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Suppose we are given a frame-indifferent isotropic function $W:GL_+(3) \to [0,\infty)$, where $GL_+(3)$ denotes the set of all real $(3\times 3)$-matrices with positive determinant. We can write $W(F)$ in terms of the principal stretches (i.e. singular values) of $F$. More specifically, there exists a function $f:[0,\infty)^3\to[0,\infty)$ such that $$ W(F) = f(\lambda_1(F), \lambda_2(F), \lambda_3(F)), $$ where $\lambda_1(F)\geq \lambda_2(F)\geq\lambda_3(F)\geq 0$ are the singular values of $F$. Assume that $W$ and $f$ are smooth. I want to write the elasticity tensor $\frac{\partial^2 W}{\partial F^2}$ in terms of $f$ and $\lambda_1,\lambda_2,\lambda_3$.

My ansatz is the following: Symmetry of $\frac{\partial^2 W}{\partial F^2}$ reduces the degrees of freedom from 81 to 45. Furthermore, every rotation $R\in SO(3)$ can be uniquely written as a product $$ R = R_z(\alpha_3)R_y(\alpha_2)R_x(\alpha_1), $$ where $R_w(\alpha_i)\in SO(3)$ with $w\in \{x,y,z\}, i\in\{1,2,3\},$ denotes the rotation around the $w$-axis by the angle $\alpha_i$. By frame-indifference and isotropy we have for every matrix $F\in GL_+(3)$ $$ W(F)=W(R_z(\alpha_6)R_y(\alpha_5)R_x(\alpha_4) F R_z(\alpha_3)R_y(\alpha_2)R_x(\alpha_1)) $$ for every $\alpha_1,\dots,\alpha_6 \in[0,2\pi)$. In particular, $$ 0 = \frac{\partial}{\partial\alpha_i} W(R_z(\alpha_6)R_y(\alpha_5)R_x(\alpha_4) F R_z(\alpha_3)R_y(\alpha_2)R_x(\alpha_1)) $$ for every $i\in\{1,\dots,6\}$. This relation yields a further reduction of the degrees of freedom of $\frac{\partial^2 W}{\partial F^2}$. However, I am not sure how to go on from here. The resulting linear system seems to be very complicated and I suspect that there is a more efficient way to determine all the entries of $\frac{\partial^2 W}{\partial F^2}$ in terms of $f,\lambda_1,\lambda_2,\lambda_3$.

For $(3\times 2)$-matrices the answer to my question was presented by Pipkin in this paper: https://academic.oup.com/imamat/article-abstract/36/1/85/762621?redirectedFrom=fulltext The result therein is a very compact expression. Unfortunately, there are no details of the computation provided.

Hence my question: Can anybody tell me how to compute $\frac{\partial^2 W}{\partial F^2}$ in terms of $f,\lambda_1,\lambda_2,\lambda_3$ (or how to improve/simplify my above ansatz)? Or does anybody know any literature where I can find the desired result?

Remark: I posted this question on math.stackexchange.com a few months ago, but it got very little attention: https://math.stackexchange.com/questions/3421016/elasticity-tensor-in-terms-of-principal-stretches

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    $\begingroup$ Isn't this, by the chain rule, mainly an issue of computing $\partial \lambda_* / \partial F$? $\endgroup$ Commented Feb 7, 2020 at 15:22
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    $\begingroup$ In the paper you cite there are only two principle stretches, and so their squares solve the characteristic equation for the square $F^* F$ (which is two-by-two). So in particular you can use the quadratic equation to write down $\lambda_*$ as algebraic expressions in terms of $F$ and find the derivatives directly. (This is the equation 6.2 in that paper) // In the $3\times 3$ case I guess you can try using the cubic formula.... $\endgroup$ Commented Feb 7, 2020 at 15:28

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