0} \boldsymbol{u}^{(2)} &= \left\{ ( \mathbb{L} : \mathbb{E} - \langle \mathbb{L} : \mathbb{E} \rangle ) + \nabla_{y} \cdot ( \mathbb{L} \cdot \mathsf{H} )\right\} \therefore \boxed{\nabla_{x} \boldsymbol{\varepsilon}^{c}} \\ &+ \sum_{\alpha=1}^{N} \left\{ \left( \mathbb{A}^{(\alpha)} - \langle \mathbb{A}^{(\alpha)} \rangle \right) + \nabla_{y} \cdot \left( \mathbb{L} \cdot \int_{\Theta^{(\alpha)}} \mathsf{h}\ \mathrm{d} y \right) \right\} \therefore \boxed{\nabla_{x} \boldsymbol{\mu}^{(0,\alpha)}} \\ &- \left( \mathbb{L} : \boldsymbol{\mu}^{(1)} \right) \cdot \nabla_{y} \end{aligned} } \]

假设位移的表达式为 \[ \boldsymbol{u}^{(2)}(x,y) = \mathbb{W} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \int_{\Theta} \mathbb{w} \therefore \nabla_{x} \boldsymbol{\mu}^{(0)} \ \mathrm{d} y + \int_{\Theta} \mathsf{h} : \boldsymbol{\mu}^{(1)} \ \mathrm{d} y \]

式中, \(\mathbb{W}\) 和 \(\mathbb{w}\) 是四阶张量, 并且 3,4 指标具有对称性. 对本征应变梯度和一阶本征应变使用相同的分片常值插值函数 \[ \boldsymbol{u}^{(2)}(x,y) \approx \mathbb{W} \therefore \nabla_{x} \boldsymbol{\varepsilon}^{c} + \sum_{\alpha=1}^{N} \int_{\Theta^{(\alpha)}} \mathbb{w} \ \mathrm{d} y \therefore \nabla_{x} \boldsymbol{\mu}^{(0, \alpha)} + \sum_{\alpha=1}^{N} \int_{\Theta^{(\alpha)}} \mathsf{h} \ \mathrm{d} y : \boldsymbol{\mu}^{(1, \alpha)} \] 代入到方程中, 再根据 \(\nabla_{x} \boldsymbol{\varepsilon}^{c}\) 和 \(\nabla_{x} \boldsymbol{\mu}^{(0,\alpha)}\) 的任意性, 得到方程组 \[ - \nabla_{y} \cdot (\mathbb{L} : \nabla_{y}^{s} \mathbb{W} \therefore \mathsf{e}) = ( \mathbb{L} : \mathbb{E} - \langle \mathbb{L} : \mathbb{E} \rangle) \therefore \mathsf{e} + \nabla_{y} \cdot ( \mathbb{L} \cdot \mathsf{H} ) \therefore \mathsf{e}, \quad \forall \mathsf{e} \in \mathbb{T}^{3} \]

\[ - \nabla_{y} \cdot \left( \mathbb{L} : \nabla_{y}^{s} \int_{\Theta^{(\alpha)}} \mathbb{w} \ \mathrm{d} y \therefore \mathsf{e} \right) = \left( \mathbb{A}^{(\alpha)} - \langle \mathbb{A}^{(\alpha)} \rangle \right) \therefore \mathsf{e} + \nabla_{y} \cdot \left( \mathbb{L} \cdot \int_{\Theta^{(\alpha)}} \mathsf{h} \ \mathrm{d} y \right) \therefore \mathsf{e}, \quad \forall \mathsf{e} \in \mathbb{T}^{3} \]

有限元离散化

\[ \left[ L_{ijkl} \left( W_{(k,{l})}^{mnp} + \delta_{lp} H_{k}^{mn} \right) \right]_{,{j}} = \overline{L}_{ijkl} I_{kl}^{mn} \delta_{jp} - L_{ijkl} E_{kl}^{mn} \delta_{jp}, \quad \boldsymbol{y} \in \Theta \]

