# Mixed Elasticity¶

## Mixed Linear¶

The boundary-value problem (Strong form) for constitutive equation (48) may be stated as follows: Given body force $$\rho \bm g$$, Dirichlet boundary $$\bar{\bm u}$$ and applied traction $$\bar{\bm t}$$, find the displacement and pressure-like variables $$(\bm u, p) \in \mathcal{V} \times \mathcal{Q}$$ (here $$\mathcal{V} = H^1(\Omega), \mathcal{Q} = L^2(\Omega)$$ ), such that:

(132)\begin{aligned} -\nabla \cdot \bm \sigma - \rho \bm g &= 0, \qquad \text{in \Omega} \\ -\nabla\cdot\bm u - \frac{p}{\bulk - \bulk_p} &= 0, \qquad \text{in \Omega} \\ \bm{u} &= \bar{\bm u}, \quad \text{on \partial \Omega^{D}} \\ \bm \sigma \cdot \bm n &= \bar{\bm t}. \qquad \text{on \partial \Omega^{N}} \end{aligned}

with $$\bm n$$ be the unit normal on the boundary and its weak formulation as:

(133)\begin{aligned} \int_{\Omega}{ \nabla \bm{v} \tcolon \bm{\sigma}} \, dv - \int_{\partial \Omega}{\bm{v} \cdot \bar{\bm t}} \, da - \int_{\Omega}{\bm{v} \cdot \rho \bm{g}} \, dv &= 0, \quad \forall \bm v \in \mathcal V \\ \int_{\Omega} q \left( -\nabla\cdot\bm u - \frac{p}{\bulk - \bulk_p} \right) \, dv & = 0. \quad \forall q \in \mathcal Q \end{aligned}

## Mixed Hyperelasticity, initial configuration¶

We can state the mixed hyperelastic strong form as follow: Given body force $$\rho_0 \bm g$$, Dirichlet boundary $$\bar{\bm u}$$ and applied traction $$\bar{\bm t}$$, find the displacement and pressure-like variables $$(\bm u, p) \in \mathcal{V} \times \mathcal{Q}$$, such that:

(134)\begin{aligned} -\nabla_X \cdot \bm P - \rho_0 \bm g &= 0. \qquad \text{in \Omega_0} \\ -\frac{\partial V}{\partial J} - \frac{p}{\bulk - \bulk_p} &= 0. \qquad \text{in \Omega} \\ \bm{u} &= \bar{\bm u}, \quad \text{on \partial \Omega^{D}_0} \\ \bm P \cdot \bm N &= \bar{\bm t}. \qquad \text{on \partial \Omega^{N}_0} \end{aligned}

and by multiplying the strong form (134) by test functions $$(\bm{v}, q)$$ and integrate by parts, we can obtain the weak form for finite-strain incompressible hyperelasticity as: find $$(\bm{u}, p) \in \mathcal{V} \times \mathcal{Q} \subset H^1 \left( \Omega_0 \right) \times L^2 \left( \Omega_0 \right)$$ such that

(135)\begin{aligned} R_u(\bm u, p) &\coloneqq \int_{\Omega_0}{\nabla_X \bm{v} \tcolon \bm{P}} \, dV - \int_{\Omega_0}{\bm{v} \cdot \rho_0 \bm{g}} \, dV - \int_{\partial \Omega_0}{\bm{v} \cdot \bar{\bm t}} \, dA = 0, \quad \forall \bm{v} \in \mathcal{V}, \\ R_p(\bm u, p) &\coloneqq \int_{\Omega_0} q \cdot L J \, dV = 0, \quad \forall q\in \mathcal{Q}, \end{aligned}

where $$L = -V' - \frac{p}{\bulk - \bulk_p}$$. This equation contains material/constitutive nonlinearities in defining $$\bm{S}(\bm{E})$$, as well as geometric nonlinearities through $$\bm{P} = \bm{F}\, \bm{S}$$, $$\bm{E}(\bm{F})$$, and the body force $$\bm{g}$$, which must be pulled back from the current configuration to the initial configuration.

