Physics-Based Simulation

Computing $\mathbf{P}$

Let's begin with the computation of $\mathbf{P}$. For isotropic materials, the first Piola-Kirchhoff stress tensor can be calculated as follows: $$\begin{array}{rl}\mathbf{P}& =\mathbf{U}\hat{\mathbf{P}}{\mathbf{V}}^T\\ \text{where}\mathbf{F}& =\mathbf{U}\mathbf{\Sigma }{\mathbf{V}}^T,\Psi (\mathbf{F})=\hat{\Psi }(\mathbf{\Sigma }),\text{and}{\hat{P}}_{ij}=\frac{\partial \hat{\Psi }}{\partial {\sigma }_i}{\delta }_{ij}.\end{array}\text{\hspace{1em}(14.2.1)}$$ This formulation leverages the property that $\mathbf{P}$ shares the same SVD space as $\mathbf{F}$, which simplifies the derivation and computation process.

Example 14.2.1. For the Neo-Hookean model (Equation (13.1.2)), we have: $${\hat{\Psi }}_{\text{NH}}(\mathbf{\Sigma })=\frac{\mu }{2}\left(\sum _i^d{\sigma }_i^2-d\right)-\mu \ln (J)+\frac{\lambda }{2}{\ln }^2(J).$$ Thus, we can first perform SVD on $\mathbf{F}=\mathbf{U}\mathbf{\Sigma }\mathbf{V}$ and derive: $${\hat{P}}_{ii}=\mu \left({\sigma }_i-\frac{1}{{\sigma }_i}\right)+\lambda \ln (J)\frac{1}{{\sigma }_i}$$ to compute $\frac{\partial \Psi }{\partial \mathbf{F}}=\mathbf{P}=\mathbf{U}\hat{\mathbf{P}}{\mathbf{V}}^T$ without symbolically deriving the derivative of $\Psi$ w.r.t. $\mathbf{F}$.

Here we provide the proof that $\mathbf{P}$ commutes with rotations in diagonal space (see Equation (14.2.1)). To demonstrate that $\mathbf{P}(\mathbf{RF})=\mathbf{RP}(\mathbf{F})$ for any rotation matrix $\mathbf{R}$, consider a generic (potentially anisotropic) material model. The key idea is that a rotation applied after deformation does not alter the material's stored energy, thus we have the identity $\Psi (\mathbf{F})=\Psi (\mathbf{RF})$. Differentiating both sides of this equation with respect to the deformation gradient $\mathbf{F}$ yields:

$$\begin{array}{rl}\delta \Psi & =\frac{\partial \Psi }{\partial \mathbf{F}}(\mathbf{F}):\delta \mathbf{F}=\frac{\partial \Psi }{\partial \mathbf{F}}(\mathbf{RF}):\delta (\mathbf{RF}),\\ \mathbf{P}(\mathbf{F}):\delta \mathbf{F}& =\mathbf{P}(\mathbf{RF}):(\mathbf{R}\delta \mathbf{F}),\\ \mathbf{P}(\mathbf{F}):\delta \mathbf{F}& =({\mathbf{R}}^T\mathbf{P}(\mathbf{RF})):\delta \mathbf{F},\\ \mathbf{P}(\mathbf{F})& ={\mathbf{R}}^T\mathbf{P}(\mathbf{RF}),\\ \mathbf{RP}(\mathbf{F})& =\mathbf{P}(\mathbf{RF}).\end{array}$$

Furthermore, for an isotropic material where $\Psi (\mathbf{FR})=\Psi (\mathbf{F})$, a similar argument shows that $\mathbf{P}(\mathbf{FR})=\mathbf{P}(\mathbf{F})\mathbf{R}$. Combining these relationships for $\mathbf{P}$ under rotation, we establish that: $$\mathbf{P}(\mathbf{F})=\mathbf{P}(\mathbf{U}\mathbf{\Sigma }{\mathbf{V}}^T)=\mathbf{UP}(\mathbf{\Sigma }){\mathbf{V}}^T=\mathbf{U}\hat{\mathbf{P}}{\mathbf{V}}^T.$$ This formulation confirms the rotational invariance of $\mathbf{P}$ in diagonal space.

