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

Derivative of The Determinant of A Second-order Tensor

The derivative of the determinant of a second order tensor is given by

$\frac{\partial }{\partial \boldsymbol{A}}\det(\boldsymbol{A}) = \det(\boldsymbol{A})~^T ~.$

In an orthonormal basis, the components of can be written as a matrix A. In that case, the right hand side corresponds the cofactors of the matrix.

Proof
Let be a second order tensor and let . Then, from the definition of the derivative of a scalar valued function of a tensor, we have $\begin{align} \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} & = \left.\cfrac{d}{d\alpha} \det(\boldsymbol{A} + \alpha~\boldsymbol{T}) \right\|_{\alpha=0} \\ & = \left.\cfrac{d}{d\alpha} \det\left \right\|_{\alpha=0} \\ & = \left.\cfrac{d}{d\alpha} \left\right\|_{\alpha=0}. \end{align}$ Recall that we can expand the determinant of a tensor in the form of a characteristic equation in terms of the invariants using (note the sign of λ) Using this expansion we can write $\begin{align} \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} & = \left.\cfrac{d}{d\alpha} \left[\alpha^3~\det(\boldsymbol{A})~ \left(\cfrac{1}{\alpha^3} + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha^2} + I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha} + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})\right) \right] \right\|_{\alpha=0} \\ & = \left.\det(\boldsymbol{A})~\cfrac{d}{d\alpha} \left[ 1 + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha + I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2 + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^3 \right] \right\|_{\alpha=0} \\ & = \left.\det(\boldsymbol{A})~\left[I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) + 2~I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha + 3~I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2 \right] \right\|_{\alpha=0} \\ & = \det(\boldsymbol{A})~I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) ~. \end{align}$ Recall that the invariant is given by Hence, Invoking the arbitrariness of we then have $\frac{\partial f}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~^T ~.$

Proof

Let be a second order tensor and let . Then, from the definition of the derivative of a scalar valued function of a tensor, we have

$\begin{align} \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} & = \left.\cfrac{d}{d\alpha} \det(\boldsymbol{A} + \alpha~\boldsymbol{T}) \right|_{\alpha=0} \\ & = \left.\cfrac{d}{d\alpha} \det\left \right|_{\alpha=0} \\ & = \left.\cfrac{d}{d\alpha} \left\right|_{\alpha=0}. \end{align}$

Recall that we can expand the determinant of a tensor in the form of a characteristic equation in terms of the invariants using (note the sign of λ)

Using this expansion we can write

$\begin{align} \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} & = \left.\cfrac{d}{d\alpha} \left[\alpha^3~\det(\boldsymbol{A})~ \left(\cfrac{1}{\alpha^3} + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha^2} + I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha} + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})\right) \right] \right|_{\alpha=0} \\ & = \left.\det(\boldsymbol{A})~\cfrac{d}{d\alpha} \left[ 1 + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha + I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2 + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^3 \right] \right|_{\alpha=0} \\ & = \left.\det(\boldsymbol{A})~\left[I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) + 2~I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha + 3~I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2 \right] \right|_{\alpha=0} \\ & = \det(\boldsymbol{A})~I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) ~. \end{align}$

Recall that the invariant is given by

Hence,

Invoking the arbitrariness of we then have

$\frac{\partial f}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~^T ~.$

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