### Newton linearization¶

As explained in Newton linearization, we need the Jacobian form of the (135) which is: find $$(\diff \bm{u}, \diff p) \in \mathcal{V} \times \mathcal{Q}$$ such that

(136)\begin{aligned} \int_{\Omega_0} \nabla_X \bm{v} \tcolon \diff \bm{P} dV &= -R_u, \quad \forall \bm{v} \in \mathcal{V}, \\ \int_{\Omega_0} q \cdot \left( \diff L J + L \diff J \right)dV &= -R_p, \quad \forall q \in \mathcal{Q} \end{aligned}

where

(137)\begin{aligned} \diff \bm{P} &= \diff \bm{F}\, \bm{S} + \bm{F} \diff \bm{S}, \\ \diff L J + L \diff J &= J \frac{\partial L}{\partial \bm{E}} \tcolon \diff \bm{E} + J \frac{\partial L}{\partial p} \diff p + L \frac{\partial J}{\partial \bm{E}} \tcolon \diff \bm{E} \end{aligned}

with $$\diff \bm{F}$$ and $$\diff \bm E$$ defined in (99). The linearization of the second equation of (137) is

(138)\begin{aligned} \diff L J + L \diff J &= \left( -V^{''} \diff J - \frac{\diff p}{\bulk - \bulk_p} \right) J + L J (\bm{C}^{-1} \tcolon \diff \bm{E}), \\ &= \left( - J^2 \, V^{''} - J \, V' - J \frac{p}{\bulk - \bulk_p} \right) \bm{C}^{-1} \tcolon \diff \bm{E} - J \frac{\diff p}{\bulk - \bulk_p} \end{aligned}

The linearization of the second Piola-Kirchhoff stress tensor, $$\diff \bm{S}$$, depends upon the material model and derived below for different hyperelastic materials.

Deriving $$\diff\bm{S}$$ for isochoric Neo-Hookean material

For the Neo-Hookean model (62), we derive split

(139)$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{vol}}^u} + \underbrace{\frac{\partial \bm{S}_{\text{iso}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{iso}}} + \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial p} \diff p}_{\diff \bm{S}_{\text{vol}}^p},$

then,

(140)\begin{aligned} \diff \bm{S}_{\text{vol}}^u &= \left( \bulk_p \diff J \, V' + \bulk_p J V^{''} \diff J - p \diff J \right) \bm{C}^{-1} + \left(\bulk_p J \, V' - p \, J \right) \diff \bm{C}^{-1}, \\ &= \left( \bulk_p J^2 V^{''} + \bulk_p J \, V' - p J \right) (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} + \left(\bulk_p J \, V' - p \, J \right) \diff \bm{C}^{-1} \end{aligned}
(141)$\diff \bm{S}_{\text{vol}}^p = -J \bm{C}^{-1} \diff p,$

and

(142)\begin{aligned} \diff \bm{S}_{\text{iso}} &= -\frac{2}{3}\mu J^{-2/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \left(\bm{I} - \frac{1}{3} \mathbb{{I}_1} \bm{C}^{-1} \right) \\ &- \frac{1}{3} \mu J^{-2/3} \left( 2 \trace \diff \bm{E} \, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right), \\ &= -\frac{4}{3}\mu J^{-2/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} \bm{E}_{\text{dev}} - \frac{1}{3} \mu J^{-2/3} \left( 2 \trace \diff \bm{E} \, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right) \end{aligned}

where

$\diff \bm{C}^{-1} = \frac{\partial \bm{C}^{-1}}{\partial \bm{E}} \tcolon \diff \bm{E} = -2 \bm{C}^{-1} \diff \bm{E} \, \bm{C}^{-1} .$

Note

If we use single field isochoric model (66) the linearization of the volumetric stress becomes

(143)$\diff \bm{S}_{\text{vol}} = \left( \bulk J^2 V^{''} + \bulk J \, V'\right) (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} + \bulk J \, V' \diff \bm{C}^{-1}.$

$$\diff \bm{S}_{\text{vol}}^p = 0$$ and we only need to solve the first equation of (135).

Deriving $$\diff\bm{S}$$ for isochoric Mooney-Rivlin material

For the Mooney-Rivlin model (74), we derive split

(144)$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{vol}}^u} + \underbrace{\frac{\partial \bm{S}_{\text{iso}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{iso}}} + \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial p} \diff p}_{\diff \bm{S}_{\text{vol}}^p},$

where $$\diff \bm{S}_{\text{vol}}^p$$ and $$\diff \bm{S}_{\text{vol}}^u$$ are the same as (141), (140) (or (143) if we use single field stress (78)) and the isochoric part is

