Tensor Derivative (continuum Mechanics) - Derivative of A Second-order Tensor With Respect To Itself

Derivative of A Second-order Tensor With Respect To Itself

Let be a second order tensor. Then

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \left_{\alpha = 0} = \boldsymbol{T} = \boldsymbol{\mathsf{I}}:\boldsymbol{T}$

Therefore,

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}$

Here is the fourth order identity tensor. In index notation with respect to an orthonormal basis

$\boldsymbol{\mathsf{I}} = \delta_{ik}~\delta_{jl}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

This result implies that

$\frac{\partial \boldsymbol{A}^T}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathsf{I}}^T:\boldsymbol{T} = \boldsymbol{T}^T$

where

$\boldsymbol{\mathsf{I}}^T = \delta_{jk}~\delta_{il}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

Therefore, if the tensor is symmetric, then the derivative is also symmetric and we get

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}^{(s)} = \frac{1}{2}~(\boldsymbol{\mathsf{I}} + \boldsymbol{\mathsf{I}}^T)$

where the symmetric fourth order identity tensor is

$\boldsymbol{\mathsf{I}}^{(s)} = \frac{1}{2}~(\delta_{ik}~\delta_{jl} + \delta_{il}~\delta_{jk}) ~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

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