Derivative of deformation gradient with respect to Green-Lagrangian strain?

Question

For hyperelastic material, the elastic energy $\Psi $ is related to the deformation gradient $F$ and other internal variables (e.g. temperature $ \theta$).

However, in many literatures (including Malvern's and Belytchko's) the derivatives (especially Hessian) are usually derived in terms of left Cauchy-Green tensor $ C = F^t F$. For example, 2nd PK stress $${S_{ij}} = \frac{{\partial \Psi }}{{\partial {E_{ij}}}} \qquad \text{and} \qquad C_{ijkl}^{SE} = \frac{{{\partial ^2}\Psi }}{{\partial {E_{ij}}\partial {E_{kl}}}} \, . $$ I can convince myself that such derivation may help simplify the steps as the materials are usually represented by tensor $C$, but what I'm having truble now is a possibility of other ways, such as:

$$ S = {F^{ - 1}}\frac{{\partial \Psi }}{{\partial F}} \qquad D = \frac{{\partial S}}{{\partial F}}\frac{{\partial F}}{{\partial E}} \, . $$

To me it looks $\frac{{\partial F}}{{\partial E}}$ should be straitforward (as both $ \frac{{\partial S}}{{\partial F}}$ and $\frac{{\partial S}}{{\partial E}}$ is attainable), but it makes me perplexed is that its inverse $\frac{{\partial E}}{{\partial F}}$ is not invertible as:

$$ \frac{{\partial E_{ij}}}{{\partial F_{kl}}}= \frac{{\partial \left( {{F_{pi}}{F_{pj}}} \right)}}{{\partial {F_{kl}}}} = \left( {{F_{ki}}{\delta _{lj}} + {F_{kj}}{\delta _{li}}} \right) \, , $$

which is a kind of Sylvestre equations. I think there is an alternative way to bridge these two equations using tensor manipulation, but I'm at a loss.

Any comments about what I am missing would be greatly appreciated.

In short, my question is whether it is possible to compute $\frac{\partial F}{\partial E}$.

It might help getting $C_{ijkl}^{SE}$ from $\frac{\partial S_{ij}}{\partial F_{ij}}$, which is sometimes conveinient when compared to $\frac{\partial S_{ij}}{\partial E_{ij}}$.

Answer 1

To my knowledge, I'm afraid it is not generally possible to compute $\frac{\partial\mathbf{F}}{\partial\mathbf{E}}$. Here's the reason:

Usually we compute the Green-Lagrange strain tensor from the deformation gradient with its definition $$ \mathbf{E}(\mathbf{F})=\frac{1}{2}(\mathbf{F}^T\mathbf{F}-\mathbf{I}) \tag{1} $$ It is easy to verify that $\mathbf{E}$ is symmetric, but $\mathbf{F}$ is not necessarily symmetric. Take the 3D space as example, one would have 9 independent components for $\mathbf{F}$, but only 6 independent components for $\mathbf{E}$. That is to say, one may not possible to obtain the inverse of Eq. (1), namely $\mathbf{F}(\mathbf{E})$, not even $\frac{\partial\mathbf{F}}{\partial\mathbf{E}}$.

Being symmetric is also the reason that people prefer to use the right Cauchy-Green tensor $\mathbf{C}$ as well as the 2nd Piola-Kirchhoff stress $\mathbf{S}$.

However, there are still formulations using the 1st Piola-Kirchhoff stress $\mathbf{P}$, which is computed as $$ \mathbf{P}=\frac{\partial\Psi}{\partial\mathbf{F}} \tag{2} $$ and the corresponding tangent modulus $$ \mathbb{A}=\frac{\partial\mathbf{P}}{\partial\mathbf{F}} \tag{3} $$ which is usually called the 1st elasticity tensor.

Maybe Eq.(2) and (3) are what you are looking for.

Answer 2

The derivative $\partial\mathbf{E}/\partial\mathbf{F}$ maps from a nine-dimensional space (the differentials d$\mathbf{F}$) to a six-dimensional space (the differentials d$\mathbf{E}$). That said, it is clear that two different d$\mathbf{F}$ can be mapped to the same d$\mathbf{E}$. So the mapping is surjective, which means it is not invertible. This image is taken from this website