(145)\begin{aligned} \diff \bm{S}_{\text{iso}} &= -\frac{4}{3}(\bm{C}^{-1} \tcolon \diff \bm{E}) \left( \mu_1 J^{-2/3} + 4 \mu_2 J^{-4/3} \right) \bm{C}^{-1} \bm{E}_{\text{dev}}\\ &- \frac{1}{3} \left( \mu_1 J^{-2/3} + 2 \mu_2 J^{-4/3} \right) \left( 2 \mathbb{{I}_1}(\diff \bm{E})\, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right) \\ &- \frac{8}{3}(\bm{C}^{-1} \tcolon \diff \bm{E}) \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{E})\bm{I} - \bm{E} \right) \\ &+2 \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{E})\bm{I} - \diff \bm{E} \right) \\ & + (c_1 + c_2) \bm{C}^{-1} - \frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{E}) + 2 \mathbb{{I}_2}(\bm{E})\right) \diff \bm{C}^{-1} \end{aligned}

where

(146)\begin{aligned} c_1 &= \frac{16}{9} \mu_2 J^{-4/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \left( \mathbb{{I}_1}(\bm{E}) + 2 \mathbb{{I}_2}(\bm{E}) \right). \\ c_2 &= -\frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{E}) + 2 \mathbb{{I}_1}(\bm{E}) \mathbb{{I}_1}(\diff \bm{E}) - 2 \bm{E} \tcolon \diff \bm{E} \right). \end{aligned}
Deriving $$\diff\bm{S}$$ for isochoric Ogden material

Similar to the Neo-Hookean model we have

(147)$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{vol}}^u} + \underbrace{\frac{\partial \bm{S}_{\text{iso}}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{\text{iso}}} + \underbrace{\frac{\partial \bm{S}_{\text{vol}}}{\partial p} \diff p}_{\diff \bm{S}_{\text{vol}}^p},$

where $$\diff \bm{S}_{\text{vol}}^p$$ and $$\diff \bm{S}_{\text{vol}}^u$$ are similar to volumetric derivative of Neo-Hookean and Mooney-Rivlin models and the isochoric part is

(148)\begin{aligned} \diff \bm{S}_{\text{iso}} &= \sum_{i=1}^3 \diff S_i^{iso} \hat{\bm{N}_i} \hat{\bm{N}_i}^T + S_i^{iso} \left( \diff \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \hat{\bm{N}_i} \diff \hat{\bm{N}_i}^T\right) \end{aligned}

For example computing $$\diff S_1^{iso}$$ from (87) gives

(149)\begin{aligned} \diff S_1^{iso} &= -\frac{2\diff \beta_1}{\beta_1^3} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &+ \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j \alpha_j}{3} \left[ 2 \diff \ell_1 \exp(\alpha_j \ell_1) - \diff \ell_2 \exp(\alpha_j \ell_2) - \diff \ell_3 \exp(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &- \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j \alpha_j}{9} \left[ 2 \operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \end{aligned}

To compute $$\diff \beta_i$$ we differentiate $$\bm{C} = \sum_{i=1}^3 \beta_i^2 \hat{\bm{N}_i} \hat{\bm{N}_i}^T$$ as

(150)$\diff \bm{C} = \sum_{i=1}^3 2 \beta_i \diff \beta_i \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \beta_i^2 \left(\diff \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \hat{\bm{N}_i} \diff \hat{\bm{N}_i}^T \right)$

and used the fact that the eigenvectors are orthonormal i.e., $$\langle \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \delta_{ij}$$ so that $$\langle \hat{\bm{N}_i}, \diff \hat{\bm{N}_i} \rangle = 0$$, we can multiply (150) from the left and right by $$\hat{\bm{N}_i}$$

(151)$\langle \hat{\bm{N}_i}, \diff \bm{C} \hat{\bm{N}_i} \rangle = 2 \beta_i \diff \beta_i \Rightarrow \diff \beta_i = \frac{1}{\beta_i} \langle \hat{\bm{N}_i}, \diff \bm{E} \hat{\bm{N}_i} \rangle$

By differentiating $$\bm{E} \hat{\bm{N}_i} = \beta_i^E \hat{\bm{N}_i}$$ we will arrive at

$\diff \bm{E} \hat{\bm{N}_i} + \bm{E} \diff \hat{\bm{N}_i}= \diff \beta_i^E \hat{\bm{N}_i} + \beta_i^E \diff \hat{\bm{N}_i}$

taking the inner product of above equation with $$\hat{\bm{N}_j}, \, j\neq i$$ simplifies to

\begin{aligned} &\quad \langle \hat{\bm{N}_j}, \diff \bm{E} \hat{\bm{N}_i} \rangle + \langle \diff \hat{\bm{N}_i}, \bm{E} \hat{\bm{N}_j} \rangle = \beta_i^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \langle \hat{\bm{N}_j}, \diff \bm{E} \hat{\bm{N}_i} \rangle + \beta_j^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \beta_i^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \frac{1}{\beta_i^E - \beta_j^E} \langle \diff \bm{E} \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \diff \hat{\bm{N}_i} = \sum_{j \neq i} \frac{1}{\beta_i^E - \beta_j^E} \langle \diff \bm{E} \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \hat{\bm{N}_j} \end{aligned}

and finally the linearization of $$\ell_i = \log \beta_i$$ is

$\diff \ell_i = \frac{\diff \beta_i}{\beta_i}.$

which complete $$\diff\bm{S}$$ for Ogden model.

