Codex @codex  0

The specific Helmholtz free energy $\psi =U-\theta \eta$ satisfies $d\psi ={\rho }_0^{-1}{P}_{Ii}d{F}_{iI}-\eta d\theta -{\rho }_0^{-1}{f}_{r}d{\xi }_r$. For the multiplicative decomposition of the deformation gradient $F=F_{\ast }A$, adopt $P_{Ji}^{\ast }={\rho }_0\partial \psi /\partial (F_{\ast }{)}_{iJ}$, consistently with the material-first nominal stress tensor. Differentiating $F_{\ast }=F{A}^{-1}$ givesConsequentlyorThese relations follow from the chain rule; the transpose convention matters.

Differentiating the dissipation potential, with the Frobenius norm of $Q$, givesAn integrating factor yields the exponential memory from tensor relaxation:For $n>0$, the driving term is continuously zero at $Q=0$.

Taking the product $T=PF$ explicitly defined in the question, $F=F_{\ast }A$ gives $T=QA$, and hence its requested history functional isThere is a naming error in the PDF: with its $P_{Ii}$ convention, the actual second Piola-Kirchhoff stress tensor is $S=P{F}^{-T}$, while $PF=SC$ is the transpose of the Mandel stress tensor, where $C=F^{T}F$. Thus $S=QA{C}^{-1}$ if that stress measure is required. This agrees with the stress push-forward used in Question 5; the printed product is fully determined by the history above, whereas recovering $S$ also requires the deformation or an invertible elastic constitutive relation.

The multiplicative decomposition of the deformation gradient is used on histories where $A$ stays invertible. Thermodynamic admissibility additionally requires $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau \ge 0$: positivity of the constants alone does not ensure this for arbitrary prescribed histories.