Variations of Tensors

As an extension to the scalar-valued explanations given in Variational Calculus, let us consider a scalar-valued function of as second-order tensor-based input argument, see Eq. (1).

(1)\[ \begin{align}\begin{aligned}\psi &= \psi(\boldsymbol{F})\\\delta \psi &= \delta \psi(\boldsymbol{F}, \delta \boldsymbol{F})\end{aligned}\end{align} \]

Let’s take the trace of a tensor product as an example. The variation is carried out in Eq. (2).

(2)\[ \begin{align}\begin{aligned}\psi &= tr(\boldsymbol{F}^T \boldsymbol{F}) = \boldsymbol{F} : \boldsymbol{F}\\\delta \psi &= \delta \boldsymbol{F} : \boldsymbol{F} + \boldsymbol{F} : \delta \boldsymbol{F} = 2 \ \boldsymbol{F} : \delta \boldsymbol{F}\end{aligned}\end{align} \]

The \(P_{ij}\) - component of the jacobian \(\boldsymbol{P}\) is now numerically evaluated by setting the respective variational component \(\delta F_{ij}\) of the tensor to one and all other components to zero. In total, \(i \cdot j\) function calls are necessary to assemble the full jacobian. For example, the \(12\) - component is evaluated as given in Eq. (3).

(3)\[\begin{split}\delta_{(12)} \psi = \frac{\partial \psi}{\partial F_{12}} &= 2 \ \boldsymbol{F} : \delta \boldsymbol{F}_{(12)} = 2 \ \boldsymbol{F} : \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\end{split}\]

The second order variation, i.e. a variation applied on another variation of a function is evaluated in the same way as a first order variation, see Eq. (4).

(4)\[\Delta \delta \psi = 2 \ \delta \boldsymbol{F} : \Delta \boldsymbol{F} + 2 \ \boldsymbol{F} : \Delta \delta \boldsymbol{F}\]

Once again, each component \(A_{ijkl}\) of the fourth-order hessian is numerically evaluated. In total, \(i \cdot j \cdot k \cdot l\) function calls are necessary to assemble the full hessian (without considering symmetry). For example, the \(1223\) - component is evaluated by setting \(\Delta \delta \boldsymbol{F} = \boldsymbol{0}\) and \(\delta \boldsymbol{F}\) as well as \(\Delta \boldsymbol{F}\) as given in Eq. (5) and Eq. (6).

(5)\[ \begin{align}\begin{aligned}\begin{split}\delta \boldsymbol{F}_{(12)} &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\end{split}\\\begin{split}\Delta \boldsymbol{F}_{(23)} &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\end{split}\\\begin{split}\Delta \delta \boldsymbol{F} &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\end{split}\end{aligned}\end{align} \]

(6)\[ \begin{align}\begin{aligned}\Delta_{(23)} \delta_{(12)} \psi &= \Delta_{(12)} \delta_{(23)} \psi = \frac{\partial^2 \psi}{\partial F_{12}\ \partial F_{23}}\\\Delta_{(23)} \delta_{(12)} \psi &= 2 \ \delta \boldsymbol{F}_{(12)} : \Delta \boldsymbol{F}_{(23)} + 2 \ \boldsymbol{F} : \Delta \delta \boldsymbol{F}\end{aligned}\end{align} \]