## Mixed Hyperelasticity, current configuration¶

Similar to what we have shown in (104), we can write the first equation of (135) in the current configuration (see (105) ) which yields to the Jacobian form

(152)\begin{aligned} \int_{\Omega_0} \nabla_x \bm{v} \tcolon \left( \diff \bm{\tau} - \bm{\tau} \left( \nabla_x \diff \bm{u} \right)^T \right) dV &= -r_u, \quad \forall \bm{v} \in \mathcal{V}, \\ \int_{\Omega_0} q \cdot \left( \diff L J + L \diff J \right)dV &= -r_p, \quad \forall q \in \mathcal{Q} \end{aligned}

where the second equation is similar to (136) and we can use (111)

$\diff\bm\tau - \bm\tau\left( \nabla_x \diff\bm u \right)^T = \left( \nabla_x \diff\bm u \right) \bm\tau + \bm F \diff\bm S \bm F^T.$

to compute the linearization of the first equation in current configuration for different material models.

Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Neo-Hookean material

Based on the split (139), we can derive

(153)$\bm{F} \diff \bm{S}_{\text{vol}}^u \bm{F}^{T} = \left( \bulk_p J^2 V^{''} + \bulk_p J \, V' - p J \right) \trace (\diff \bm \epsilon) \bm{I} - 2 \left(\bulk_p J \, V' - p \, J \right) \diff \bm \epsilon.$
(154)$\bm{F} \diff \bm{S}_{\text{vol}}^p \bm{F}^{T} = -J \diff p \bm{I},$

where $$\diff \bm \epsilon$$ is defined in (109) and we have used (116) and (117). The isochoric part can be simplified as

(155)\begin{aligned} \bm{F} \diff \bm{S}_{\text{iso}} \bm{F}^{T} &= -\frac{2}{3} \mu J^{-2/3} \left( 2 \trace (\diff \bm \epsilon) \bm{e}_{\text{dev}} + \trace(\diff \bm e) \bm{I} - \mathbb{{I}_1} \diff \bm \epsilon \right), \end{aligned}

where $$\mathbb{{I}_1} = \trace(\bm{b})$$.

Note

If we use single field isochoric model the linearization of the volumetric stress becomes

(156)$\bm{F} \diff \bm{S}_{\text{vol}} \bm{F}^{T} = \left( \bulk J^2 V^{''} + \bulk J \, V'\right) \trace (\diff \bm \epsilon) \bm{I} -2 \bulk J \, V' \diff \bm \epsilon.$

$$\diff \bm{S}_{\text{vol}}^p = 0$$ and we only need to solve the first equation of (152).

Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Mooney-Rivlin material

The $$\bm{F} \diff \bm{S}_{\text{vol}}^p \bm{F}^T$$ and $$\bm{F} \diff \bm{S}_{\text{vol}}^u \bm{F}^T$$ in (144) are the same as (154), (153) (or (156) if we use single field formulation ). The isochoric part can be derived as

(157)\begin{aligned} \bm{F} \diff \bm{S}_{\text{iso}} \bm{F}^T &= -\frac{4}{3}(\trace (\diff \bm \epsilon)) \left( \mu_1 J^{-2/3} + 4 \mu_2 J^{-4/3} \right) \bm{e}_{\text{dev}}\\ &+ \frac{2}{3} \left( \mu_1 J^{-2/3} + 2 \mu_2 J^{-4/3} \right) \left( \mathbb{{I}_1} \diff \bm \epsilon -\mathbb{{I}_1}(\diff \bm{e})\, \bm{I} \right) \\ &- \frac{8}{3}(\trace (\diff \bm \epsilon)) \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{e})\bm{b} - \bm{b} \bm{e} \right) \\ &+2 \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{e})\bm{b} - \bm{b} \diff \bm \epsilon \bm{b} \right) \\ & + (c_1 + c_2) \bm{I} + \frac{8}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{e}) + 2 \mathbb{{I}_2}(\bm{e})\right) \diff \bm \epsilon \end{aligned}

where

(158)\begin{aligned} c_1 &= \frac{16}{9} \mu_2 J^{-4/3} (\trace (\diff \bm \epsilon)) \left( \mathbb{{I}_1}(\bm{e}) + 2 \mathbb{{I}_2}(\bm{e}) \right). \\ c_2 &= -\frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{e}) + 2 \mathbb{{I}_1}(\bm{e}) \mathbb{{I}_1}(\diff \bm{e}) - 2 \trace (\bm{b} \bm{e} \diff \bm \epsilon) \right). \end{aligned}