Additional Proof for $\mathbf{P}(\mathbf{\Sigma })=\hat{\mathbf{P}}=\frac{\partial \hat{\Psi }}{\partial \Sigma }$

In the above, the last equality comes from the fact that $$\mathbf{P}(\mathbf{F}=\mathbf{\Sigma })=\frac{\partial \hat{\Psi }}{\partial \mathbf{\Sigma }}.$$ Here we show why this is true.

(1) First, we claim that $\mathbf{P}(\mathbf{\Sigma })$ is diagonal. This can be seen by realizing that for isotropic elasticity, $$\mathbf{P}(\mathbf{F})=\sum _k\frac{\partial \Psi }{\partial I_k}(\mathbf{F})\frac{\partial I_k}{\partial \mathbf{F}}(\mathbf{F}),$$ where $I_k$ is the isotropic invariants. Following [Sifakis & Barbic 2022] (page 23), we can observe that $\frac{\partial I_k}{\partial \mathbf{F}}(\mathbf{F})$ when the argument $\mathbf{F}$ is diagonal, must be diagonal. Therefore, $\mathbf{P}(\mathbf{F})$ is diagonal when $\mathbf{F}$ is diagonal.

(2) Next, we claim that $$\text{diag}\left(\frac{\partial \Sigma }{\partial F_{ij}}\right)=\text{diag}\left({\mathbf{U}}^T\frac{\partial \mathbf{F}}{\partial F_{ij}}\mathbf{F}\right).$$ This is proven in [Xu et al. 2015] (Equation 7).

(3) Based on (2), we know that for any $ij$, after substituting $\mathbf{F}=\mathbf{\Sigma }$, we have $$\text{diag}\left(\frac{\partial \Sigma }{\partial F_{ij}}(\Sigma )\right)=\text{diag}\left({\mathbf{I}}^T\frac{\partial \mathbf{F}}{\partial F_{ij}}(\Sigma )\mathbf{I}\right),$$ using this we can write out the cases for $ij=11,ij=22,ij=33$. For example, for $ij=11$, we have $$\frac{\partial \Sigma }{\partial F_{11}}(\Sigma )=\left(\begin{array}{ccc}1& \ast & \ast \\ \ast & 0& \ast \\ \ast & \ast & 0\end{array}\right)$$

(4) Finally, let's derive $\mathbf{P}(\Sigma )$. Since we know it is diagonal from (1), we just need to derive its diagonal entry. Let's use $11$ entry as an example: $$\begin{array}{rl}{P}_{ab}(\Sigma )& =\frac{\partial \hat{\Psi }}{\partial F_{ab}}(\Sigma )=\frac{\partial \hat{\Psi }}{\partial \Sigma }(\Sigma ):\frac{\partial \Sigma }{\partial F_{ab}}(\Sigma )\\ P_{11}(\Sigma )& =\frac{\partial \hat{\Psi }}{\partial \Sigma }(\Sigma ):\frac{\partial \Sigma }{\partial F_{11}}(\Sigma )\\ P_{11}(\Sigma )& =\left(\begin{array}{ccc}\frac{\partial \hat{\Psi }}{\partial {\sigma }_1}& & \\ & \frac{\partial \hat{\Psi }}{\partial {\sigma }_1}& \\ & & \frac{\partial \hat{\Psi }}{\partial {\sigma }_1}\end{array}\right):\left(\begin{array}{ccc}1& \ast & \ast \\ \ast & 0& \ast \\ \ast & \ast & 0\end{array}\right)=\frac{\partial \hat{\Psi }}{\partial {\sigma }_1}\end{array}$$ Now are are done with the final proof.