推导该方程的弱形式, 方程两边乘函数 \(\delta v_{i}\), 然后再单胞上进行积分得到 \[ \int_{\Theta} \delta v_{i} \left[ L_{ijkl} \left( W_{(k,{l})}^{mnp} + \delta_{lp} H_{k}^{mn}\right) \right]_{,{j}} \mathrm{d} \boldsymbol{y}\\ = \int_{\Theta} \delta v_{i} \overline{L}_{ijkl} I_{kl}^{mn} \delta_{jp} - \delta v_{i} L_{ijkl} E_{kl}^{mn} \delta_{jp} \mathrm{d} \boldsymbol{y} \] 方程的左端项使用分部积分公式得到 \[ \mathrm{LHS} = - \int_{\Theta} \delta v_{i,j} L_{ijkl} \left( W_{(k,{l})}^{mnp} + \delta_{lp} H_{k}^{mn}\right) \mathrm{d} \boldsymbol{y} \\ + \int_{\partial \Theta} \delta v_{i} n_{j} L_{ijkl} \left( W_{(k,{l})}^{mnp} + \delta_{lp} H_{k}^{mn}\right) \mathrm{d} \boldsymbol{y} \] 在边界上的积分项可以根据周期性边界条件消去, 这就得到 \[ \int_{\Theta} \delta v_{i,j} L_{ijkl} W_{(k,{l})}^{mnp} \mathrm{d} \boldsymbol{y} = - \int_{\Theta} \delta v_{i,j} L_{ijkl} \delta_{lp} H_{k}^{mn} \mathrm{d} \boldsymbol{y} \\ + \int_{\Theta} \delta v_{i} \left( L_{ijkl} E_{kl}^{mn} - \overline{L}_{ijkl} I_{kl}^{mn} \right) \delta_{jp} \mathrm{d} \boldsymbol{y} \]

\[ \mathrm{find} \ \boldsymbol{H}_{i} \in \mathcal{W}_{\mathrm{per}}(\Theta) \quad \mathrm{s.t.} \\ \int_{\Theta} \nabla_{y} \boldsymbol{v} : \mathbb{L} : \nabla_{y} \boldsymbol{H}_{i} \ \mathrm{d} y = - \int_{\Theta} \nabla_{y} \boldsymbol{v} : \mathbb{L} : \boldsymbol{e}_{i} \ \mathrm{d}y, \quad \forall \boldsymbol{v} \in \mathcal{W}_{\mathrm{per}}(\Theta) \]

\[ \mathrm{find} \ \boldsymbol{h}_{i}^{(\alpha)} \in \mathcal{W}_{\mathrm{per}}(\Theta) \quad \mathrm{s.t.} \\ \int_{\Theta} \nabla_{y} \boldsymbol{v} : \mathbb{L} : \nabla_{y} \boldsymbol{h}_{i}^{(\alpha)} \ \mathrm{d} y = \int_{\Theta} \nabla_{y} \boldsymbol{v} : \mathbb{L} : \mathbb{I}^{(\alpha)} : \boldsymbol{e}_{i} \ \mathrm{d}y, \quad \forall \boldsymbol{v} \in \mathcal{W}_{\mathrm{per}}(\Theta) \]

考虑待求解的影响函数 \(W_{k}^{mnp}\) 可以被离散为 \[ \boldsymbol{W}^{mnp}(\boldsymbol{y}) = \mathbf{N}(\boldsymbol{y}) \mathbf{d}^{mnp}, \quad \delta \boldsymbol{v}(\boldsymbol{y}) = \mathbf{N}(\boldsymbol{y}) \delta \mathbf{d} \] 函数 \(H_{k}^{mn}\), \(E_{kl}^{mn}\) 是一阶多尺度方法求解得到的影响函数, 可以被形函数离散为: \[ H_{k}^{mn}(\boldsymbol{y}) = \mathbf{N}(\boldsymbol{y}) \mathbf{d}^{mn}, \quad E_{kl}^{mn} = \mathbf{B}(\boldsymbol{y}) \mathbf{d}^{mn} + \mathbf{I}_{6\times 6} \] 式中, \(\mathbf{d}^{mn}\) 是 3nen x 6 的已知矩阵, \(\mathbf{N}(\boldsymbol{y})\) 是 3 x 3nen 形函数矩阵, \(\mathbf{B}(\boldsymbol{y})\) 是 6 x 3nen 形函数关于坐标偏导数的矩阵.