and we have used (119), (120), and

\begin{aligned} \bm{F} \bm{C}^{-1} \bm{E}_{\text{dev}} \bm{F}^T &= \bm{F}^{-T} \left( \bm{E} - \frac{1}{3} \mathbb{{I}_1}(\bm{E}) \bm{I} \right) \bm{F}^T = \bm{e}_{\text{dev}}. \\ \bm{F} \bm{E} \bm{F}^T &= \frac{1}{2} \left( \bm{F} \bm{C} \bm{F}^T - \bm{F} \bm{F}^T \right) = \bm{b} \bm{e}. \\ \diff \bm{E} &= \frac{1}{2} \left( \bm{F}^T \diff \bm{F} \bm{F}^{-1} \bm{F} + \bm{F}^T \bm{F}^{-T} \diff \bm{F}^T \bm{F} \right) = \bm{F}^T \diff \bm \epsilon \bm{F}. \\ \bm{E} \tcolon \diff \bm{E} &= \trace(\bm{E} \diff \bm{E}) = \trace(\bm{E} \bm{F}^T \diff \epsilon \bm{F}) = \trace (\bm{F} \bm{E} \bm{F}^T \diff \epsilon) = \trace(\bm{b} \bm{e} \diff \bm \epsilon ). \end{aligned}
Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Ogden material

Based on the split (147), the $$\bm{F} \diff \bm{S}_{\text{vol}}^p \bm{F}^T$$ and $$\bm{F} \diff \bm{S}_{\text{vol}}^u \bm{F}^T$$ are the same as (154), (153) (or (156) if we use single field formulation ) and the isochoric part is

(159)\begin{aligned} \bm{F} \diff \bm{S}_{\text{iso}} \bm{F}^{T} &= \sum_{i=1}^3 \beta^2_i \diff S_i^{iso} \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \beta_i S_i^{iso} \left( \bm{F} \diff \hat{\bm{N}_i} \hat{\bm{n}_i}^T + \hat{\bm{n}_i} \diff \hat{\bm{N}_i}^T \bm{F}^T \right) \\ &= \sum_{i=1}^3 \left(\diff \tau_i^{iso} - 2 \frac{\diff \beta_i}{\beta_i} \tau_i^{iso} \right) \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \tau_i^{iso} \bm{A}_i \end{aligned}

where for example $$\diff \tau_1^{iso}$$ can be written

\begin{aligned} \diff \tau_1^{iso} &= \sum_{j=1}^N \frac{m_j \alpha_j}{3} \left[ 2 \diff \ell_1 \exp(\alpha_j \ell_1) - \diff \ell_2 \exp(\alpha_j \ell_2) - \diff \ell_3 \exp(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &- \sum_{j=1}^N \frac{m_j \alpha_j}{9} \left[ 2 \operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \trace \left(\diff \bm \epsilon \right) \end{aligned}

and

$\bm{A}_i = 2 \frac{\diff \beta_i}{\beta_i} \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \diff \left( \hat{\bm{n}_i} \hat{\bm{n}_i}^T \right) + \left( \left( \nabla_x \diff \bm{u} \right) \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \hat{\bm{n}_i} \hat{\bm{n}_i}^T \left( \nabla_x \diff \bm{u} \right)^T \right)$

where $$\tau_i^{iso}$$ is defined in (90) and we have used $$\bm{F} \hat{\bm{N}_i} = \beta_i \hat{\bm{n}_i}$$ and $$\tau_i^{iso} = \beta_i^2 S_i^{iso}$$. Note that the linearization of principal stretches, $$\beta_i$$, and eigenvectors, $$\hat{\bm{n}_i}$$, are similar to initial configuration but are written in terms of Green-Euler tensor i.e.,

$\diff \beta_i = \frac{1}{\beta_i} \langle \hat{\bm{n}_i}, \diff \bm{e} \hat{\bm{n}_i} \rangle.$
$\quad \diff \hat{\bm{n}_i} = \sum_{j \neq i} \frac{1}{\beta_i^e - \beta_j^e} \langle \diff \bm{e} \hat{\bm{n}_i}, \hat{\bm{n}_j} \rangle \hat{\bm{n}_j}.$

## Perturbed Lagrange-multiplier method¶

In the previous sections we introduced the strong form of general mixed formulation for small and finite strain and by considering test functions $$(\bm v, q)$$ as derived in (133) and (135). Alternatively, we can derive the mixed $$\bm u- p$$ weak formulation based on minimization of two fields functional $$\Pi(\bm u, p)$$, which it is known as the perturbed Lagrangian approach. For the mixed linear case, we can write

