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

Derivatives of The Invariants of A Second-order Tensor

The principal invariants of a second order tensor are

$\begin{align} I_1(\boldsymbol{A}) & = \text{tr}{\boldsymbol{A}} \\ I_2(\boldsymbol{A}) & = \frac{1}{2} \left \\ I_3(\boldsymbol{A}) & = \det(\boldsymbol{A}) \end{align}$

The derivatives of these three invariants with respect to are

$\begin{align} \frac{\partial I_1}{\partial \boldsymbol{A}} & = \boldsymbol{\mathit{1}} \\ \frac{\partial I_2}{\partial \boldsymbol{A}} & = I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T \\ \frac{\partial I_3}{\partial \boldsymbol{A}} & = \det(\boldsymbol{A})~^T = I_2~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T~(I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T) = (\boldsymbol{A}^2 - I_1~\boldsymbol{A} + I_2~\boldsymbol{\mathit{1}})^T \end{align}$

Proof
From the derivative of the determinant we know that $\frac{\partial I_3}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~^T ~.$ For the derivatives of the other two invariants, let us go back to the characteristic equation $\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) = \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) ~.$ Using the same approach as for the determinant of a tensor, we can show that $\frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~^T ~.$ Now the left hand side can be expanded as $\begin{align} \frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) & = \frac{\partial }{\partial \boldsymbol{A}}\left[ \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) \right] \\ & = \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}~. \end{align}$ Hence $\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}} = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~^T$ or, $(\lambda~\boldsymbol{\mathit{1}}+\boldsymbol{A})^T\cdot\left[ \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}\right] = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~\boldsymbol{\mathit{1}} ~.$ Expanding the right hand side and separating terms on the left hand side gives $(\lambda~\boldsymbol{\mathit{1}} +\boldsymbol{A}^T)\cdot\left[ \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}\right] = \left \boldsymbol{\mathit{1}}$ or, $\begin{align} \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.& \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda\right]\boldsymbol{\mathit{1}} + \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\ & = \left \boldsymbol{\mathit{1}} ~. \end{align}$ If we define and, we can write the above as $\begin{align} \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.& \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_4}{\partial \boldsymbol{A}}\right]\boldsymbol{\mathit{1}} + \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}~\lambda^3 + \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\ &= \left \boldsymbol{\mathit{1}} ~. \end{align}$ Collecting terms containing various powers of λ, we get $\begin{align} \lambda^3&\left(I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}\right) + \lambda^2\left(I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}\right) + \\ &\qquad \qquad\lambda\left(I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}\right) + \left(I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}}\right) = 0 ~. \end{align}$ Then, invoking the arbitrariness of λ, we have $\begin{align} I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}} & = 0 \\ I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}} & = 0 \\ I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} & = 0 ~. \end{align}$ This implies that $\begin{align} \frac{\partial I_1}{\partial \boldsymbol{A}} &= \boldsymbol{\mathit{1}} \\ \frac{\partial I_2}{\partial \boldsymbol{A}} & = I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\\ \frac{\partial I_3}{\partial \boldsymbol{A}} & = I_2~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T~(I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T) = (\boldsymbol{A}^2 -I_1~\boldsymbol{A} + I_2~\boldsymbol{\mathit{1}})^T \end{align}$

Proof

From the derivative of the determinant we know that

$\frac{\partial I_3}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~^T ~.$

For the derivatives of the other two invariants, let us go back to the characteristic equation

$\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) = \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) ~.$

Using the same approach as for the determinant of a tensor, we can show that

$\frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~^T ~.$

Now the left hand side can be expanded as

$\begin{align} \frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) & = \frac{\partial }{\partial \boldsymbol{A}}\left[ \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) \right] \\ & = \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}~. \end{align}$

Hence

$\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}} = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~^T$

or,

$(\lambda~\boldsymbol{\mathit{1}}+\boldsymbol{A})^T\cdot\left[ \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}\right] = \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~\boldsymbol{\mathit{1}} ~.$

Expanding the right hand side and separating terms on the left hand side gives

$(\lambda~\boldsymbol{\mathit{1}} +\boldsymbol{A}^T)\cdot\left[ \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_3}{\partial \boldsymbol{A}}\right] = \left \boldsymbol{\mathit{1}}$

or,

$\begin{align} \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.& \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda\right]\boldsymbol{\mathit{1}} + \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\ & = \left \boldsymbol{\mathit{1}} ~. \end{align}$

If we define and, we can write the above as

$\begin{align} \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.& \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_4}{\partial \boldsymbol{A}}\right]\boldsymbol{\mathit{1}} + \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}~\lambda^3 + \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda + \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\ &= \left \boldsymbol{\mathit{1}} ~. \end{align}$

Collecting terms containing various powers of λ, we get

$\begin{align} \lambda^3&\left(I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}\right) + \lambda^2\left(I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}\right) + \\ &\qquad \qquad\lambda\left(I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}\right) + \left(I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}}\right) = 0 ~. \end{align}$

Then, invoking the arbitrariness of λ, we have

$\begin{align} I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}} & = 0 \\ I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}} & = 0 \\ I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} & = 0 ~. \end{align}$

This implies that

$\begin{align} \frac{\partial I_1}{\partial \boldsymbol{A}} &= \boldsymbol{\mathit{1}} \\ \frac{\partial I_2}{\partial \boldsymbol{A}} & = I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\\ \frac{\partial I_3}{\partial \boldsymbol{A}} & = I_2~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T~(I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T) = (\boldsymbol{A}^2 -I_1~\boldsymbol{A} + I_2~\boldsymbol{\mathit{1}})^T \end{align}$

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