Addendum to “fourth-order tensor calculus operations and application to continuum mechanics”

Abstract

The authors provided a novel explicit expression for the dot derivative of an isotropic tensor function in the work by Kellermann et al. An alternate proof is provided in this addendum without the power series assumption.

The authors provided a novel explicit expression for the dot derivative of an isotropic tensor function in the work by Kellermann et al. [1] whose proof relied on the power series expression for an isotropic tensor function. The derivative expression is rewritten here:

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }=[\mathbf{F}\left(\mathbf{A}\right)\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}\left(\mathbf{A}\right)+\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}:{\mathbb{H}}_{3\odot }]:{{\mathbb{R}}_{3\odot }}^{-\odot }}\\ \multicolumn{1}{c}{=\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}+[\mathbf{F}\left(\mathbf{A}\right)\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}\left(\mathbf{A}\right)\mathbf{-}\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}:{\mathbb{S}}_{3\odot }]:{{\mathbb{R}}_{3\odot }}^{-\odot }}\end{array}$$

(1)

with

$${\mathbb{H}}_{3\odot }=(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2){\mathbb{S}}_{3\odot }=(\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A}){\mathbb{R}}_{3\odot }={\mathbb{S}}_{3\odot }+{\mathbb{H}}_{3\odot }$$

(2)

with the terminology defined in the work by Kellermann et al. [1]. An alternate proof is provided in this addendum without the power series assumption. The aim is to show

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:(\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A}+\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2)}\\ \multicolumn{1}{c}{=[\mathbf{F}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}+\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}:(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2)].}\end{array}$$

(3)

This equation is examined in two parts:

$$\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:(\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A})=\mathbf{F}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}.$$

(4)

$$\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2)=\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}:(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2).$$

(5)

An isotropic tensor function $\mathbf{F}(\mathbf{A})$ by definition commutes with $\mathbf{A}$ , i.e.,

$$\mathbf{F}·\mathbf{A}=\mathbf{A}·\mathbf{F}.$$

(6)

If $\mathbf{A}$ is a diagonalizable second-order tensor $\mathbf{A}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:\mathbf{\Lambda }$ with $\mathbf{\Lambda }$ the diagonal eigenvalue tensor and $\mathbf{Q}$ the eigenvector tensor, an isotropic tensor function $\mathbf{F}(\mathbf{A})$ must be diagonalizable and have the same eigenvectors:

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{F}\left(\mathbf{A}\right)=\mathbf{Q}·\mathbf{F}\left(\mathbf{\Lambda }\right)·{\mathbf{Q}}^{-1}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:\mathbf{F}\left(\mathbf{\Lambda }\right)=\mathbf{F}\left(\mathbf{\Lambda }\right):{\mathbf{Q}}^{\mathrm{T}}\odot {\mathbf{Q}}^{-\mathrm{T}}}\\ \multicolumn{1}{c}{=\mathbb{Q}:\mathbf{F}\left(\mathbf{\Lambda }\right)=\mathbf{F}\left(\mathbf{\Lambda }\right):{\mathbb{Q}}^{\mathrm{T}\odot }}\end{array}$$

(7)

with $\mathbf{F}(\mathbf{\Lambda })$ being diagonal. Taking the dot derivative of both sides of equation (6) and using the product rule $(\mathbf{F}·\mathbf{G}{)}_{,\mathbf{X}\odot }={\mathbf{F}}_{,\mathbf{X}\odot }·\mathbf{G}+\mathbf{F}·{\mathbf{G}}_{,\mathbf{X}\odot }$ , gives