(160)\begin{aligned} \Pi (\bm u, p) &= \int_{\Omega} \left[ \mu \, \bm \varepsilon_{\text{dev}} \tcolon \bm \varepsilon_{\text{dev}} - p\, \trace \bm \varepsilon + \frac{\bulk_p}{2} \left(\trace \bm \varepsilon \right)^2 - \frac{p^2}{2 (\bulk-\bulk_p)} \right] \, dv - \Pi_{\text{ext}} (\bm u) \\ \Pi_{\text{ext}} (\bm u) &= \int_{\Omega}{\bm{u} \cdot \rho \bm{g}} \, dv + \int_{\partial \Omega}{\bm{u} \cdot \bar{\bm{t}}} \, da \end{aligned}

and by invoking the stationarity of $$\Pi$$ with respect to $$\bm u$$ and $$p$$, we obtain

(161)\begin{aligned} \int_{\Omega} \nabla \delta \bm u \tcolon \left[ 2 \mu \, \bm \varepsilon_{\text{dev}} + \left(\bulk_p \trace \bm \varepsilon - p \right) \bm{I} \right] \, dv - \bm L_{\text{ext}}(\delta \bm u) &=0, \\ \int_{\Omega} \delta p \left( -\trace \bm \varepsilon - \frac{p}{\bulk-\bulk_p} \right) \, dv &= 0, \end{aligned}

where $$\bm L_{\text{ext}}(\delta \bm u) = \int_{\Omega}{\delta \bm{u} \cdot \rho \bm{g}} \, dv + \int_{\partial \Omega}{\delta \bm{u} \cdot \bar{\bm{t}}} \, da$$, and $$\delta \bm u$$ and $$\delta p$$ are virtual displacement and pressure and can be seen as the test functions $$\bm v, q$$, i.e., $$\delta \bm u = \bm v, \, \delta p = q$$. It is clear that the weak form (161) agrees with what we obtained in (133). However, the hyperelastic weak form (135), can not be derived by minimizing any functional and its linearization is not symmetric.

To write the mixed functional $$\Pi(\bm u, p)$$ for hyperelastic, we need to consider the strain energy of the form

(162)$\psi \left(\bm{C} \right) = \psi_{\text{vol}}(J) + \psi_{\text{iso}}(\bar{\bm{C}}) = \frac{\bulk}{2} \left(U(J)\right)^2 + \psi_{\text{iso}}(\bar{\bm{C}})$

we can write a two fields energy functional as

(163)\begin{aligned} \Pi (\bm u, p) &= \int_{\Omega_0} \left[ \psi_{\text{iso}}(\bar{\bm{C}}) - p \, U(J) - \frac{1}{2}\frac{p^2}{\bulk - \bulk_p} + \frac{\bulk_p}{2} U^2\right] \, dV - \Pi_{\text{ext}} (\bm u) \\ \Pi_{\text{ext}} (\bm u) &= \int_{\Omega_0}{\bm{u} \cdot \rho_0 \bm{g}} \, dV + \int_{\partial \Omega_0}{\bm{u} \cdot \bar{\bm{t}}} \, dA \end{aligned}

Finding the stationary conditions with respect to $$\bm{u}$$ and $$p$$ by taking Gateaux derivative gives the weak form

(164)\begin{aligned} \int_{\Omega_0} \bm F \underbrace{\left( \bm{S}_{\text{iso}} + (\bulk_p U - p) \, J \, U' \, \bm{C}^{-1} \right)}_{\bm S} \tcolon \nabla_X \bm v \, dV &= L_{\text{ext}} (\bm v) \\ \int_{\Omega_0} \left(- U(J) - \frac{p}{\bulk - \bulk_p} \right) q \, dV &= 0 \end{aligned}

where, $$L_{\text{ext}} (\bm v) = \int_{\Omega_0}{\bm{v} \cdot \rho_0 \bm{g}} \, dV + \int_{\partial \Omega_0}{\bm{v} \cdot \bar{\bm{t}}} \, dA$$ and we have used

\begin{aligned} \frac{\partial \psi_{\text{iso}}}{\partial \bm u} \cdot {\delta \bm u} &= \frac{\partial \psi_{\text{iso}}}{\partial \bm E} \tcolon \frac{\partial \bm E}{ \partial \bm u}{\delta \bm u} = \bm{S}_{\text{iso}} \tcolon {\delta \bm E} = \bm{S}_{\text{iso}} \tcolon \text{sym} \left(\bm F^T \delta \bm F \right) \\ \frac{\partial U}{\partial \bm u} \cdot {\delta \bm u} &= \frac{\partial U}{\partial J} \frac{\partial J}{\partial \bm u} {\delta \bm u} = U' {\delta J} = J \, U' \, \bm{C}^{-1} \tcolon \tcolon {\delta \bm E} = J \, U' \, \bm{C}^{-1} \tcolon \text{sym} \left(\bm F^T \delta \bm F \right) \end{aligned}

where $$\delta \bm{F} = \nabla_X \delta \bm u = \nabla_X \bm v$$. The Jacobian for problem (164) can be written as

