Tensor Derivative (continuum Mechanics) - Derivative of The Inverse of A Second-order Tensor

Derivative of The Inverse of A Second-order Tensor

Let and be two second order tensors, then

\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}

In index notation with respect to an orthonormal basis

\frac{\partial A^{-1}_{ij}}{\partial A_{kl}}~T_{kl} = - A^{-1}_{ik}~T_{kl}~A^{-1}_{lj} \implies \frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = - A^{-1}_{ik}~A^{-1}_{lj}

We also have

\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-T}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-T}\cdot\boldsymbol{T}^{T}\cdot\boldsymbol{A}^{-T}

In index notation

\frac{\partial A^{-1}_{ji}}{\partial A_{kl}}~T_{kl} = - A^{-1}_{jk}~T_{kl}~A^{-1}_{li} \implies \frac{\partial A^{-1}_{ji}}{\partial A_{kl}} = - A^{-1}_{li}~A^{-1}_{jk}

If the tensor is symmetric then

\frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = -\cfrac{1}{2}\left(A^{-1}_{ik}~A^{-1}_{jl} + A^{-1}_{il}~A^{-1}_{jk}\right)
Proof
Recall that
\frac{\partial \boldsymbol{\mathit{1}}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathit{0}}

Since, we can write

\frac{\partial }{\partial \boldsymbol{A}}(\boldsymbol{A}^{-1}\cdot\boldsymbol{A}):\boldsymbol{T} = \boldsymbol{\mathit{0}}

Using the product rule for second order tensors

\frac{\partial }{\partial \boldsymbol{S}}:\boldsymbol{T} = \left(\frac{\partial \boldsymbol{F}_1}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)\cdot\boldsymbol{F}_2 + \boldsymbol{F}_1\cdot\left(\frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)

we get

\frac{\partial }{\partial \boldsymbol{A}}(\boldsymbol{A}^{-1}\cdot\boldsymbol{A}):\boldsymbol{T} = \left(\frac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right)\cdot\boldsymbol{A} + \boldsymbol{A}^{-1}\cdot\left(\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right) = \boldsymbol{\mathit{0}}

or,

\left(\frac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right)\cdot\boldsymbol{A} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}

Therefore,

\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}

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