$$\begin{array}{c}{(\mathbf{F}·\mathbf{A})}_{,\mathbf{A}\odot }={(\mathbf{A}·\mathbf{F})}_{,\mathbf{A}\odot }\mathbf{=}{\mathbf{F}}_{,\mathbf{A}\odot }·\mathbf{A}+\mathbf{F}·\underset{\odot }{\mathbb{I}}=\mathbf{A}·{\mathbf{F}}_{,\mathbf{A}\odot }+\underset{\odot }{\mathbb{I}}·\mathbf{F}\hfill \\ \therefore \mathbf{I}·{\mathbf{F}}_{,\mathbf{A}\odot }·\mathbf{A}-\mathbf{A}·{\mathbf{F}}_{,\mathbf{A}\odot }·\mathbf{I}=\underset{\odot }{\mathbb{I}}·\mathbf{F}\mathbf{-}\mathbf{F}·\underset{\odot }{\mathbb{I}}\hfill \\ \therefore (\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A}):{\mathbf{F}}_{,\mathbf{A}\odot }=(\mathbf{F}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}).\hfill \end{array}$$

(8)

The eigendecomposition of the dot derivative tensor $\mathbf{F}(\mathbf{A}{)}_{,\mathbf{A}\odot }$ can be inferred from the form of the dot derivative for integer powers of $\mathbf{A}$ (see Kellermann et al. [1]) and is of the form given in equation (83) of our paper [1]:

$$\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }=(\mathbf{Q}\odot {\mathbf{Q}}^{-1}):\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:({\mathbf{Q}}^{-1}\odot \mathbf{Q})=\mathbb{Q}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbb{Q}}^{-\odot }.$$

(9)

With $\mathbf{F}(\mathbf{\Lambda }{)}_{,\mathbf{\Lambda }\odot }$ being diagonal when flatten in the matrix dot representation as defined in the work by Kellermann et al. [2]. The derivative of a tensor-valued function using the Gateaux directional derivative definition (see Itskov [3]) is written here with respect to the dot derivative:

$${\frac{d\mathbf{F}(\mathbf{A}+t\mathbf{H})}{\mathrm{dt}}|}_{t=0}=\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:\mathbf{H}\mathrm{for}\mathrm{all}\mathbf{H}$$

(10)

where t is a scalar and $\mathbf{H}$ being from the same domain as $\mathbf{A}$ . Consider the special case where $\mathbf{F}(\mathbf{A}+t\mathbf{M})$ is a diagonalizable isotropic tensor function and $\mathbf{M}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:{\mathbf{\Lambda }}_{\mathbf{M}}$ is also diagonalizable, with the same eigenvectors as $\mathbf{A}$ and commutes with $\mathbf{A}$ .

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:\mathbf{M}={\frac{d\mathbf{F}(\mathbf{A}+t\mathbf{M})}{\mathrm{dt}}|}_{t=0}\mathrm{for}\mathbf{M}·\mathbf{A}=\mathbf{A}·\mathbf{M}}\\ =(\mathbf{Q}\odot {\mathbf{Q}}^{-1}):{\frac{d\mathbf{F}(\mathbf{\Lambda }+t{\mathbf{\Lambda }}_{\mathbf{M}})}{\mathrm{dt}}|}_{t=0}=(\mathbf{Q}\odot {\mathbf{Q}}^{-1}):\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbf{\Lambda }}_{\mathbf{M}}\hfill \\ =\{(\mathbf{Q}\odot {\mathbf{Q}}^{-1}):\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:({\mathbf{Q}}^{-1}\odot \mathbf{Q})\}:\mathbf{M}\therefore \mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }=\mathbb{Q}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbb{Q}}^{-\odot }.\hfill \end{array}$$

(11)

Confirming equation (9). Substituting equation (9) into equation (8), it follows that

$$\begin{array}{c}(\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A}):{\mathbf{F}}_{,\mathbf{A}\odot }=\hfill \\ \mathbb{Q}:(\mathbf{\Lambda }\odot \mathbf{I}-\mathbf{I}\odot \mathbf{\Lambda }):\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbb{Q}}^{-\odot }=\mathbb{Q}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:(\mathbf{\Lambda }\odot \mathbf{I}-\mathbf{I}\odot \mathbf{\Lambda }):{\mathbb{Q}}^{-\odot }\hfill \\ \therefore {\mathbf{F}}_{,\mathbf{A}\odot }:(\mathbf{A}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{A})=(\mathbf{F}\odot \mathbf{I}-\mathbf{I}\odot \mathbf{F}).\hfill \end{array}$$

(12)

Providing a proof for equation (4) subject to the assumption equation (9).