(165)\begin{aligned} \int_{\Omega_0} \nabla_X \bm{v} \tcolon \left(\bm F \diff \bm{S} + \diff \bm{F} \bm{S} \right) dV &= -R_u^{PL}, \\ \int_{\Omega_0} q \left( -\diff U - \frac{\diff p}{\bulk - \bulk_p} \right) dV &= -R_p^{PL}, \end{aligned}

where

(166)\begin{aligned} \diff \bm{S} &= \diff \bm{S}_{\text{iso}} + \diff \bm{S}_{\text{vol}}^u + \diff \bm{S}_{\text{vol}}^p, \\ \diff \bm{S}_{\text{vol}}^u &= \left(\bulk_p U' \diff J \right) J U' \bm{C}^{-1} + \left(\bulk_p U - p \right) \left(\diff J \, U' \bm{C}^{-1} + J\, U^{''} \diff J \bm{C}^{-1} + J U' \diff \bm{C}^{-1} \right), \\ &= \left[\bulk_p \left( J U' \right)^2 + \left(\bulk_p U - p \right) \left( J U' + J^2 U^{''} \right) \right] \left( \bm{C}^{-1} \tcolon \diff \bm E \right) \bm{C}^{-1} \\ &+ \left(\bulk_p U - p \right) J U' \, \diff \bm{C}^{-1}\\ \diff \bm{S}_{\text{vol}}^p &= - dp J U' \, \bm{C}^{-1} \\ \diff U &= U' \diff J = J \, U' \, \bm{C}^{-1} \tcolon \diff \bm E, \end{aligned}

where $$\diff \bm{S}_{\text{iso}}$$ is derived for Neo-Hookean, Mooney-Rivlin and Ogden in (142), (145) and (148). To complete the derivation we only need $$U(J)$$ function with condition

$U(J) = 0 \quad \text{if and only if} \quad J = 1.$

For the volumetric strain energy function of the form given in (59), if we choose $$V(J) = \frac{1}{4} \left(J^2 - 1 - 2 \log J \right)$$, from $$\bulk V(J) = \frac{\bulk}{2} U^2(J)$$, we will have

$U(J) = \pm \sqrt{2 V} = \frac{\sign(J-1)}{\sqrt{2}} ( \underbrace{J^2 - 1 - 2 \log J }_{A})^{1/2}$

where the derivatives are

\begin{aligned} U' &= \frac{\partial U}{\partial J} = \sign(J-1) \frac{J^2 - 1}{J \sqrt{2}} A^{-1/2}, \\ U{''} &= \frac{\partial^2 U}{\partial J^2} = \sign(J-1) \frac{1}{J^2 \sqrt{2}} \left((J^2 + 1) A^{-1/2} - (J^2 - 1)^2 A^{-3/2} \right). \end{aligned}

## Matrix-free implementation¶

Similar to Matrix-free implementation for single field, we present mathematical formulation of matrix-free method for a general Dirichlet problem for general mixed problem $$\bm{R}(\bm{u}, p) = [\bm{R}_u(\bm{u}, p) ~~ \bm{R}_p(\bm{u}, p)]^T = \bm 0$$: Find $$(\bm u, p) \in \mathcal{V} \times \mathcal{Q}$$, such that

(167)\begin{aligned} \langle\bm{v}, \bm{R}_u(\bm{u}, p) \rangle &= \int_{\Omega_0}{\bm{v} \cdot \bm{f}_0 + \nabla \bm{v} \tcolon \bm{f}_1} \, dV = 0, \quad \forall \bm{v} \in \mathcal{V}, \\ \langle q, \bm{R}_p(\bm{u}, p) \rangle &= \int_{\Omega_0}{q \cdot \bm{g}_0 + \nabla q \tcolon \bm{g}_1 } \, dV = 0, \quad \forall q \in \mathcal{Q}, \end{aligned}

where the operators $$\bm{f}_0, \bm{f}_1, \bm{g}_0, \bm{g}_1$$ contain all possible sources in the problem i.e., $$\bm{u}, \nabla \bm{u}, p, \nabla p$$. It should be noted that the gradient in the (167) depends on the configuration system and it could be with respect to initial configuration $$\bm{X}$$ i.e., $$\nabla_X \bm{v}$$ (Lagrangian approach) or current configuration $$\bm{x}$$ i.e., $$\nabla_x \bm{v}$$ (Eulerian approach).

In order to solve (167) with Newton-Krylov iterative solvers we need its Jacobian form: Find $$(\diff \bm{u}, \diff p) \in \mathcal{V} \times \mathcal{Q}$$ such that

(168)\begin{aligned} \langle\bm{v}, \bm{J}_u(\bm{u}, p) \diff \bm{U} \rangle &= \int_{\Omega_0}{\bm{v} \cdot \diff \bm{f}_0 + \nabla \bm{v} \tcolon \diff \bm{f}_1} \, dV, \\ \langle q, \bm{J}_p(\bm{u}, p) \diff \bm{U} \rangle &= \int_{\Omega_0}{q \cdot \diff \bm{g}_0 + \nabla q \tcolon \diff \bm{g}_1} \, dV. \end{aligned}

where $$\diff \bm{U} = [\diff \bm{u} ~~ \diff p]^T$$ and the linearization of operators $$\bm{f}_0, \bm{f}_1, \bm{g}_0, \bm{g}_1$$ are

\begin{aligned} \diff \bm{f}_i &= \frac{\partial \bm{f}_i}{\partial \bm{u}} \diff \bm{u} + \frac{\partial \bm{f}_i}{\partial \nabla \bm{u}} \nabla \diff \bm{u} + \frac{\partial \bm{f}_i}{\partial p} \diff p + \frac{\partial \bm{f}_i}{\partial \nabla p} \nabla \diff p, \quad i=0,1 \\ \diff \bm{g}_i &= \frac{\partial \bm{g}_i}{\partial \bm{u}} \diff \bm{u} + \frac{\partial \bm{g}_i}{\partial \nabla \bm{u}} \nabla \diff \bm{u} + \frac{\partial \bm{g}_i}{\partial p} \diff p + \frac{\partial \bm{g}_i}{\partial \nabla p} \nabla \diff p, \quad i=0,1 \end{aligned}

Compare with governing equations derived in pervious sections for linear and large deformation, it is easy to verify that

• For mixed linear elasticity described in (133) we have

\begin{aligned} \bm f_0 &= -\rho \bm g, \quad \bm f_1 = \bm{\sigma}(\bm u, p), \quad \bm g_0 =-\trace \bm \varepsilon - \frac{p}{\bulk-\bulk_p}, \quad \bm g_1 = \bm 0, \\ \diff \bm f_1 &= \bm f_1(\diff \bm{u}, \diff p), ~ ~ \diff \bm g_0 = \bm g_0(\diff \bm{u}, \diff p) \end{aligned}
• For mixed hyperelastic in initial configuration described in (135) and (136) we have

\begin{aligned} \bm f_0 &= -\rho_0 \bm g, \quad \bm f_1 = \bm F \bm S, \quad \bm g_0 = L J, \quad \bm g_1 = \bm 0, \\ \diff \bm f_1 &= \diff \bm F \bm S + \bm F \diff \bm S, ~ ~ \diff \bm g_0 = \diff L J + L \diff J \end{aligned}
• For mixed hyperelastic in current configuration derived in (152)

\begin{aligned} \bm f_0 &= -\rho_0 \bm g, \quad \bm f_1 = \bm \tau, \quad \bm g_0 = L J, \quad \bm g_1 = \bm 0, \\ \diff \bm f_1 &= \diff \bm{\tau} - \bm{\tau} \left( \nabla_x \diff \bm{u} \right)^T, ~ ~ \diff \bm g_0 = \diff L J + L \diff J \end{aligned}
• For mixed hyperelastic with variational approach in initial configuration given in (164) and (165)

\begin{aligned} \bm f_0 &= -\rho_0 \bm g, \quad \bm f_1 = \bm F \bm S, \quad \bm g_0 = -{U}(J) - \frac{p}{\bulk - \bulk_p}, \quad \bm g_1 = \bm 0, \\ \diff \bm f_1 &= \diff \bm F \bm S + \bm F \diff \bm S, ~ ~ \diff \bm g_0 = -\diff {U}(J) - \frac{\diff p}{\bulk - \bulk_p} \end{aligned}

We solve the above mixed equations using static and quasistatic solver in Ratel as explained in elasticity static and quasistatic section. For mixed elastodynamic, we only need to add the same terms (see (128) and (129)) to the residual and jacobian of the first equation of mixed matrix-free formulation described in (167) and (168), respectively.