To examine the second equation (5), we rewrite equation (11), thus,

$$\begin{array}{c}\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:\mathbf{M}={\frac{d\mathbf{F}(\mathbf{A}+t\mathbf{M})}{\mathrm{dt}}|}_{t=0}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:{\frac{d\mathbf{F}(\mathbf{\Lambda }+t{\mathbf{\Lambda }}_{\mathbf{M}})}{\mathrm{dt}}|}_{t=0}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:\mathbf{F}\prime \left(\mathbf{\Lambda }\right)·{\mathbf{\Lambda }}_{\mathbf{M}}.\hfill \\ \therefore \mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:\mathbf{M}=\mathbf{F}\prime \left(\mathbf{A}\right)·\mathbf{M}\hfill \end{array}$$

(13)

It therefore follows that

$$\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:{\mathbf{A}}^k=\mathbf{F}\prime \left(\mathbf{A}\right)·{\mathbf{A}}^k.$$

(14)

Also, equation (13) can be rewritten as follows:

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }:\mathbf{M}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:{\frac{d\mathbf{F}(\mathbf{\Lambda }+t{\mathbf{\Lambda }}_{\mathbf{M}})}{\mathrm{dt}}|}_{t=0}=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbf{\Lambda }}_{\mathbf{M}}}\\ \multicolumn{1}{c}{=\mathbf{Q}\odot {\mathbf{Q}}^{-1}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbf{Q}}^{-1}\odot \mathbf{Q}:\mathbf{Q}\odot {\mathbf{Q}}^{-1}:{\mathbf{\Lambda }}_{\mathbf{M}}=\mathbb{Q}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbb{Q}}^{-\odot }:\mathbf{M}}\\ \multicolumn{1}{c}{\therefore \mathbf{F}{\left(\mathbf{A}\right)}_{,\mathbf{A}\odot }=\mathbb{Q}:\mathbf{F}{\left(\mathbf{\Lambda }\right)}_{,\mathbf{\Lambda }\odot }:{\mathbb{Q}}^{-\odot }.}\end{array}$$

(15)

Confirming the form of equation (9). Now we examine the second expression involving $\mathbf{F}\prime (\mathbf{A})\odot \mathbf{I}$ in equation (5). Using the identity in equation (14) we show that

$$\begin{array}{c}\mathbf{F}\prime \left(\mathbf{A}\right)\odot \mathbf{I}:(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2)=\mathbf{F}\prime \left(\mathbf{A}\right)\mathbf{A}\otimes \mathbf{I}-\mathbf{F}\prime \left(\mathbf{A}\right)\otimes \mathbf{A}+\mathbf{F}\prime \left(\mathbf{A}\right){\mathbf{A}}^2\otimes {\mathbf{A}}^2\hfill \\ =({\mathbf{F}}_{,\mathbf{A}\odot }:\mathbf{A})\otimes \mathbf{I}-({\mathbf{F}}_{,\mathbf{A}\odot }:\mathbf{I})\otimes \mathbf{A}+({\mathbf{F}}_{,\mathbf{A}\odot }:{\mathbf{A}}^2)\otimes {\mathbf{A}}^2={\mathbf{F}}_{,\mathbf{A}\odot }:(\mathbf{A}\otimes \mathbf{I}-\mathbf{I}\otimes \mathbf{A}+{\mathbf{A}}^2\otimes {\mathbf{A}}^2).\hfill \end{array}$$

(16)

Combining equations (12) and (16) gives equation (3), without relying on the power series representation for an isotropic tensor function but relies on assumption equation (9).